Problems

Problems

1

Year 1993 Olympiad Level A Problem 1. Denote by S(x) the sum of the digits of a positive integer x. Solve: (a) x + S(x) + S(S(x)) = 1993. (b) x + S(x) + S(S(x)) + S(S(S(x))) = 1993. Problem 2*. Suppose n is the sum of the squares of three positive integers. Prove that n2 is also the sum of the squares of three positive integers. Problem 3. A red and a blue poker chip are stacked, the red one on top. Suppose one can carry out only the following operations: (a) adding two chips of the same color to the stack together, in any position; and (b) removing any two neighboring chips of the same color. After finitely many operations, is it possible to end up with only two chips left, the blue one on top of the red one? Problem 4. At the court of Tsar Gorokh, the royal astrologer built a clock remarkably similar to modern (analog) ones, with hands for hours, minutes, and seconds, all moving smoothly around the same point. He calls a moment of time lucky if the three hands of his clock, counting clockwise from the hour hand, appear in the order hours/minutes/seconds, and unlucky if they appear in the order hours/seconds/minutes. Is the amount of lucky time in a 24-hour day more or less than the amount of unlucky time?

Adaptation

Remark. Tsar Gorokh (King Pea) is a character from Russian folklore. “In the time of Tsar Gorokh” is a Russian idiom meaning “a very long time ago”.

∗

Problem 5. Prove or disprove: There is a finite string of letters of the alphabet such that there are no two identical adjacent substrings, yet a pair of identical adjacent substrings appears as soon as one adds any letter of the alphabet at the beginning or at the end of the string.

∗

Problem 6. A circle centered at D passes through points A, B, and the excenter O of the triangle ABC relative to side BC (that is, O is the center of the circle tangent to BC and to the extensions of sides AB and AC). Prove that A, B, C, and D lie on a circle.

4

PROBLEMS

Level B Problem 1. For two distinct points A and B in the plane, find the locus of points C such that the triangle ABC is acute and the value of its angle at A is intermediate among the triangle’s angles (meaning that \B ≤ \A ≤ \C or \C ≤ \A ≤ \B). Problem 2. Let x1 = 4, x2 = 6, and define xn for n ≥ 3 to be the least nonprime greater than 2xn−1 − xn−2 . Find x1000 .

Problem 3. A paper triangle with angles of 20◦ , 20◦ , and 140◦ is cut along one of its bisectors into two triangles; one of these triangles is also cut along one of its bisectors, and so on. Can we obtain a triangle similar to the initial one after several cuts?

Problem 4. In Pete’s class, there are 28 students besides him. Each of the 28 has a different number of friends in the class. How many friends does Pete have in this class? Problem 5. To every pair of numbers x and y we assign a number x ∗ y. Find 1993 ∗ 1935 if it is known that x ∗ x = 0 and

x ∗ (y ∗ z) = (x ∗ y) + z

for any x, y, z.

Problem 6. Given a convex quadrilateral ABM C with \BAM = 30◦ , \ACM = 150◦ , and AB = BC, prove that AM is the bisector of \BM C. Level C Problem 1. In the decimal representation of two numbers A and B, the minimal periods have lengths 6 and 12, respectively. What are the possibilities for the length of the minimal period of A + B? Problem 2. The grandfather of Baron von M¨ unchhausen built a castle with a square floor plan. He divided the castle into 9 equal square areas, and placed the arsenal in the center square. The Baron’s father divided each of the remaining 8 areas into 9 equal square halls and built a greenhouse in each central hall. The Baron himself divided each of the 64 empty halls into 9 equal square rooms and placed a swimming pool in each of the central rooms. He then furnished the other rooms lavishly and connected each pair of adjacent furnished rooms by a door, locking all other doors. The Baron boasts that he can tour all his furnished rooms, visiting each exactly once and returning to the starting point. Can this be true? Problem 3. From any point on either bank of a river one can reach the other bank by swimming a distance of no more than 1 km. (a) Is it always possible to pilot a boat along the whole length of the river while remaining within 700 m of both banks? (b) * Same question with 800 m.

∗

YEAR 1993 OLYMPIAD

5

Remark. Both the answer and the degree of difficulty of the problem depend on adaptation what additional assumptions are made. Naturally the boat is to be considered a point. The original problem had a note saying so, and also that “the river joins two round lakes, each 10 km in radius, and the river banks consist of straight line segments and arcs of circle.” But in spite of this precision, ambiguities remain. Can there be islands in the river? And does “within 700 m” refer to the straight-line distance or the swimming distance? (See figure.) Warning: Part (b) is surprisingly difficult, unless islands are allowed, in which case both parts are easy.

Problem 4. Given real numbers a and b, define pn = [2{an+b}], where {x} denotes the fractional part of x and [x] the integer part. (a) Can all possible quadruples of 0s and 1s occur as substrings of the sequence p0 , p1 , p2 , . . . , if we are allowed to vary a and b? (b) Can all possible 5-uples of 0s and 1s occur? Problem 5. In a botanical classifier, a plant is identified by 100 features. Each feature can either be present or absent. A classifier is considered to be good if any two plants have less than half of their features in common. Prove that a good classifier cannot describe more than 50 plants. Problem 6. On the side AB of a triangle ABC, a square is constructed outwards; let its center be O. Let M and N be the midpoints of AC and BC, and let the lengths of these sides be a and b. Find the maximum of the sum OM + ON as the angle ACB varies. Level D Problem 1. Knowing that tan α + tan β = p and cot α + cot β = q, find tan(α + β). Problem 2. The unit square is divided into finitely many smaller squares, not necessarily of the same size. Consider the small squares that overlap (possibly at a corner) with the main diagonal. Is it possible for the sum of their perimeters to exceed 1993? Problem 3. We are given n points in the plane, no three of which lie on a line. Through each pair of points a line is drawn. What is the least possible number of pairwise nonparallel lines among these lines? Problem 4. We start with a number of boxes, each with some marbles in them. At each step, we select a number k and divide the marbles in each box into groups of size k with a remainder of less than k; we then remove all but one marble from each group, leaving the remainders intact. Is it possible to ensure that in 5 steps each box is left with a single marble, if initially each box has at most (a) 460 marbles, (b) 461 marbles? Problem 5. (a) It is known that the domain of a function f is the segment [−1, 1], and f (f (x)) = −x for all x; also, the graph of f is the union of

∗

6

PROBLEMS

finitely many points and straight line segments (with or without endpoints). Draw a possible graph for f . (b) Is it possible to draw the graph of f if the domain of f is (−1, 1)? the whole real line? Problem 6*. A fly lives inside a regular tetrahedron with edge a. What is the shortest length of a flight the fly could make to visit every face and return to the initial spot?

∗

Year 1994 Olympiad Level A Problem 1. A local business gets apple and grape juices in identical jugs and makes a mixed drink, which it packages in same-size bottles. One jug of apple juice used to be enough for exactly 6 bottles of the mix, and one jug of grape juice for exactly 10 bottles of the mix. Then the recipe was changed, and one jug of apple juice is now sufficient for exactly 5 bottles of the mix. How many bottles of mix is one jug of the grape juice good for now? (The drink is not diluted with water.) Problem 2. A student multiplying two three-digit numbers noticed that if he wrote the two numbers next to one another, the resulting six-digit number would be seven times greater than the product. Find the two factors. Problem 3. In a triangle ABC, let P and Q be the bases of the perpendiculars dropped from B to the bisectors of A and C. Prove that P Q k AC. Problem 4. Four grasshoppers sit at the vertices of a square. Every now and then one of them hops over another, landing at a point symmetric, with respect to the jumped-over grasshopper, to where it jumped from. Prove that at no time can the grasshoppers occupy the vertices of a bigger square than the original one. Problem 5. The royal astrologer considers a moment in time favorable if the hour, the minute and the second hands of the clock all lie on the same side of some diameter of the clock face. All other times are considered unfavorable. The hands all turn smoothly around the same point. Is the amount of favorable time in a 24-hour day more or less than the amount of unfavorable time? Problem 6. Two people play a game on a piece of graph paper, 19 × 94 squares in size. Each, in turn, colors a square of any desired size, so long as its edges coincide with lines of the grid and no part of it has been colored yet. The player who colors the last square wins. Who is the winner under optimal play, and what is the winning strategy?

∗

8

PROBLEMS

Level B Problem 1. Is there a nonconvex pentagon no two of whose five diagonals intersect, other than at a vertex? Problem 2. Sue starts with a line segment of length k, and Leo with one of length l. First Sue divides her segment into three parts, then Leo divides his into three parts. If it is possible to build two triangles from the six segments obtained, Leo wins; otherwise Sue wins. Depending on the ratio k/l, which player can be sure to win, and what should the winning strategy be? Problem 3. Prove that the equation x2 + y 2 + z 2 = x3 + y 3 + z 3 has infinitely many integer solutions. Problem 4. Two circles intersect at the points A and B. Tangents are drawn to both circles through A. The tangents intersect the circles at the points M and N . The lines BM and BN intersect the circles again at the points P (on BM ) and Q (on BN ). Prove that the segments M P and N Q have the same length. Problem 5*. Dropping one of the digits of a certain positive integer leaves another number that divides the first. The dropped digit is not the leftmost one. What is the highest possible value for the first number, assuming it does not end in 0? Problem 6. In a variation on the game of Battleship, ten ships must be placed in a 10 × 10 square of graph paper: one ship has dimensions 1 × 4, two are 1 × 3, three are 1 × 2, and four are 1 × 1. The ships cannot touch even at a corner, but they can be placed at the sides of the square. (a) Prove that if the ships are placed from largest to smallest, there is always room for all of them, regardless of where earlier ones were placed. (b)* Show by an example that if the ships are laid down from smallest to largest, this may not be the case. Level C Problem 1. A student multiplying two seven-digit numbers noticed that if she wrote the two numbers next to one another, the resulting 14-digit number would be three times greater than the product. Find the two factors. Problem 2. An infinite sequence of numbers xn , where n runs over positive integers, is defined by the condition xn+1 = 1 − |1 − 2xn |, where 0 ≤ x1 ≤ 1. Prove that the sequence is eventually periodic if and only if x1 is rational.

∗

YEAR 1994 OLYMPIAD

9

Problem 3. Each of the 1994 members of the Parliament of Dunces has slapped exactly one colleague in the face. Prove that it is possible to draw from this Parliament a 665-member committee none of whom have slapped one another. Problem 4. Let D be a point on side BC of triangle ABC. Circles are inscribed in the triangles ABD and ACD, and a common outer tangent to the circles (distinct from BC) is drawn. Let its intersection with AD be called K. Prove that the length of AK does not depend on the position of D on BC. Problem 5. Consider an arbitrary polygon, not necessarily convex. (a) Is there always a chord of the polygon that divides it into two pieces of equal area? (b) Prove that the polygon can be divided by a chord into pieces whose areas are each at least one-third the total area of the polygon. (By a chord of a polygon we mean a line segment whose endpoints belong to the polygon’s perimeter while the segment itself lies entirely on the polygon, including its perimeter.) Problem 6*. Can a polynomial P (x) have a negative coefficient if all the powers P n (x), for n > 1, only have positive coefficients? Level D Problem 1. Devise a polyhedron such that no three of its faces all have the same number of edges. Problem 2. See Problem 2 of Level C. Problem 3. A spherical cherry of radius r is dropped into a goblet whose axial cross-section is the graph of the function y = x4 . What is the largest value of r that allows the cherry to touch the very bottom of the goblet? (In other words, what is the radius of the largest circle contained in the region y ≥ x4 and touching the origin?) Problem 4. A convex polyhedron has nine vertices, one of which is A. The translations that send A into each of the other vertices form eight congruent polyhedra. Prove that at least two of these eight polyhedra have an interior point in their intersection. Problem 5*. The extensions of the sides AB and CD of a convex quadrilateral ABCD intersect at a point P ; the extensions of the sides BC and AD intersect at Q. Consider three pairs of bisectors: those of the outer angles of the quadrilateral at vertices A and C, those of the outer angles at vertices B and D, and those of the outer angles at vertices P and Q of the triangles QAB and P BC. Prove that if each of these three pairs of lines intersects, the intersection points are collinear.

10

PROBLEMS

Problem 6. Prove that for any k > 1, there exists a power of 2 such that among its last k digits, there are at least as many 9s as other digits. (For example: 212 = . . . 96, 253 = . . . 992.)

Year 1995 Olympiad Level A Problem 1. Isaac Newton spent a groat a day for a loaf of bread and a tankard of ale. When prices went up by 20%, he started buying half a loaf of bread and the same ale for a groat. Will a groat be enough to buy the ale if prices rise by 20% again?

∗ ∗

Problem 2. Prove that all the numbers of the form 10017, 100117, 1001117, and so on are divisible by 53. Problem 3. Consider a convex quadrilateral and a point O inside it such that \AOB = \COD = 120◦ , AO = OB and CO = OD. Let K, L and M be the midpoints of the sides AB, BC and CD, respectively. Prove that (a) KL = LM ; (b) the triangle KLM is equilateral. Problem 4. To manufacture a closed box of volume at least 1995 cubic units in the shape of a parallelepiped we have (a) 962, (b) 960, (c) 958 square units of material. Is the material sufficient? Problem 5. Roads lead from a city to several nearby villages; there is no direct communication between villages. A truck loaded with goods to be delivered to all the villages starts out from the city. The amount of fuel spent on a leg of the trip is proportional to the current load weight and the distance. Suppose the load to be delivered to each village weighs, in some units, the same as the distance from the city to the item’s destination. Prove that the fuel cost does not depend on the order in which the deliveries are made. Problem 6. A straight line cuts a triangle AKN off of the regular hexagon ABCDEF so that AK + AN = AB. Find the sum of the angles at which the segment KN is seen from the vertices of the hexagon, i.e., find \KAN + \KBN + \KCN + \KDN + \KEN + \KF N. Level B Problem 1. Prove that if we insert any number of digits 3 between the 0s in 12008, we get a number divisible by 19.

∗

12

PROBLEMS

Problem 2. Consider an equilateral triangle ABC. For an arbitrary point P inside the triangle, consider the intersection A′ of the line AP with BC and the intersection C ′ of CP with BA. Find the locus of points P for which the segments AA′ and CC ′ have equal length. Problem 3. A strip is a rectangle of size 1 × k, where k is some positive integer. For what integers n can one cut a 1995 × n rectangle into strips all of different lengths?

∗

Problem 4. Suppose ab = cd, where a, b, c, d are positive integers. Can a + b + c + d be a prime?

∗

Problem 5. We start with four identical right triangles. In one move we can cut one of the triangles along the altitude perpendicular to the hypotenuse into two triangles. Prove that, after any number of moves, there are two identical triangles among the whole lot. Problem 6. A team of geologists took 80 cans of food on an expedition. The weights of the cans were all different and were listed precisely in an inventory. After a while the labels became unreadable and only the cook knew which can was which. She boasted that she could prove the identity of all the cans simultaneously using only the inventory and a few weighing operations. Her balance has two pans and a pointer that shows the difference in weight between the contents of the pans. (a) Show how the cook can prove her claim using four weighings. (b) Can she do it in only three? Level C Problem 1. The number sin α is known. What is the largest number of possible values for (a) sin α2 ? (b)* sin α3 ? Problem 2. See Problem 58.9.2.

REF 58.9.2

Problem 3. The diagonals of a trapezoid ABCD meet at a point K. Let two circles be constructed, each having one of the lateral sides of the trapezoid as a diameter. Supposing that K lies outside both circles, prove that the tangents from K to these circles have equal lengths. Problem 4. See Problem 58.9.5.

REF 58.9.5

Problem 5*. Prove that if a, b and c are integers such that a/b + b/c + c/a and a/c + c/b + b/a are also integers, then |a| = |b| = |c|.

Problem 6. A panel has a number of buttons and a number of light bulbs. Each button is connected to some of the lights, and pressing the button flips the state of the lights it’s connected with from on to off or vice versa. Some of the lights are on at the start. It is known that, for any set of lights, there is a button that flips an odd number of lights from this set. Prove that one can switch off all the lights by pressing buttons.

∗

YEAR 1995 OLYMPIAD

13

Level D Problem 1. Prove that |x| + |y| + |z| ≤ |x + y − z| + |x − y + z| + |−x + y + z|

for all real numbers x, y, z.

Problem 2. Is it possible to color the edges of an n-angled prism with three colors so that all three colors occur among the edges of any face and among the edges meeting at any vertex? Answer for (a) n = 1995 and (b) n = 1996. Problem 3. In a triangle ABC, consider the median AA1 , the bisector AA2 , and a point K on AA1 such that KA2 k AC. Prove that AA2 ⊥ KC. Problem 4*. (a) Divide the interval [−1, 1] into black and white intervals so that the integral of any linear function over intervals of one color is the same for both colors. (b) Same problem with quadratic trinomials instead of linear functions. Problem 5. Find the largest value of n such that there are two-sided infinite strings A and B satisfying the following conditions: • Any subtring of B of length at most n is contained in A. • A is periodic with minimal period 1995, but B is not (it may be aperiodic or have a different period). The strings can contain arbitrary symbols. Problem 6*. Prove that there exist infinitely many nonprime values of n such that 3n−1 − 2n−1 is divisible by n. Problem 7. Is there a polyhedron and a point outside it such that from this point none of the polyhedron’s vertices is visible?

∗ ∗

Year 1996 Olympiad Level A Problem 1. If a + b2 /a = b + a2 /b, is it true that a = b? Problem 2. Ten iron weights are arranged around a circle. A small bronze ball is placed between each pair of neighboring weights. The mass of each ball equals the difference of the masses of the neighboring weights. Prove that the balls can be placed on the two pans of a balance so as to bring it into equilibrium. Problem 3. Each intersection of a square grid is home to a gardener, and there are flowers planted everywhere. Each flower must be looked after by the three gardeners closest to it. Draw the area that the gardener at the origin must look after. Problem 4. The side BC of an equilateral triangle ABC is divided into three equal parts by the points K and L, and point M divides the side AC in the ratio AM : M C = 1 : 2. Prove that the sum of angles AKM and ALM is 30◦ . Problem 5. A rook is at the corner of an n × n chessboard. It makes n2 successive moves (of any length), alternating between horizontal and vertical motion. For what values of n can this sequence of moves be chosen so that the rook stops on all the squares of the board and finally returns to its original position? (Note that each move must be followed by one in a perpendicular direction.) Problem 6. (a) Eight students were given eight problems to work on. It turns out that each problem was solved by five students. Prove that there are two students such that each problem was solved by at least one of them. (b) If each problem was solved by four students, show by a counterexample that two students satisfying the condition in part (a) need not exist. Level B Problem 1. Prove that any convex polygon has at most 35 angles less than 170◦ .

16

PROBLEMS

Problem 2. Suppose the numbers a, b, and c satisfy the inequalities |a − b| ≥ |c|,

|b − c| ≥ |a|,

|c − a| ≥ |b|.

Prove that one of these numbers is the sum of the other two. Problem 3. Consider two points A and B on the circle circumscribed to a triangle ABC (the circumcircle), and assume the tangents to the circle through A and B meet at a point M . Choose a point N on the side BC so that the line M N is parallel to the side AC. Prove that AN = N C. Problem 4. The integers from 1 to n are written in a row. The same numbers are written under them in a different order. Is it possible to write the second row so that all the sums of pairs of corresponding numbers in the two rows are perfect squares when (a) n = 9; (b) n = 11; (c) n = 1996? Problem 5. Two points A and B on a circle divide it into two arcs. Consider all the chords joining a point on one arc AB to a point on the other. What is the locus of midpoints of these chords? Remark. The statement is ambiguous on whether the endpoints A and B should be regarded as part of the arcs. For definiteness, assume they are not.

Problem 6. Ali Baba and a thief must share a trove of 100 gold coins. They arrange them into 10 piles of 10 coins each. Ali Baba takes four cups, places them next to four piles of his choice, and transfers from each of these piles into the corresponding cup some number of coins — at least one, but not the entire pile. The number of moved coins need not be the same for the four piles. After that, the thief must permute the cups without leaving them all in place, and then move the coins from each cup onto the pile the cup is next to now. This process is repeated: Ali Baba again places the empty cups near any four of the ten piles, and so on. At any time, Ali Baba can take three piles of his choice and leave; his opponent gets all the rest. What is the greatest number of coins that Ali Baba can walk away with if the thief also works to get as many coins as possible? Level C Problem 1. The positive numbers a, b, and c satisfy the equation a2 + b2 − ab = c2 . Prove that (a − c)(b − c) ≤ 0. Problem 2. A hundred points are marked on a piece of paper in a 10 × 10 square array. How many lines not parallel to either side of the square are required if we want lines going through all 100 points?

Adaptation

YEAR 1996 OLYMPIAD

17 A

Problem 3. Divide side BC of an equilateral triangle ABC into n equal parts with points P1 , P2 , . . . , Pn−1 as in the figure, so BP1 = P1 P2 = · · · = Pn−1 C. Choose a point M on the side AC so that AM = BP1 . Prove that \AP1 M +\AP2 M +· · ·+\APn−1 M = 30◦ if (a) n = 3; B P1 (b) n is an arbitrary positive integer.

P2

M

P n−1 C

Problem 4. A rook is on a corner square of an m × n chessboard. Two players take turns moving it, either straight across or up and down, any number of squares each time; but the rook is not allowed to land on, or even cross, a square on which it has already landed or through which it has passed. The player who cannot make a move loses. Which player can force a to win: the one who plays first or the second one? Describe a winning strategy. Problem 5. Two laws are in force in a certain country: (1) A person may play basketball only if he/she is taller than most of his/her neighbors. (2) A person gets free bus rides only if she/he is shorter than most of her/his neighbors. By a person’s “neighbor” is meant anyone whose house (regarded as a point) lies inside a certain circle whose center is the person’s house and whose radius is chosen by the person — and a different choice of radius is allowed for the purposes of the first law and the second law! Is it possible that 90% of people or more are allowed to play basketball and that 90% or more are allowed free travel? Problem 6. Prove that for any polynomial P (x) of degree n with positive integer coefficients, there exists an integer k such that P (k), P (k + 1), . . . , P (k+1996) are composite numbers if (a) n = 1; (b) n is an arbitrary positive integer. Level D Problem 1. See Problem 1 on page 16. Problem p √ 2. Find a polynomial with integer coefficients having 5 2− 3 as a root.

p √ 5 2+ 3 +

Problem 3. A point is chosen on each of eight evenly spaced parallel planes. Can these points be the vertices of a cube? Problem 4. Prove that there exist infinitely many positive integers n with the property that n is the sum of two perfect squares, but n − 1 and n + 1 are not. Problem 5. A point X lies outside two disjoint circles ω1 and ω2 , and the tangent segments drawn from X to ω1 and ω2 are equal. Prove that the intersection of the diagonals of the quadrilateral formed by the four tangency

18

PROBLEMS

points coincides with the intersection of the two internal common tangents of ω1 and ω2 . Problem 6. The 2n possible strings of length n containing only the numbers 1 and −1 form a table with 2n rows and n columns. Suppose that a number of entries in the table are replaced by 0. Prove that one can select certain rows so that their sum (defined as the string obtained by adding each column separately) is a string of zeros.

Year 1997 Olympiad Level A Problem 1. Chess pieces are placed on the squares of a chessboard in such a way that each row has at least one piece and the number of squares occupied in each row is different for every row. Prove that one can choose eight occupied squares so that every row and every column contains exactly one of the chosen squares. Problem 2. The walk from an observation station to the top of Mount Stromboli takes 4 hours along the road plus 4 hours along a path. There are two craters at the top. The first erupts for 1 hour, then lies quiet for 17 hours, then erupts for 1 hour again, and so on. The second crater erupts for 1 hour, lies quiet for 9 hours, erupts for 1 hour and so on. It is dangerous to walk either along the path or along the road during eruptions of the first crater, but when the second crater erupts, only the path is dangerous. Professor Garibaldi, a volcanologist, sees both craters start erupting simultaneously, exactly at noon. Will Garibaldi ever be able to walk up to the top of the volcano and back without endangering his life? Problem 3. Points M and N are chosen inside an acute angle XOY so that \XON = \Y OM . Points Q and P are chosen on the line segments OX and OY , respectively, so that \N QO = \M QX and \N P O = \M P Y . Prove that the paths M P N and M QN have the same length. Problem 4. Prove that there exists a positive integer that yields a composite number whenever any three adjacent digits are replaced by arbitrary digits. Is there a 1997-digit number with this property? Problem 5. In a rhombus ABCD, the angle B measures 40◦ , a point E is the midpoint of BC, and F is the foot of the perpendicular dropped from A to DE. Find the measure of angle DF C. Problem 6. A banker found out that one of a number of seemingly identical coins is counterfeit (lighter than the others). He asked an expert to find the lighter coin. To make matters interesting, the expert decided to do this

20

PROBLEMS

using only a two-pan balance, without weights, and to use each coin in at most two weighings. What is the greatest number of coins such that the expert can be sure of being able to determine the counterfeit coin after n weighings? Level B Problem 1. The length of one side of a triangle is one-third the sum of the lengths of the other two. Prove that the angle opposite this side is the smallest angle of the triangle. Problem 2. You have nine pieces of cheese, all of different weights. Can you always cut one of them into two parts so that the ten pieces thus obtained can be divided into two groups that weigh the same, each with five pieces? Problem 3. A convex hexagon AC1 BA1 CB1 satisfies the relations AB1 = AC1 , BC1 = BA1 , CA1 = CB1 , and \A + \B + \C = \A1 + \B1 + \C1 . Prove that the area of the triangle ABC is half that of the hexagon. Problem 4. Imagine n metro trains running in the same direction at equal intervals along a circular line. Stations A, B, and C are located along the line in this order, and the distances from A to B, from B to C and from C to A are all the same. Ira and Lyosha enter stations A and B, respectively, at the same time, planning to take the next train. It is known that if they enter their stations at the moment when Roma, the conductor of one of the trains, is going through an underground portion of the line, then Ira will get on a train before Lyosha; at any other time Lyosha will be the first to board, or they will both board simultaneously. What portion of the line runs underground? Problem 5. Two round-robin tournaments had the same 2n participants. (In a round-robin tournament every participant plays against every other once.) A win is worth one point, a tie half a point, and a loss zero. Prove that if the scores of each player in the two tournaments differ by at least n points, they must differ by exactly n points. Problem 6. Suppose that 1 + x + x2 + · · · + xn = F (x)G(x), where F and G are polynomials whose coefficients are only zeros and ones. Prove that one of the polynomials F (x) and G(x) is representable in the form (1 + x + x2 + · · · + xk )T (x), where k > 0 and T is also a polynomial with coefficients 0 and 1. Level C Problem 1. Is there a convex solid, other than a ball, whose orthogonal projections onto three pairwise perpendicular planes are disks?

← In the original, the solution assumes the weights are all different. The translation I got tried to relax that assumption. But I think that doesn’t work since the problem insists that one piece must be *divided* into two and this would not be possible with weights 1,1,...,1,8. So I’m clarifying that the weights have to be different. Adaptation

YEAR 1997 OLYMPIAD

21

Problem 2. Prove that the minimum perimeter of a quadrilateral with given diagonal lengths and angle between diagonals is attained by a parallelogram. Problem 3. (a) As the perimeter of a quadrilateral is traced clockwise, each of its sides is extended by its length. The endpoints of the extensions turn out to be vertices of a square. Prove that the original quadrilateral is a square. (b) Prove that if a similar construction applied to an n-gon yields a regular n-gon, the original n-gon is also regular. Problem 4. Given real numbers a1 ≤ a2 ≤ a3 and b1 ≤ b2 ≤ b3 such that a1 + a2 + a3 = b 1 + b 2 + b 3

and a1 a2 + a2 a3 + a1 a3 = b 1 b 2 + b 2 b 3 + b 1 b 3 , prove that if a1 ≤ b1 , then a3 ≤ b3 . Problem 5. In a round-robin tournament everyone plays everyone else; a win scores 1 point and a loss 0. (We assume there are no ties.) Suppose we compute a “superscore” for each player by adding up the scores of the opponents who lost to that player. If the superscores of all players are the same and there are more than two players, prove that the scores of the players are also all the same. Problem 6. Consider the powers of five (1, 5, 25, 125, 625, . . . ) and take the sequence of their first digits: 1, 5, 2, 1, 6, . . . Prove that any segment of this sequence written in reverse order occurs in the sequence of first digits of powers of two (1, 2, 4, 8, 1, 3, 6, 1, . . . ). Level D Problem 1. Points C ′ , A′ , and B ′ are taken on the sides AB, BC, and CA of a triangle ABC, respectively (see figure). Prove that the triangle A′ B ′ C ′ has area C′ ′ ′ ′ ′ ′ ′ AB · BC · CA + AC · CB · BA , 4R where R is the circumradius of the triangle ABC. Problem 2. Calculate the integral Z π/2 cos2 (cos x) + sin2 (sin x) dx.

A

B A′

B′

C

0

Problem 3. Three functions, 1 f1 (x) = x + , f2 (x) = x2 , and f3 (x) = (x − 1)2 , x are written on the blackboard. You are allowed to add, subtract, and multiply these functions (and so also square them, cube them, . . . ), multiply

22

PROBLEMS

them by an arbitrary number, add an arbitrary number to them, and perform the same operations with the expressions thus obtained. Construct the function 1/x by means of these operations. Prove that if we erase any of the functions f1 , f2 , f3 from the blackboard, the function 1/x will no longer be constructible. Problem 4. Is it possible to cut a regular tetrahedron with a unit edge into regular tetrahedrons and octahedrons with edges of length less than 1/100? Problem 5. Let a, b, and c be positive numbers such that abc = 1. Prove the inequality 1 1 1 + + ≤ 1. (1) 1+a+b 1+b+c 1+c+a Problem 6. A disk of radius 1 and a finite number of (infinite) strips whose widths add up to 100 are given on the plane. Prove that each of the strips can be translated so that together they cover the disk completely.

Year 1998 Olympiad Level A Problem 1. Are there positive integers x, y, and z that satisfy the equation 28x + 30y + 31z = 365? Problem 2. Is it possible to find eight positive integers such that none of them is divisible by any other, but their squares are all divisible by each of the eight initial numbers? Problem 3. The dialgonals AC and BD of a parallelogram ABCD meet at a point O. Point M lies on the line AB and \AM O = \M AD. Prove that M is equidistant from C and D. Problem 4. Some of the numbers a1 , a2 , . . . , a200 are written in blue, the rest in red. If we erase every red number, the remaining numbers will be all the positive integers from 1 to 100, written in ascending order. If we erase every blue number, the remaining numbers will be all positive integers from 100 to 1 written in descending order. Prove that the set {a1 , a2 , . . . , a100 } coincides with the set of integers from 1 to 100. Problem 5. Several guests are seated around a round table. Some of them are acquainted with each other; each acquaintance is mutual. All the acquaintances of each person (including the person himself or herself) are seated at equal intervals. For different people these intervals may be different. It is known that any two guests have at least one common acquaintance at the table. Prove that everybody is acquainted with everybody else. Problem 6. A red square is covered by 100 white squares. All 101 squares are congruent and the sides of each white square are parallel to the sides of the red square. Is it always possible to remove one of the white squares so that the remaining white squares will still completely cover the red square? Level B Problem 1. Is 49 + 610 + 320 a prime number?

24

PROBLEMS

Problem 2. Two altitudes, AD and CE, are drawn in an acute triangle ABC. Then a square ACP Q and two rectangles CDM N and AEKL are drawn, with AL = AB and CN = CB. Prove that the area of ACP Q is the sum of the areas of AEKL and CDM N . Problem 3. Each of the inhabitants of a village either always tells the truth or always lies. The villagers formed a circle, and a traveler asked each of them whether his or her neighbor to the right is honest or lies. Based on their answers, the traveler managed to determine the exact fraction of truthful people among the villagers. Your task is to find this fraction without having heard the answers. Problem 4. A network of military bases in Mathistan is connected by roads, each road starting and ending at a base. A set of roads is said to be essential if, when these roads are closed, there are bases no longer connected by roads. An essential set is said to be strategic if it does not contain a smaller essential set. Let two different strategic sets V and W be given. Prove that the set of roads that belong to exactly one of V and W is essential. Problem 5. A point O lies inside a rhombus ABCD. The angle DAB measures 110◦ , while \AOD and \BOC measure 80◦ and 100◦ , respectively. What values can the measure of angle AOB take? Problem 6. A number of distinct points are marked on the interval [0, 1]. Each of these points is either halfway between two other marked points (not necessarily its neighbors) or halfway between a marked point and an endpoint of the interval. Prove that all the marked points are rational. Level C Problem 1. Let a, b, and c be nonnegative integers such that 28a + 30b + 31c = 365. Prove that a + b + c = 12. Problem 2. A square of side 1 is cut up into rectangles. One side of each rectangle is marked. Prove that the sum of lengths of all the marked sides is at least 1. Problem 3. A road of length 1 km is entirely lit by a certain number of street lamps, each of which lights a segment of road 1 m long. Suppose that turning off any of the lamps causes the road no longer to be lit in its entirety. What is the highest possible number of lamps? Problem 4. Is there a positive integer divisible by 1998 whose digits add up to less than 27? Problem 5. An equilateral triangle ABC is cut out of plywood and placed on the floor. Three nails are hammered into the floor, one touching each side of the triangle, so that it is impossible to turn the triangle without lifting it

YEAR 1998 OLYMPIAD

25

off the floor. The first nail divides the side AB in the ratio 1 : 3, measuring from A, another one divides BC in the ratio 2 : 1, measuring from B. In what ratio does the third nail divide the side AC? Problem 6. The positive integers from 1 to n are arbitrarily arranged in a sequence (the number n is fixed). An arrangement is said to be bad if we can select 10 of its numbers (not necessarily consecutive) so that they form a decreasing sequence. All other arrangements are said to be good. Prove that the number of good arrangements does not exceed 81n . Level D Problem 1. Three numbers x, y, z satisfy the equation x + y + z − 2(xy + yz + xz) + 4xyz = 21 .

Prove that at least one of them is equal to 12 .

Problem 2. It is known that a continuous function f : (1) is defined on the entire real axis; (2) has a derivative at each point (thus, the graph of f has a unique tangent at each point); (3) is such that its graph does not contain points one of whose coordinates is rational and the other one irrational. Does it follow that the graph of f is a straight line? Problem 3. Two medians, AK and BL, are drawn in a scalene triangle ABC. The angles \BAK and \CBL measure 30◦ . Find the angles of the triangle ABC. Problem 4. Find all the solutions of 3x + 4y = 5z in positive integers. Problem 5. Is it possible to make a closed chain of 61 identical coherently rotating cogwheels in space in such a way that the angles between meshed cogwheels are at least 150◦ ? Assume that: (1) the cogwheels are disks (just for simplicity); (2) two cogwheels are meshed if the corresponding circles have a common tangent at their common point; (3) the angle between two meshed cogwheels is the angle between the radii of their circles drawn at the point of tangency; (4) the first cogwheel must be meshed with the second, the second with the third, . . . , the 61st with the first; no other pair of cogwheels should have a common point. Problem 6. See Problem 6 on page 25.

Answers

Year 1993 Olympiad Level A 1. (a) No solutions. (b) x = 1963. 3. No. 4. The amounts of lucky and unlucky time are the same. 5. There is always such a string. Level B 1. The locus is given by the shaded domains and solid curves in the figure. (The points on the dotted curves do not belong to the locus.) 2. x1000 = 21 · 1000 · 1003 = 501500. 3. No. 4. Pete has 14 friends. 5. 58.

A

B

Level C 1. 4 or 12. 2. Yes. 3. 700 m is not always possible, but 800 m is (if islands are not allowed). Here we’re considering the swimming distance 4. (a) Yes. (b) No. 6. The rather than the straight-line distance. √ maximum equals 12 (1 + 2)(a + b) and is attained when \ACB = 135◦ . Level D 1. pq/(q−p), unless p = q, in which case the answer is either 0 (if p = 0) or undefined (if p 6= 0). 2. Yes. 3. n lines if n > 2; one line if n = 2. 4. (a) Yes. (b) No. 5. (a) One solution is shown in the figure. (b) No. 6. √410 a.

1 1 2

−1

−

1 2

1 2

0

−

1 2

−1

1

∗

Year 1994 Olympiad Level A 1. Fifteen bottles. 2. 143 and 143. 5. There is more favorable time than unfavorable time. 6. The first player wins. Level B 1. Yes. 2. If k > l, Sue wins; otherwise Leo wins. 180625 becomes 17 times smaller if 8 is stricken out. Level C 1. 3333334 and 1666667. Level D √ 3. r = 34 3 2.

5. (a) Not always.

5. The number

6. Yes.

Year 1995 Olympiad Level A 1. Yes.

4. Yes in all three cases.

6. 240◦ .

∗

B

Level B 2. The locus is the union of the altitude of △ABC from vertex B with an arc of circle AC of measure 120◦ lying in △ABC; see figure. 3. For n ≤ 998 and for n ≥ 3989. A 4. No. 6. (b) No, she can’t. Level C 1. (a) 4; (b) 3. Level D 2. (a) yes; (b) no.

5. n = 1994.

7. Yes.

∗ C

∗

Year 1996 Olympiad Level A 1. Yes, it is true.

3.

5. For even n.

A Level B 4. (a) Yes; (b) no; (c) yes. 5. Let O be the center of the circle. Consider the circles with diameters AO and BO (see figure). The desired locus is the interior of these two circles minus their intersection. 6. 72 coins.

B

O

Level C 2. 18. 4. For m = n = 1 player 2 wins. In all other cases, player 1 wins. 5. Yes, it is possible. Level D 2. x5 − 5x3 + 5x − 4.

3. Yes, they can.

Year 1997 Olympiad Level A 2. Yes; Garibaldi can leave at the beginning of the 38th hour, say. 6. 2n2 + 1 coins.

5. 110◦.

Level B 2. Yes, it is always possible. 4. If n is divisible by 3, there are no underground portions; if n = 3k + 1 or n = 3k + 2 for k ∈ Z, two thirds or one third, respectively, of the track runs underground. Level C 1. Yes, such a solid exists. Level D 2. π/2. 3. 1/x = f1 (x) −

1 2


f2 (x) − f3 (x) + 1 .

4. Yes.

Year 1998 Olympiad Level A 1. Yes. For example, we can take x = 1, y = 4, and z = 7. 6. No. Level B 1. No. 3. Level C 3. 1998.

1 2.

4. No.

2. Yes.

5. 80◦ or 100◦ .

5. In the ratio 5 : 7, measuring from vertex A.

Level D 5 2. No. 3. \B = 120◦ , \C = arccos 4√ , \A = arccos √27 . 7 solution: x = y = z = 2. 5. Yes.

3. Unique

Hints

Year 1993 Olympiad Level A 1. (a) Use divisibility by 3. (b) Estimate x from below and find the remainder after division of x by 9. 2. Rewrite the expression (a2 + b2 + c2 )2 in a different form. 3. Consider all pairs of chips with a red chip above a blue chip. 4. Favorable and unfavorable times are interchanged by reflection. 5. Use induction on the number of letters in the alphabet. 6. Prove that \ADB = \ACB. Level B 2. Write down xn − xn−1 for n small. 3. From what triangles can a triangle with the initial angles be obtained? 4. Consider the most friendly and the least friendly of Pete’s classmates. 6. Consider the points symmetric to B with respect to AM . Level C 1. The digits of A + B must repeat in blocks of 12. 4. Draw the points {an + b} on the circle of unit circumference. 5. Estimate the total number of distinctions between all possible pairs of plants with respect to all features. Level D 4. Let there be 1, 2, . . . , n stones in the boxes. Express the maximal number of stones after one move in terms of n and k. Investigate for which k this number is minimal. 5. The graph of such a function would be left unchanged by a 90◦ rotation around the origin. 6. Let P and Q be the midpoints of the sides KL and MN of a space quadrilateral KLMN . Then P Q ≤ 21 (KN + LM ).

Year 1994 Olympiad Level A 1. Express the volume of the bottle in terms of the volume of the jug. 2. If x and y are the three-digit numbers to be found, then 7xy = 1000x+y. 3. Extend the perpendiculars until they intersect the line AC. 4. The grasshoppers can only hop to points in a square grid. 5. After an integer number of hours, the second hand and the minute hand return to their positions. 6. Use the axial symmetry of the rectangle. Level B 2. Leo should try to make isosceles triangles. 3. Set z = −x. 4. It suffices to prove that the triangles AQN and AM P are congruent. Prove that they are similar and that the arcs AQ and M A are equal. 6. (b) To prove that it is possible to place the next 1 × 3 ship, draw 8 auxiliary 1 × 3 ships in such a way that a 1 × 4 or 1 × 3 ship would touch not more than two of these auxiliary ships. Level C 1. Let x and y be the 7-digit numbers required. Then 3xy = 107 x + y. 2. (a) Observe how the denominators of the terms of the sequence change. (b) Fix n. One can write xn+k = bk xn + ak . What is |bk |? 3. Prove that some member of Parliament was slapped at most once. 4. AK = 12 (AB + AC − BC). 5. (b) Draw a chord and move it so it remains parallel to itself. (If the chord goes through a concave vertex of the polygon, it may split the polygon into more than two parts; do not forget to account for this case.) 6. Note that the condition “for all n > 1” can be replaced by “for n = 2 and n = 3”. If P has a term with a zero coefficient while all the coefficients of P 2 and P 3 are positive, it is possible to “jiggle” P a bit so it satisfies the condition of the problem.

46

HINTS

Level D 1. Try to construct a polyhedron with 6 faces. 3. What is the smallest value of r for which the circle of radius r with center (0, r) intersects the curve y = x4 other than at the origin? 4. Show that the polyhedron obtained by applying a homothety with center A and factor 2 contains the eight translated polyhedra. 5. Use Fact 25. 6. It is enough to find m such that 2m is just a bit less than a power of 10. For this, it suffices to find n such that 2n + 1 is divisible by a sufficiently large power of 5.

Year 1995 Olympiad Level A 2. Consider the difference between consecutive numbers. 3. Consider a 120◦ rotation about the point O. 5. You can either compare the fuel costs for deliveries having almost the same route, or deduce an explicit expression for the cost for any delivery. 6. Rotate all the angles of interest so their vertices coincide with A. Level B 1. Consider the difference between consecutive numbers. 2. For a fixed A′ , there exist precisely two points C ′ such that AA′ = CC ′ . 3. Start with a similar problem for the 5 × n rectangle. 5. The order in which we cut the triangles does not matter. 6. (a) The cook should first put the 27 lightest cans on the left pan. (b) Use the pigeonhole principle (Fact 1). Level C 1. (b) Express sin 3x in terms of sin x. 3. Use the tangent-secant theorem (power of a point) for an appropriate secant. 5. Let p be a prime divisor of a. Reduce the fractions to the least common denominator, and consider the highest powers of p dividing the numerator and denominator. 6. Use induction. (Assume it’s been proved that all the bulbs, except an arbitrary fixed one, can be switched off.) Level D 1. |a + b| ≤ |a| + |b|. 3. Complete the triangle BAC to a parallelogram and use similarity. 4. If f is a polynomial of degree n, then f (x) − f (x−d) is a polynomial of degree n − 1, for any fixed d 6= 0. 6. Look for n of the form 3b − 2b .

∗

∗

Year 1996 Olympiad Level A 1. A nonzero factor can be canceled from both sides of an equation. 2. Let mi be the mass of the ith weight; then (m1 − m2 ) + (m2 − m3 ) + · · · + (m9 − m10 ) + (m10 − m1 ) = 0. 3. Consider a flower and find out which gardeners look after it. 6. (a) Consider two cases: (1) there is a student who has solved six problems, and (2) each student has solved at most five problems. Level B 1. Use the formula for the sum of angles of a convex polygon. 3. Use Fact 15 (page 105). 4. (c) Reduce the problem to a similar one with a smaller n. 6. First, prove that the thief can always ensure that there are no piles of less than four coins. Then show that Ali Baba can ensure that there are seven piles of at most four coins each. Level C 2. Consider the 36 unit squares adjacent to the sides of the big square. 3. Express the angles in question in terms of angles at the points A and M . 5. Consider a country with 10 inhabitants whose houses are placed along a straght line. 6. If k divides x − y, then k divides P (x) − P (y). Level D p √ p √ 5 5 2. Compute 2 + 3 · 2 − 3. 3. Write the equations of parallel planes drawn through the vertices of the unit cube in such a way that the distances between any two neighboring planes are the same. 5. Use the law of sines and Fact 17 (page 105).

Year 1997 Olympiad Level A 2. Start the walk right after the first crater erupts and at such a time that four hours later the second crater has just calmed down. 3. Consider the reflections of M in the lines OX and OY . 5. Extend the lines DE and AB until they meet. 6. Consider the analogous problem under the condition that each coin can be used in only one weighing, rather than two. Level B 1. The smallest angle of a triangle is opposite the shortest side. 2. Cut the largest piece. 4. If the distance between two points is a multiple of the interval between trains, then either both points are underground or both are above ground. 5. Divide the participants of the tournament into two groups: those whose score in the second tournament is greater than in the first tournament, and those whose score in the first tournament is greater than in the second tournament. 6. Suppose that the coefficient of x in the polynomial F (x) is equal to one. Consider the smallest m such that the coefficient of xm in F (x) is zero. Then take k = m − 1. Level C −→ 2. Translate the quadrilateral ABCD by the vector AC . 3. Suppose that the construction in question applied to a polygon A1 . . . An yields a polygon B1 . . . Bn . Prove that the polygon A1 . . . An can be recovered from B1 . . . Bn . 4. Use the relations between the coefficients of a cubic polynomial and its roots. 5. Assuming that the statement is not true, consider the participants with the greatest number of wins and those with the smallest number of wins.

52

HINTS

6. Prove that for any k there exists a power of 2 whose decimal representation starts with a 1 followed by k zeros. Level D 2. Write the integral as the sum of two integrals and apply the substitution y = π/2 − x to one of them. 3. Applying the allowed operations to polynomials, we can obtain only a polynomial; therefore, we can’t do without f1 . To prove that f2 is indispensable, consider the derivatives of functions at x = 1. The indispensability of f3 is proved by means of complex numbers. 5. Set a = x3 , b = y 3 , and c = z 3 .

Year 1998 Olympiad Level A 1. Notice that 365 is the number of days in a year. 2. Use prime factorization. 3. Join the midpoints of the sides AB and CD of the parallelogram. 5. If a person has acquaintances sitting next to each other, this person is acquainted with all the guests.

Level B 1. (a + b)2 = a2 + 2ab + b2 . 3. Imagine that all truthful villagers become liars, and all liars are reformed. 4. If we close all the roads in a strategic set, the bases break up into exactly two disconnected sets such that no road joining two bases in one of these two sets belongs to the strategic set of roads. 5. Prove that there are exactly two such points O. 6. This problem requires a certain knowledge of linear algebra (namely, the Gaussian elimination algorithm); see Field2Elem and also Fact 25.

Level C 1. Assume that the statement is not true and derive a contradiction. 2. Write out the sum of the areas of all the rectangles. 3. If two segments of a road lighted by two lamps overlap, then these lamps are neighboring. 4. Invent a criterion for divisibility by 999. 5. Suppose there is one nail hammered on a side of the triangle. Find all the points in the plane such that the triangle can be rotated clockwise around any of these points and all the centers of possible counterclockwise rotations. 6. If an arrangement is good, the numbers can be colored in nine colors so that all the numbers of each of these colors form an increasing sequence.

CITE Field2Elem

54

HINTS

Level D 1. Factor the difference between the left- and right-hand sides of the given equation. 2. Define the function for x > 1 by one formula and for x ≤ 1 by another formula. 4. Analyze the parity of the variables.

Solutions page 95

Year 1993 Olympiad Level A Problem 1. (a) Since x, S(x) and S(S(x)) have the same remainder upon division by 3 (by Fact 6), the sum x + S(x) + S(S(x)) is divisible by 3. Since 1993 is not divisible by 3, there is no solution. (b) Clearly x < 1993. It is easy to see that 1899, 1989 and 999 have the largest sum of digits among the numbers from 1 to 1993. Thus S(x) ≤ 27. Further, S(S(x)) ≤ S(19) = 10 and S(S(S(x))) ≤ 9. The equation to be solved implies that x = 1993 − S(x) − S(S(x)) − S(S(S(x))) ≥ 1993 − 27 − 10 − 9 = 1947. As in part (a), we know that x, S(x), S(S(x)) and S(S(S(x))) leave the same remainders when divided by 9; let’s call this remainder r. Then 4r has remainder 4 when divided by 9 (because that’s the remainder of 1993). In other words, 4r − 4 is divisible by 9, which implies that r − 1 is divisible by 9 (since 4 and 9 are coprime; see Fact 9). So r = 1. Now list the numbers from 1947 through 1993 that have remainder 1 upon division by 9: they are 1954, 1963, 1972, 1981 and 1990. We can verify directly that only 1963 satisfies the equation. Problem 2. Let n = a2 + b2 + c2 . Expand the square and rearrange: (a2 +b2 +c2 )2 = a4 +b4 +c4 +2a2 b2 +2a2 c2 +2b2 c2 = (a4 +b4 +c4 +2a2 b2 −2a2 c2 −2b2 c2 )+(2ac)2 +(2bc)2

= (a2 +b2 −c2 )2 +(2ac)2 +(2bc)2 .

We can assume that a ≥ b ≥ c > 0, so a2 + b2 − c2 > 0. Thus we have expressed n2 as the sum of squares of three positive integers. Remarks. 1. Try to prove a similar statement for the sum of four and more squares. 2. For the sum of two squares the analogous statement is not true: (12 + 12 )2 = 4 cannot be represented as the sum of two squares of positive integers, although an analogous identity is true: (a2 + b2 )2 = (a2 − b2 )2 + (2ab)2 .

page 96

58

SOLUTIONS For more on sums of two squares, see the solution to Problem 59.11.4 on page 4 REF 59.11.4 and Remark 2 thereto.

3. A famous theorem of Lagrange says that any positive integer can be represented as the sum of four integer squares. (An integer square, or simply a square, means the same as the square of an integer. This includes 0 as well as the squares of positive integers.) 4. The number 7 is not a sum of three squares. It turns out that a positive integer is not representable as a sum of three squares if and only if it is of the form (8k + 7) · 4m .

∗

5. Suppose that instead of specific numbers we consider sums of squares of arbitrary variables. An identity going back at least to the third-century Greek mathematician Diophantus of Alexandria says that (a2 + b2 )(c2 + d 2 ) = (ac + bd)2 + (ad − bc)2 , which of course can also be written as (a21 + a22 )(b21 + b22 ) = (a1 b1 + a2 b2 ) + (a1 b2 − a2 b1 )2 .

This elegant equality was extended by Euler in the eighteenth century to the case of four pairs of variables: (a21 +a22 +a23 +a24 )(b21 +b22 +b23 +b24 ) =

(a1 b1 −a2 b2 −a3 b3 −a4 b4 )2 + (a1 b2 +a2 b1 +a3 b4 −a4 b3 )2

+(a1 b3 +a3 b1 +a4 b2 −a2 b4 )2 + (a1 b4 +a4 b1 +a2 b3 −a3 b2 )2 . A similar identity also exists representing (a21 + a22 + · · · + a2k )(b21 + b22 + · · · + b2k )

(1)

as a sum of squares of polynomials when k = 8. However, for no other k — apart from the already discussed k = 2, 4, 8 and the trivial case k = 1 — is the product (1) writable as a sum of squares of polynomials! The proof of this fact is far from elementary; to intrigue the reader, we mention only that the identity page 97 for k = 2 is related to complex numbers, for k = 4 to quaternions, and for k = 8 to Cayley numbers, also called octonions. More about this in the book Cantor-Solodovnikov. CITE Cantor-Solodovnikov

Problem 3. First solution. Consider all pairs of chips in the stack, not necessarily adjacent. There are four possibilities for such pairs, in terms of the ordering of colors: RR, RB (meaning a red chip above a blue R chip), BR and BB. For instance, in the stack RBRB of the figure, B R there are three RB pairs (first and second chips from the top, first B and last, third and last). The reader can check that the parity (see Fact 23) of the number of RB pairs cannot change under our operations. For example, suppose we insert two red chips at a spot that has k blue chips below it. It is easy to see that this adds exactly 2k new RB pairs, so the parity remains the same. The other operations — insertion of two blue chips and removal of two blue or two red chips — can be analyzed similarly.

YEAR 1993 OLYMPIAD

59

Now, in the initial position, there is exactly one RB pair. In the desired final position, there are none. Since 1 and 0 have different parities, it’s not possible to reach one state from the other. Remark. The parity of the number of RB pairs in this solution is an invariant (see Fact 2), since it remains unchanged no matter what happens to the stack, under the rules of the problem. One could alternatively consider the number of red chips below which there is an odd number of blue chips; the parity of this number is also invariant. (Check it!) The next solution also relies on an invariant, but of a different type: a transformation of the real line into itself. (A transformation is a map that is one-to-one and onto, and therefore has an inverse.) Recall that two maps from a set to itself can be composed, that is, applied successively, producing a third map. If f : R → R and g : R → R are maps of the real line, the composition f g (also written f ◦ g) is the map obtained by applying g, then f :

∗ removed “Though

more abstract... final state.”

I’m leaving “Recall...” here since it’s background material (f g)(x) = f (g(x)). and not part of the The composition of any map f with the identity (the map Id such that Id(x) = x solution proper. for all x) is of course f again. The composition of several maps depends on the order in which they are applied, but not on the grouping: in symbols, ∗ (f g)h = f (gh), because both of these maps take x to f (g(h(x))). page 100

Second solution. Consider two transformations of the real line, r and b, defined by r(x) = 1 − x (this is a reflection in the point 21 of the line) and b(x) = −1 − x (a reflection in the point − 21 ). Obviously, composing r with r brings each point of the line back to itself, and so gives the identity map; in algebraic notation (see preceding remark), rr = Id. Similarly, bb = Id. The composition of two reflections of the line is, in general, a translation. For instance, applying b, then r, has the overall effect of a translation to the right by 2: (rb)(x) = r(b(x)) = 1 − (−1 − x) = x + 2.

(1)

Now let’s associate to red chips R the reflection r, and to blue chips B the reflection b. For any stack of chips, imagine reading off the colors from the bottom of the stack up and applying consecutively the reflection corresponding to each color. The overall effect of all these reflections r and b, in the given order, is again a map of the real line. The initial stack in the problem, RB, corresponds to applying b then r; this, according to equation (1), is the translation by 2. But if we consider the reverse stack BR, the composition gives a different result altogether: b(r(x)) = −1 − (1 − x) = x − 2. The composition of transformations is not, in general, commutative! Now the key observation is this: two adjacent chips of the same color cancel each other out from the point of view of composition: their combined

∗

60

SOLUTIONS

effect, as we have seen, is the identity. This means that inserting or removing two adjacent chips of the same color has no effect on the overall map represented by the stack of chips; for example, rb = bbrb = brrbrb. Therefore no sequence of such insertions of removals can lead from the stack RB to the stack BR, as they represent different overall maps. Remarks. 1. (Follow-up on the second solution, for more advanced readers familiar with group theory language.) The set of compositions is the group of transformations of the real line generated by r and b, and these generators satisfy the relations r2 = 1 and b2 = 1. One can also define an abstract group with generators r and b — these letters now being regarded just as symbols — subject to the relations r2 = 1 and b2 = 1, meaning that whenever bb or rr appear in a product of generators, they can be erased. By interpreting multiplication as composition we obtain a map from the abstract group to the group of transformations; it can be checked that this is an isomorphism (that is, the group of transformations has no additional relations beyond those in the abstract group). Mathematicians call such an isomorphism a faithful representation of the abstract group in R. This abstract group can be shown to be isomorphic to Z o Z/2, the semidirect product of Z and Z/2. 2. The problem we’ve solved (twice!)comes up in a proof of an amusing and geometrically intuitive fact: Two continuous paths inside a square, each of which joins a pair of opposite vertices, must intersect. We sketch the proof in the case where the paths are polygonal. (The reader acquainted with continuity might try to prove that this implies the general case, using two facts from topology: the image of a continuous path [0, 1] → R2 is compact; and if two compact sets don’t intersect, there is some minimum positive distance between points in one set and points in the other.) So let’s consider two nonintersecting polygonal paths as described, one blue and one red, inside a square. We make the square’s sides vertical and horizontal. Now move a vertical line continuously from the left edge of the square to the right edge, keeping track of the intersections of this line with the red and blue paths. The pattern of red and blue intersection points corresponds to the stack of chips in the problem. (Since the paths don’t intersect we can assume that none of the polygonal segments is vertical: just jiggle a vertex a tiny bit if it is. Therefore the intersection of either path with a vertical line is a finite set.) The intersection points always appear and disappear in pairs of a single color, as in the problem. (See the diagram on the right for an appearing pair and a disappearing one.) Therefore if on the left edge of the square we have the red dot above the blue dot, we cannot have the inverse situation on the right edge, contradicting the assumption that the paths join opposite corners. 3. The geometric result just proved is closely related to the Jordan curve theorem, which states that every closed plane curve without self-intersections divides the plane into two parts, an inside and an outside. This intuitively clear statement is difficult to prove, since the curve may have no smooth pieces. But in the case of polygonal closed curves an argument very similar to the one above provides a proof.

∗

∗

YEAR 1993 OLYMPIAD

61 page 101

Problem 4. The key idea is that the mirror image of a lucky pattern of hands is unlucky, and vice versa. Consider the position of the hands at two distinct times: T seconds before noon and T seconds after noon. The patterns formed by the hands of the clock at those moments are mirror images of each other, with respect to the axis of symmetry formed by a vertical diameter. (Why?) For example, here is a clock showing 1 hour, 15 minutes, and 22 seconds after noon (01:15:22 PM) and another showing the same amount of time before noon (10:44:38 AM): 12

12

9

3 6

6

Now, a moment’s thought shows that reflection interchanges lucky and unlucky patterns of hands. Thus, to each lucky moment before noon there corresponds an unlucky moment after noon. An interval of lucky moments in the morning is matched by an interval of unlucky moments in the afternoon or evening, and both intervals have the same length. Similarly, an interval of unlucky moments before noon is matched by an equal interval of lucky moments after noon. Hence the total amount of lucky time in the day equals the total amount of unlucky time.

∗

Remarks. 1. It’s not enough to establish a one-to-one correspondence between lucky and unlucky moments; we need to consider the length of the intervals. This question is discussed further in Remark 2 after the solution to Problem 57.8.5, which deals with a similar situtation (see page 83). REF 57.8.5 2. It is also not enough to argue that each lucky pattern of hands has a mirror image that is an unlucky pattern of hands, because the mirror image might conceivably not correspond to an actual moment in time. Some combinations of hand positions simply cannot occur on a properly functioning clock. (Find examples!) So one must establish a correspondence between actual intervals of time.

Problem 5. Consider the sequence of strings A, ABA, ABACABA, ABACABADABACABA, . . . each string being obtained from the previous one by writing it twice and inserting the first unused letter between the two copies. When we run out of letters, the last string will provide an affirmative answer to the problem. We will prove this by complete induction (Fact 24). For convenience, we will denote the n-th string of the sequence by Zn , so Z1 = A, Z2 = ABA, etc. We also denote the n-th letter of the alphabet by xn , so x1 = A, x2 = B, . . . , x26 = Z. Then the sequence of strings is defined

page 102

62

SOLUTIONS

by the properties Z1 = x1

and

Zn+1 = Zn xn+1 Zn .

The “multiplication” on the right-hand side simply means that we are concatenating (writing one after the other) the string denoted by Zn , the letter xn+1 , and again the string Zn , is this order. Obviously this process stops when we run out of letters, that is, after n = 26. We claim that for any n ≤ 26, (a) the string Zn doesn’t have identical adjacent substrings, but (b) a pair of identical adjacent substrings appears as soon as one writes any one of the first n letters of the alphabet either at the beginning or at the end of Zn . Base of the induction. For n = 1, the statement is obvious. Induction step. Suppose (a) and (b) are true for Z1 , . . . , Zn−1 , and consider the n-th string, Zn = Zn−1 xn Zn−1 . Suppose Zn has two identical adjacent substrings. They cannot contain the central letter xn , since there is only one copy of it. Therefore they lie both to the left or both to the right of the central letter; that is, they’re identical adjacent substrings of Zn−1 . But such substrings cannot exist, by the induction assumption. This proves statement (a) for Zn . To prove (b), again we proceed by cases. Suppose we write after Zn one of the first n letters of the alphabet. If the letter we wrote is the central letter xn , the result is two copies of Zn−1 xn . If we wrote any other letter xk , the string now ends with Zn−1 xk , with k < n. But the induction assumption says that Zn−1 xk contains two identical adjacent substrings somewhere. Either way, we’ve found the desired identical substrings. The argument also applies if we write a letter to the left of Zn instead of to the right. Having proved the induction, we now just take n = 26, so all the letters I prefer to leave the of the alphabet are allowed. (The word Zn thus constructed has 226 − 1 ∗ 50 million here for letters, or more than 50 million!) esthetic reasons, but Remark. This construction is important in combinatorics and the theory of semi- I’ve put it in groups. Try to prove that in any infinite sequence of letters of the alphabet, parentheses. there must be somewhere a string in the pattern of Zn , for any n. By a string in the pattern of Zn we mean one that is obtained from Zn by replacing each letter xi by a fixed nonempty string Xi ; for instance “abracadabra” is a string in the pattern of Z2 (take X1 = abra and X2 = cad).

Problem 6. The circle tangent to BC and to the extensions of sides AB and AC is an excircle of the triangle ABC. Its center O is the intersection of the bisectors of the angle A and the exterior angle at vertex B; see Fact 16. Let α, β, γ be the angles of △ABC. Then \BAO = α/2 and \CBO = (180◦ − β)/2, and B

\ABO = β + \CBO = 90◦ + β/2.

O

β

A

α/2 α/2

γ C D

page 103

YEAR 1993 OLYMPIAD

63

From △AOB we see that


\AOB = 180◦ − α/2 − 90◦ + β/2 = 90◦ − α/2 − β/2 = γ/2,

since α + β + γ = 180◦ . On the other hand, we have \AOB = 21 \ADB, because this is the angle inscribed in the circle centered at D. Hence \ADB = γ. Thus, \ADB = \ACB. By the converse of the theorem on inscribed angles, the points A, B, C and D lie on one circle. Remarks. 1. The points O, C, D lie on a straight line since \AOC = β/2 and \AOD = β/2 (prove it). 2. The statement is true for the inscribed circle as well (prove it).

Level B

C C C Problem 1. We first draw the perpendicular to AB C C ◦ through the endpoint A. Clearly, \A < 90 if and only if B and C lie on the same side of this line. Applying the same argument to B, we see that the locus of points C such that \A < 90◦ and \B < 90◦ A B is the strip bounded by the perpendiculars to AB at both A and B. (See figure on the right.) Next we construct the circle with diameter AB. From Fact 14 we know that a point C is outside this circle if and only if it satisfies \ACB < 90◦ . Taking the intersection with the strip already found, we see A B that the locus of points C such that the ABC is acute is the shaded set to the left. Finally we study the condition that the angle A is intermediate between the others, that is, either \B ≤ \A ≤ \C

or

\C ≤ \A ≤ \B.

(1)

Since the greater angle of a triangle subtends the longer side, the condition \B ≤ \A ≤ \C is equivalent to AC ≤ BC ≤ AB. The points equidistant from A and B are those on the perpendicular bisector L of the segment AB; therefore AC ≤ BC if and only if C is in the half-plane determined by L and containing A. At the same time, BC ≤ AB if and only if C lies in the circle of center in B and radius AB. Thus, the locus of points satisfying AC ≤ BC ≤ AB

page 104

64

SOLUTIONS

is the shaded set in diagram (a) below:

A

B

A

(a)

B

A

(b)

B

(c)

Similarly, the condition \C ≤ \A ≤ \B is equivalent to AB ≤ BC ≤ AC, and the corresponding locus is depicted in diagram (b) of the previous page. The union of the sets in (a) and (b), shown in diagram (c), is therefore the locus of points C satisfying (1). There remains to draw the intersection of this locus and the one calculated immediately before (1). Problem 2. We look at the first several terms of the sequence, in search of patterns. We find x3 = 9, the least nonprime greater than 2x2 − x1 = 8; then x4 = 14 is the least nonprime greater than 2x3 − x2 = 12 (because 13 is prime). Continuing we obtain x5 = 20, x6 = 27, and so on, where each time — after that exceptional 13 — the least nonprime greater than 2xn−1 − xn−2 appears to be just 2xn−1 − xn−2 + 1, a composite number already. So we will conjecture that, for n > 4, the number 2xn−1 − xn−2 + 1 is composite and so equals xn . If this is so, we can write, for n > 4,

page 105

xn − xn−1 = xn−1 − xn−2 + 1. That is, the differences between successive elements of the series increase by one each time, forming an arithmetic progression: x4 = 14 = x3 + 5, x5 = 20 = x4 + 6, x6 = 27 = x5 + 7, . . . , or, in compact form, xn = xn−1 + n + 1

∗

for n ≥ 4.

If the differences form an arithmetic progression, the numbers xn themselves can be calculated by summing up the progression. This would give xn = x3 + 5 + 6 + 7 + · · · + (n + 1)


= 2 + 3 + 4 + 5 + 6 + 7 + · · · + (n + 1) = 21 n 2 + (n + 1) = 12 n(n + 3). {z } |

∗

n terms

(Here we broke x3 down as 2 + 3 + 4 to make the calculation less messy.) Conjecturally, then, we conclude that xn = 12 n(n + 3) for n ≥ 4.

(1)

There remains to prove that this is really the answer. We do it by complete induction. Base of the induction. For n = 4, the equality (1) is true.

∗ ∗

YEAR 1993 OLYMPIAD

65

Induction step. Let (1) be true for x4 , . . . , xn . We need to prove that xn+1 = 12 (n+1)(n+4). We have 2xn − xn−1 = 2 · 21 n(n + 3) − 21 (n − 1)(n + 2) = 12 (n + 1)(n + 4) − 1.

Thus, xn+1 will be equal to 21 (n+1)(n+4) if this number is composite. This is indeed so: if n is even, 12 (n+1)(n+4) has a factor 21 (n+4) > 1, and if n is odd, it has a factor 12 (n+1) > 1. Now we just have to apply formula (1) to n = 1000, obtaining x1000 = 501500. Problem 3. Assume that at some step we get a triangle similar to the initial one. All its angles are multiples of 20◦ . Lemma. All angles of the preceding triangle, and, more generally, of all preceding triangles are multiples of 20◦ . Proof. Let the triangle with angles α, β, γ be a child of the triangle with angles α1 , β1 , γ1 , obtained by cutting along an angle bisector. Arrange the labeling so that α1 = 2α and β = β1 , as in the figure. Since the sum of the β=β1 angles of any triangle is 180◦ , we deduce that

α α

γ

γ1

γ1 = 180◦ − α1 − β1 = (α + β + γ) − 2α − β = γ − α.

Clearly, then, if α, β, γ are multiples of 20◦ , so are α1 , β1 , γ1 , the angles of the parent triangle (see Fact 5). But then the angles of the grandparent triangle are also multiples of 20◦ , and so on. ˜ This, however, cannot be true even after the first cut of the initial triangle: if we start with an angle of 20◦ , we get an angle of 10◦ , and if we start with the angle of 140◦ we get an angle of 70◦ . The contradiction shows that it is impossible to get a triangle similar to the initial one. Remarks. 1. For any positive integer n, it is possible to find an initial triangle so the construction leads to a similar triangle after n appropriate bisector cuts, and no sooner. 2. For a similar situation, see Problem 64.9.4. REF 64.9.4

Problem 4. A classmate of Pete’s can have between 0 and 28 friends. Of these 29 possibilities, we know that 28 occur. Thus, either there is a classmate who has 28 friends, or there is a classmate who has no friends. But if a classmate is friends with everyone, then everyone has at least one friend. So it’s not possible for both 0 and 28 to occur: either the tally of friends is 1, . . . , 28 for the various classmates, or it is 0, . . . , 27. Denote the classmate of Pete’s with the most friends by A and the one with the least friends by B. In the first case just considered, A is everybody’s friend, while B has only one friend, A. In the second case B has no friends, while A is friends with everyone except B. In either case, A is a friend of Pete’s, and B is not.

page 107

66

SOLUTIONS

Now let’s send A and B to another class. Then Pete is left with 26 classmates and everybody has one fewer friend in this class than before. Thus each classmate still has a different number of friends in this class. We again send the classmate with most friends and the one with least friends to another class. We can keep doing this until we have sent away 14 pairs of classmates. Each pair included exactly one friend of Pete’s, so Pete had 14 friends in his class. Remarks. 1. Several ideas work together toward the solution: friendship is assumed to be a symmetric relation; it’s useful to look at extreme cases; and we are able to apply inductive descent. 2. There is one very short, but wrong solution: Let x be the number of Pete’s friends. Now replace all friendships by nonfriendships, and all nonfriendships by friendships. Then Pete’s classmates will again each have a different number of friends, so the conditions of the problem are still satisfied, meaning that Pete will again have x friends. But at the same time, we know that Pete now has 28 − x friends (his nonfriends in the original situation). Therefore x = 28 − x. Where is the mistake? (Hint: Have we shown that the problem has a unique solution?) Although wrong, this argument can point the way to the answer. 3. Solve the same problem if Pete has 27 classmates. 4. This is an extension of a problem you may have heard before: Prove that in any group of more than one person, there are two people with the same number of friends within the group (the number can be zero).

Problem 5. In the second identity set y = z. Then we get (x ∗ y) + y = x ∗ (y ∗ y) = x ∗ 0. Thus, x ∗ y = x ∗ 0 − y. It remains to compute x ∗ 0. For this, set x = y = z in the second identity; we get x ∗ 0 = x ∗ (x ∗ x) = (x ∗ x) + x = 0 + x = x. Thus, x ∗ y = x ∗ 0 − y = x − y, so 1993 ∗ 1935 = 1993 − 1935 = 58. Remark. Check that with x ∗ y = x − y, both identities are indeed verified. B

page 108

Problem 6. First solution. (See figure.) Let B ′ be the M reflection of B in the line AM . Since AB = AB ′ and 60◦ \BAB ′ = 2\BAM = 60◦ , the triangle ABB ′ is A C 30◦ equilateral. Hence, the points A, C, B ′ lie on a circle of center B. Since the inscribed angle is B′ ∗ “the” → “a”; half the central angle, it follows that \ACB ′ = 30◦ . Then, removed “of radius since \M CA = 150◦ , the points C, B ′ , M lie on the same line. Now, by AB” construction, AM is the bisector of \BMB ′ , and so also of \BM C.

YEAR 1993 OLYMPIAD

67

Second solution. Through the points A, C and M , draw a circle with center O (see figure). Since \ACM = 150◦ , the arc AM has measure 60◦ , so the triangle MAO is equilateral. Point B lies on a symmetry axis of △MAO, since \BAM = 30◦ . It follows that \AMB = \AOB = 21 \AOC = \AM C,

where the second equality is a corollary of the congruence △ABO = △CBO.

O

B M

A

C

Level C Problem 1. The least period length of a decimal divides any other period length of that decimal; see Fact 4. (We regard any terminating decimal as having minimal period length 1). Lemma. If k is a (not necessarily minimal ) period length for each of two decimals P and Q, then k is also a period length for P + Q and P − Q. Proof. Recall (Fact 13) that a recurring decimal P with period k can be written in the form X , P = l k 10 (10 − 1) where X is an integer. Similarly, we can write Q=

Y 10m (10k

− 1)

,

where Y is an integer. Without loss of generality we may assume that l ≥ m. We obtain X ± 10l−m Y P ±Q= , 10l (10k − 1) where again the numerator is an integer, so the decimals corresponding to P + Q and P − Q are recur with periods of length k. ˜ Now we can solve the problem. We know that A has least period 6 and B has least period 12. The lemma implies that 12 is a period length of A + B, so the divisors of 12 are the candidates for the least period of A + B. But 6 cannot be a period length of A + B, otherwise B = (A + B) − A would have a period of length 6, contradicting the assumption. Hence the least period of A + B cannot be a divisor of 6. Two possibilities remain for the least period of A + B: 12 and 4. Both options are possible: A = 0.(000001),

B = 0.(000000000001),

A + B = 0.(000001000002);

A = 0.(000001),

B = 0.(011100110110),

A + B = 0.(0111).

page 109

68

SOLUTIONS

Remarks. 1. How would someone come up with this last example? By working backwards! Take any decimal of least period 4 and call it A + B. Now subtract any decimal B of least period 6. The result has least period 12. (Why?) 2. We can find all possible least periods of a sum of two decimals. Let m, n, k be the least periods of the decimals P , Q and P + Q, respectively. Then k divides the least common multiple lcm(m, n) of m and n, by the lemma; but at the same time m divides lcm(n, k), and n divides lcm(m, k). Let p1 , . . . , ps be all the prime divisors of lcm(m, n), and write αs 1 m = pα 1 . . . ps ,

n = pβ1 1 . . . pβs s ,

where the exponent are allowed to be 0; see Fact 10. The preceding arguments imply that k = pγ11 . . . pγs s , where  max(αi , βi ) if αi 6= βi , γi = any number from 0 through αi if αi = βi . It can be shown that the opposite is also true: any such k can be a least period page 110 of a sum of two decimals whose least periods are m and n, respectively.

Problem 2. We can certainly tour the eight rooms in any make a tour of → hall, going clockwise, say. We can also combine tours of ∗ tour (to compensate two adjacent halls (top diagram) by using two of the three for next addition) doors that separate the two halls (middle diagram). Now the key observation is that we can always add a hall to our tour, so long as it is shares a wall with a hall already on the tour. For example, the bottom diagram adds a third hall to the first two. We can continue adding one hall at a time, until the tour includes all the halls of the castle. Problem 3. The answer to (a) is that it is not always possible; see figure on the right for a counterexample. In it, AC = 1000 m, AB > 1400 m, CD = 1 m. The segment AB divides the river into two parts, and boat going down the river end to end must cross this segment at some point. The distance from any point of AB to one of the banks exceeds 700 m.

A D C B

Part (b) turned out to be unexpectedly difficult, unless islands are allowed, in which case a counterexample is not hard to find (see figure on the right; the river has width just under 1 km). The intention of the authors of the problem was to not allow islands, and we (the authors of the book) were at first unable to solve it under this constraint. A contest was announced, and a solution was found through the joint efforts of A. Akopyan, V. Kleptsyn, M. Prokhorova, and the authors. It turned out to be more of a research-level mathematics problem than an Olympiad problem!

page 111

YEAR 1993 OLYMPIAD

69

The solution is given in full in Appendix B, but the main steps — each of which requires a whole argument in itself — are these: • Show that no disc of radius 750 m and center in the river (that is, not

in a lake) lies entirely on water. • Deduce that if a point in the river is not within 750 m of one bank, it lies within 750 m of the other bank. • Use the (possibly very complicated) boundary of the set of points that lie within 750 m of the left bank to prove that there is a path that remains within 800 m of both banks.

∗

page 112

Problem 4. Since we’re taking fractional parts of numbers, it’s a good idea to visualize the real line “rolled up” into a circle of unit length, so numbers with the same fractional part correspond to the same point a+b b on the circle. (See figure on the right, and compare the re- 2a+b 1 mark after the solution of Problem 60.10.6.) It is clear that 0 2 pn = 0 if xn = {an + b} lies on the upper semicircle [0, 12 ), and pn = 1 if xn lies on the lower semicircle. Furthermore, xn = {an + b} is the point on the circle obtained from {b} by n consecutive rotations through the arc {a}.

REF 60.10.6

(a) A rational value of a leads to a periodic pattern for the sequence pn . (Why?) Trying out b = 0 and the simplest possible fractions for a, we see that all 4-tuples of 0s and 1s can occur: • The “bigon” with a = 12 gives the sequence 010101. . . , so 0101 can

occur. (We only list 4-tuples starting with 0, since we know the complementary ones can be obtained by choosing b = 21 instead of b = 0, leading to the replacement of pn by 1 − pn .) • The equilateral triangle with a = 31 gives the sequence 001001. . . , so we get 0010 and 0100. • The square with a = 14 gives 0011 and 0110. • The octagon with a = 18 gives 0000, 0001, 0011 again, and 0111.

(b) First solution. We will prove that the string 00010 cannot be realized for any a and b. (In the second solution below we give a general condition necessary for a string to be realizable.) Consider three consecutive terms of the sequence: xn , xn+1 and xn+2 . If the points xn and xn+2 are diametrically opposed, each point is obtained from the previous one by a 90◦ rotation, and we’re in the situation of the square in part (a); the sequence pn cannot contain three zeros in a row. If the points xn and xn+2 are not diametrically opposed, they divide the circle into two distinct arcs, and either xn+1 lies on the longer arc, as in

∗

70

SOLUTIONS

diagram (a), or xn+1 lies on the shorter arc, as in (b). xn

xn+2

xn+1 xn

xn+1

(a)

xn+2

(b)

Suppose xn+1 lies on the longer arc. Then any other three consecutive points xm , xm+1 and xm+2 are similarly situated, being obtained from the points xn , xn+1 and xn+2 by the same rotation. This means that three such points cannot appear in the upper half-circle; that is, the substring 000 cannot appear in the sequence pn . Now suppose instead that xn+1 lies on the shorter arc xn xn+2 . Then any other three consecutive points xm , xm+1 and xm+2 are similarly situated. In particular, if xm and xm+2 belong to the upper half-circle, so does xm+1 ; that is, the substring 010 cannot appear in the sequence pn . We have considered all the scenarios and proved that the string 00010 cannot be encountered. Second solution. Let’s look at runs of consecutive 0s and 1s in the sequence p0 , p1 , p2 , . . . . We claim that all such runs have either the same length or lengths differing by 1, except that the first run may be short (it can start in the middle, so to speak). In particular, the string 00010 can never occur for any a and b, because it would mean the sequence has a run of 0s of length at least 3, but also a run of 1s of length 1. The reason the runs have almost uniform length is that the spacing between xn and xn+1 around the circle is the same for all n. Indeed, suppose first that 0 < a ≤ 21 and let i ≥ 1 be the integer uniquely defined by 1 i≤ (1) < i+1, or equivalently, ia ≤ 12 < (i+1)a. 2a We now show that a change from 0 to 1 must be followed by a change from 1 to 0 after exactly i or i+1 entries. More formally, suppose pn = 0 and pn+1 = 1; that is, xn belongs to the upper semicircle [0, 21 ), while xn+1 = xn + a belongs to the lower semicircle [ 21 , 1). The third inequality in (1) gives a < 2a < · · · < ia ≤ 21 . It follows that xn+1 , xn+2 , . . . , xn+i all lie in [ 12 , 1), giving i consecutive 1s. If we had pn+i+1 = pn+i+2 = 1, there would be i + 1 consecutive intervals of length a inside the lower semicircle: from xn to xn+1 , then to xn+2 , and so on up to xn+i+2 . But this is impossible by the last inequality in (1). Therefore the run of 1s stops at length i or i + 1. A completely analogous argument shows that a change from l to 0 must be followed by a change from 0 to 1 after exactly i or i+1 entries. This proves our claim for a ∈ (0, 12 ]. If a ∈ ( 12 , 1) we replace a by 1−a, which corresponds to a rotation through the same angle but in the opposite direction. We needn’t consider values of a outside [0, 1) because only the fractional part of a matters. Finally, for a = 0, there is only one run, of infinite length. So our claim is proved in all cases.

∗

∗ ∗ ∗

YEAR 1993 OLYMPIAD

71

Remarks. 1. This problem is relevant in symbolic dynamics; see Smale-horseshoe. CITE Smale-horseshoe 2. The original Olympiad problem said “. . . the sequence determined by some a and b”, which is perhaps ambiguous: the question might be whether all possible quadruples of 0s and 1s can occur for some fixed choice of a and b. The second solution to part (b) shows that the answer is no: any values of a and b that allow the string 0000 cannot allow the string 0100.

∗

Problem 5. Let m be the number of plants in a certain good classifier. Let us estimate the total number S of distinctions between all pairs of plants with respect to all features. There are 21 m(m−1) pairs of plants, and each pair differs in at least 51 features, so S ≥ 51 · 12 m(m−1). There is another way to look at S. Let mi be the number of plants having feature i. The number of pairs of plants that can be distinguished by means of feature i is (m−mi )mi . Summing over all the features, we obtain the total number S of distinctions: S=

100 X

(m−mi )mi .

i=1

Now, the arithmetic mean of m−mi and mi is 12 m, so the inequality between the arithmetic and geometric means (Fact 26) gives (m − mi )mi ≤ 41 m2 . Therefore S ≤ 100 · 14 m2 = 25m2 . Combining this with the earlier bound, we obtain 51 · 12 m(m−1) ≤ S ≤ 25m2 . (1) Subtracting 25m2 from the first and last expressions and simplifying we get 1 2 m(m−51) ≤ 0. Hence m ≤ 51. It remains to prove that m 6= 51. If m = 51, we have a strict inequality mi (m − mi ) < 41 m2

(since we have an integer on the left and a fraction on the right). That means the second inequality in (1) is strict, implying m < 51. This contradiction implies that a good classifier cannot describe more than 50 plants. Remarks. 1. One might be tempted to guess that a good classifier can describe 50 plants, but with a bit more work one can show that this is far from true. What we do is extend the classifier with one extra feature, which we declare present if and only if an even number of the original 100 features were present. This new classifier has 101 features and any pair differs by at least 52 of them: if a pair differs in 51 of the original features, it must also differ in the new feature. (Why?) Now the same arguments as above yield 52 · 21 m(m−1) ≤ S ≤ 101 · 41 m2 , which leads to m ≤ 34. Thus, the new classifier (and hence the initial one) can describe at most 34 plants.

page 113

72

SOLUTIONS

2. This problem is related to error correcting codes. Replace plants by messages and descriptions of features by length-n strings of bits (0s and 1s). The classifier — which is now a collection of m strings — is said to be a code of length n. The minimum number d of differences between two sequences in the code is the code distance; in our problem d = 51. If we take a message in the code and distort it arbitrarily by flipping no more than 21 (d − 1) positions, we are still able to recover the original message, simply by selecting the message that shares the most bits with the distorted one. (There will be at most 12 (d−1) differences, whereas the comparison with any other string will give at least d − 12 (d−1) > 21 (d−1) differences.) This is why the code is said to be error-correcting. page 114 One important open problem of code theory is finding the maximal size (number of different messages) of a length-n error-correcting code with code distance d, for arbitrary n and d. A famous result in this direction is the Plotkin–Levenshtein theorem, which establishes an upper bound (called the Plotkin bound ) in the case d > 12 n, and provides certain natural conditions that guarantee the bound can be achieved. In our problem we have n = 100 and d = 51, so Plotkin’s bound applies, and its value is 34. This bound is achievable: there does exist a code with 34 messages. C

Problem 6. Let the remaining vertices of the square on N side AB be D and E, so the square is ABDE, and set γ = M γ \ACB. (See figure on the right.) Applying the intercept A B theorem to the triangle ADC we see that CD = 2 OM ; similarly, CE = 2 ON . Therefore it suffices to find the O maximum of CD + CE = 2(OM + ON ). First solution. On side BC of △ABC, construct a D square CBD′ E ′ outwards. (See figure below.) Triangles E ′ ′ ABD and DBC have two sides and the included angle equal, so CD = AD √ . ′ ′ In the triangle ACD , two sides are known: AC = b and CD = a 2. Moreover, \ACD′ = γ + 45◦ . The side AD′ attains E′ its maximal value when the triangle degenerates into a segment, so √ a 2 C √ D′ max(CD) = max(AD′ ) = b + a 2 a b √ for γ = 135◦ . Similarly, we have max(CE) = a+b 2, A B c again for γ = 135◦ . Thus, each of OM and ON attains its maximum when γ = 135◦ , and so does c their sum: √ 1+ 2 E D max(OM + ON ) = (a + b). 2 Second solution. Let \CAB = α, \ABC = β, c = AB, d = CD, e = CE. The law of cosines for △ABC gives c2 = a2 + b2 − 2ab cos γ.

page 115

YEAR 1993 OLYMPIAD

73

Next we apply the law of cosines to △AEC:

e2 = b2 + c2 − 2bc cos(90◦ + α) = b2 + c2 + 2bc sin α,

since cos(90◦ + α) = − sin α. Substituting c2 from the first formula into the second, we get e2 = 2b2 + a2 − 2ab cos γ + 2bc sin α. The law of sines for △ABC implies that sin α = (a/c) sin γ. Therefore e2 = 2b2 + a2 + 2ab (sin γ − cos γ).

Similarly, d2 = 2a2 + b2 + 2ab (sin γ − cos γ). Hence both e and d attain their maximum values when sin γ −cos γ does, which is to say, when γ = 135◦ . Hence √ the maximum of e + d is also attained for γ = 135◦ , and it equals 21 (1 + 2)(a + b). Level D Problem 1. If tan(α + β) is defined, then p tan α + tan β = . (1) tan (α + β) = 1 − tan α · tan β 1 − tan α · tan β The product of tangents is related with p and q as follows: 1 1 tan α + tan β p q= + = = . (2) tan α tan β tan α · tan β tan α · tan β We deduce from (2) that either p and q are both zero or they are both nonzero; therefore we have only these cases to consider: 1. If p = 0 = q, then (1) implies tan(α + β) = 0. We have to verify here that the denominator of (1) does not vanish. Indeed, since p = 0, we have tan α = − tan β,

so

1 − tan α · tan β = 1 + tan2 α > 0.

2. If p 6= 0, q 6= 0 and p 6= q, then (2) implies tan α · tan β = p/q, so (1) implies tan(α + β) = pq/(q − p). 3. If p 6= 0, q 6= 0 but p = q, then tan(α + β) is not defined. Problem 2. We will construct a subdivision into squares satisfying the condition of the problem. We first divide the unit square into four equal squares. The little squares that intersect the main di3 2 agonal only at a vertex will be called level-1 squares. 18 3 1 3 We subdivide each of the remaining squares into four 1 2 4 3 equal squares of side 41 . The little squares of side 41 3 2 that intersect the main diagonal only at a vertex will 1 3 1 3 be called level-2 squares. We continue in this way (see 2 2 3 figure) until we have 500 levels of squares. There are 2k squares at level k, each with side 2−k . Hence the total perimeter of all level-k squares is 4, and the total perimeter of all squares intersecting the diagonal is 4 · 500 > 1993.

page 116

74

SOLUTIONS

Remarks. 1. A stronger result is in fact true: The unit square can be partitioned into squares in such a way that the sum of perimeters of squares intersecting the main diagonal in a segment exceeds any given number. The construction is a modification of the previous one; we make the under-thediagonal level-1 square have side 35 , say, instead of 21 , while the complement is subdivided into two squares of side 52 along the diagonal, plus an irregular area that we leave aside, knowing that it can be subdivided into tiny squares (since 35 is a rational number). We then repeat the procedure on the two new squares along the diagonal to get level-2 squares, and so on. All level-k squares now have side ( 52 )k , instead of ( 12 )k as before, and there are 2k−1 of them, instead of 2k . So now the total perimeter of level-k squares is 2 · ( 45 )k ; that is, it decreases in a geometric progression instead of being the same for all levels. But the ratio of the progression can be made arbitrarily close to 1, by replacing the number 53 by some rational number very close to 12 . So the sum of perimeters can be made as large as desired. 2. This problem arose during a lecture of the illustrious mathematician N. N. Luzin, when he wanted to shorten the proof of a theorem of Cauchy (Luzin loved to improvise). Luzin conjectured: Fix a curve of a bounded length in the unit square and consider a partition of the square into little squares. The total perimeter of the little squares that intersect the curve is bounded by a constant depending on the curve only. A. N. Kolmogorov, who was to become just as famous a mathematician, was at the lecture and soon constructed a counterexample.

Problem 3. For n = 2 the answer is obviously 1. So assume n ≥ 3. The desired number is at most n, because we can exhibit an arrangement of points generating only n pairwise nonparallel lines: the vertices of a regular n-gon, We prove this by showing that there are as many nonparallel lines as there are axes of symmetry of the polygon. To each side and each diagonal, we assign an axis of symmetry: the perpendicular bisector of this side or diagonal. Two sides or diagonals have the same perpendicular bisector if and only if they are parallel. Therefore we just need to count the axes of symmetry of a regular n-gon. For n odd, each axis of symmetry passes through a unique vertex. Hence, the total number of axes of symmetry is n. For n even, an axis of symmetry passes either through a pair of opposite vertices or through the midpoints of opposite sides. There are n/2 axes of either type. Hence, the total number of axes of symmetry is again n. Now we prove the converse: The desired number is at least n. That is, for any arrangement of n points, no three of which lie on a line, we can always find n pairwise nonparallel lines. It is easy to find n − 1 lines: just take a point and consider all the lines from it to the other points. It is a bit harder to construct the n-th line. One method is to introduce cartesian coordinates on the plane. Among the n points, take one, say O, having highest y-coordinate, and move the

page 117

YEAR 1993 OLYMPIAD

75

origin there. Among the other n−1 points, choose A such that the ray OA makes the biggest possible angle with the positive x-axis, and O B such that the ray OB makes the least possible angle with the positive x-axis. All the rays connecting O with the B other points lie inside the angle AOB, by our choice A of A and B (see figure). Thus they must intersect the segment AB and hence cannot be parallel to it. Now we just take AB as the n-th line. Remarks. 1. The points A, O and B are adjacent vertices of the convex hull of the n points, that is to say, the smallest convex set containing all the points. (It can be shown that the convex hull of a finite set is a polygon whose vertices are contained in the set. The convex hull of the seven points in the preceding figure has five vertices; see figure on the right.)

∗ ∗

∗ ∗

2. The given noncollinearity condition cannot be replaced by the weaker one that the points do not all lie on a single line. For example, for the set consisting of the vertices of a regular 2k-gon and its center, there are only 2k nonparallel lines.

Problem 4. Clearly, the worst-case scenario is when all marble populations occur at the start; that is, we have n boxes — where n = 460 in part (a) and n = 461 in part (b) — and there is a box with j marbles for every j = 1, 2, . . . , n. So from now on we assume this is the situation. We start with the observation that a box having m = qk + r marbles (0 ≤ r < k) before a step with group size k with be left with q + r marbles after that step; see Fact 7. Lemma 1. After the first step, with group size k, there is a number f (n, k) such that the marble populations are exactly all the numbers in the range 1, 2, . . . , f (n, k), and no others. page 118

Proof. Let f (n, k) be the highest marble population in a box at the end of the step. We show by reverse induction on j (Fact 24) that there is a box with exactly j marbles, for all j = 1, . . . , f (n, k). Suppose this is true for some j. Then there exist numbers m (starting population), q (quotient) and r (remainder) such that 1 ≤ m ≤ n, 0 ≤ r < k, m = qk + r, and j = q + r. If r > 0, we look at the box that started off with m−1 marbles. If r = 0, we take the one that had m − k marbles. Either way, we have found a box holding exactly j − 1 marbles at the end of the step. This completes the induction step and proves the lemma. ˜ This lemma effectively reduces the situation at the end of the first step to the original situation, with a smaller number of boxes (we just ignore boxes with duplicate populations). There remains to select k so the new largest population f (n, k) is as low as possible.

76

SOLUTIONS

Lemma 2. The largest population f (n, k) after the first step is given by h i n+1 f (n, k) = + k − 2. (1) k Proof. The function “population of a box at the end of the first step” grows by 1 when its first argument (the initial population) grows by 1, except when the argument is 1 less than a multiple of k, in which case the function drops by k−2 (see figure, where k = 5). Thus the maximum of the function is always achieved for an argument value that is 1 less than a multiple of k, and that is as large as possible under this condition. So let Q = [(n+1)/k] be the quotient of the division of n+1 by k. The box that started off with Qk − 1 = (Q−1)k + (k−1) marbles achieves the maximum, and its new population is Q + k − 2 = [(n+1)/k] + k − 2 marbles. This proves (1). ˜ Lemma 3. For√ a given value of n, the function f (n, k) achieves its minin+1 + 1. mum when k =

Proof. To find the value of k that minimizes f (n, k), we first write h i n+1 f (n, k) = + k − 2. k √ The function inside the √ brackets decreases in the interval (0, n+1 ) and increases in the interval ( n+1, n]. Since [x] the function  decrease,√  √does not n+1 or at k = n+1 + 1. f (n, k) attains its maximum either at k = It remains to show that we always have √ √ f (n, [ n+1 ] + 1) ≤ f (n, [ n+1 ]). √ Let [ n+1 ] = s. Then s2 ≤ n+1 < (s+1)2 . (2)

Thanks to (1) it suffices to prove that i h i h n+1 n+1 < . s+1 s Equation (2) implies that h i h i n+1 n+1 ≤ s and ≥ s. s+1 s Therefore, it suffices to prove it is not possible for both sides to equal s. But h h i i n+1 n+1 n+1 n+1 < s+1 =⇒ < s =⇒ = s =⇒ < s. s s s+1 s+1 The lemma is proved. ˜ We are now ready to find the minimum number of steps required for any starting value of n. We simply apply repeatedly the function √ g(n) = f (n, [ n+1 ] + 1),

page 119

YEAR 1993 OLYMPIAD

77

which represents the maximum population after one step, with an optimal choice of k. We verify by direct computation that, after five applications, g(g(g(g(g(460))))) = 1 but

g(g(g(g(g(461))))) = 2.

The sequences of steps for n = 460 and n = 461 are as follows: n 460 40 10 4 2 1

step k 1 2 3 4 5

n

22 461 7 41 4 11 5 3 3 2 2

To recapitulate, we spell out the details of the argument for n = 461. The three lemmas say that if the boxes start with all populations from 1 through 461, then after step 1 there remain all populations from 1 through g(461) = f (461, 22) = 41; after step 2 there remain all populations from 1 through g(41) = 11; after step 3, all populations from 1 through 5; after step 4, all populations from 1 through 3; and after step 5, populations 1 and 2. Thus, for n = 461 it is not always possible to be left with a single marble in each box. Remarks. 1. Instead of proving that √ √ f (n, [ n+1 ] + 1) ≤ f (n, [ n+1 ]),

one could just check case by case. 2. This problem originated in computer programming. A text containing one or more blank spaces between words had to be processed so as to leave precisely one blank between words. The programmers solved the problem by recursively applying the operation of selecting a positive integer k and replacing every group of k consecutive blanks by a single blank.

Problem 5. (a) A possible solution is shown in the figure. The function is defined by  1 1 − 2 − x for − 1 ≤ x < − 12 ,     1 1 1  for − 2 ≤ x < 0,  1 2 x − 2 1 2 f (x) = 0 for x = 0, 0 1 − 1   − 1  2 x + 21 for 0 < x ≤ 12 , −  2  1 1 − x for < x ≤ 1. − 1 2 2

(b) We will show that a function satisfying the conditions of the problem cannot exist on the interval (−1, 1). The case of the function defined on the whole real axis is analogous. Step 1. Suppose, to the contrary, that such a function f (x) exists. Its graph is mapped to itself under a clockwise 90◦ rotation: if (x, y) is a point

∗

page 120

∗

78

SOLUTIONS

on the graph, we have y = f (x), so f (y) = −x by assumption, and the point (y, −x) also belongs to the graph; but this is precisely the image of (x, y) under the specified rotation. By applying the 90◦ rotation repeatedly we see that the graph maps to itself also under 180◦ and 270◦ rotations. Step 2. This implies that the coordinate axes cannot intercept the graph except at the origin: any other intersection would imply three more intersections (obtained by 90◦ , 180◦ and 270◦ rotations), and in particular there would be two distinct intersections of the graph with the y-axis, which is impossible. Step 3. Now consider the portion of the graph that lies within the open first quadrant {(x, y) : x > 0, y > 0}. Recalling the assumption that the graph is a union of finitely many points and line segments, we can write this intersection as L1 ∪ L2 ∪ · · · ∪ Ln ∪ P1 ∪ P2 ∪ · · · ∪ Pm ,

where the Lk are line segments and the Pj are points. We may assume that the line segments Lk are pairwise disjoint and open (meaning the endpoints are excluded) and that the points Pj are distinct and do not belong to any of the line segments Lk . For each k, let Jk be the line segment obtained from Lk by a clockwise 90◦ rotation. Next, for each j, let Qj be the point obtained from Pj by the same rotation. All these points and segments lie on the open fourth quadrant, and in fact J1 ∪ J2 ∪ · · · ∪ Jn ∪ Q1 ∪ Q2 ∪ · · · ∪ Qm

is precisely the intersection of the graph with the fourth quadrant. (Why?) Since we already know the graph does not intersect the positive x-axis, we conclude that the intersection of the graph with the half-plane x > 0 consists of 2n line segments (the Lk and Jk ) and 2m points (the Pj and Qj ). None of the line segments can be vertical. Thus the projections of all these line segments and points on the x-axis partition the interval (0, 1) into 2n open intervals and 2m points. But it is impossible to divide an open interval into an even number of subintervals using an even number of points! We have reached a contradiction. D Problem 6. Clearly a shortest flight touches each face H of the tetrahedron exactly once. Let the tetrahedron have vertices ABCD, and let the shortest flight be F G the space quadrilateral EF GH, where E ∈ △ABC, C F ∈ △BCD, G ∈ △ABD, and H ∈ △ACD A E (see figure). Our job is to find the smallest possible B perimeter for the quadrilateral EF GH. Draw the symmetry plane of the tetrahedron containing CD; note that it is perpendicular to AB. Let E1 F1 G1 H1 be the reflection of EF GH in this plane (so E1 and G1 lie on the same faces as E and G, respectively,

page 121

YEAR 1993 OLYMPIAD

79

while F1 lies on the same face as H, and H1 lies on the same face as F ). The quadrilaterals EF GH and E1 F1 G1 H1 have the same perimeter.

page 122

Lemma. In any space quadrilateral, the distance between the midpoints of two opposite edges is less than or equal to the mean of the lengths of the remaining edges. M Proof. Let KLMN be the quadrilateral and let P and Q be the midpoints of KL and MN , respectively (see figure). Denote by R the midpoint of the diagonal LN . We have P R = 12 KN and RQ = 12 LM . Hence K ˜ P Q ≤ P R + RQ = 21 (KN + LM ).

Q

N P

R L

Denote the midpoints of the segments EE1 , F H1 , GG1 , and HF1 by E2 , F2 , G2 , and H2 , respectively. These points also lie on the faces of the tetrahedron. By the lemma, the perimeter of E2 F2 G2 H2 is no greater than that of EF GH. Moreover, E2 and G2 lie on the symmetry plane of the tetrahedron containing CD; that is, they lie respectively on the medians CT and DT of the faces ABC and ABD, where T is the midpoint of AB. Next we repeat this symmetrization operation for another symmetry plane. That is, we reflect E2 F2 G2 H2 in the symmetry plane of ABCD that contains AB, obtaining a quadrilateral E3 F3 G3 H3 , and then we take the midpoints of the segments E2 G3 , F2 F3 , G2 E3 , and H2 H3 , D obtaining a quadrilateral E4 F4 G4 H4 , whose vertices all belong to one of the two planes of symmetry of ABCD considered S so far (one through AB and the other through CD). Specifically, the vertices E4 and G4 lie on CT and DT , C while F4 and H4 lie on the medians AS and BS of the A faces BCD and ACD, where S is the midpoint of CD. T B (See figure on the right.) Again, the perimeter of E4 F4 G4 H4 is no longer than that of E2 F2 G2 H2 , which as we know is no longer than that of EF GH. Hence, the perimeter of EF GH is at least 4d, where d is the distance between CT and BS. It remains to construct a path of length 4d and find d. Let the common perpendicular to CT and BS intersect CT at E0 and BS at F0 . Let G0 be the reflection of E0 in the plane ABS. It follows from symmetry that F0 G0 is the common perpendicular to BS and DT . Similarly we construct the point H0 such that G0 H0 is the common perpendicular to DT and AS and H0 E0 is the common perpendiculars to AS and CT . The perimeter of E0 F0 G0 H0 is 4d. We also have to prove that the bases of our common perpendiculars lie on the faces of the tetrahedron, rather than on their extensions. This will be checked below (we still have to calculate d). Draw the plane through AB perpendicular to CT and take the projection of the tetrahedron on this plane. pWe obtain triangle ABD′ , in which AB = a and D′ T = a 2/3, by

page 123

D′

80

SOLUTIONS S′

the formula for the length of the altitude of the regular A B T tetrahedron. (See figure on the right.) The projection sends S to S ′ , the midpoint of D′ T . Hence, d equals the distance between T and the line BS ′ , because the common perpendicular is parallel to the plane of projection. It is also obvious that the base of the perpendicular dropped from T onto the line BS ′ lies on the segment BS ′ and not on its extension; hence F0 lies on the segment BS. We similarly prove that the remaining vertices of the quadrilateral lie on the medians, not on their extensions. q In the right triangle BT S ′ , the legs BT = 12 a and T S ′ = 21 a 23 are known. Hence q BT · T S ′ a ′ 1 d= BS = 2 a 53 , =√ . ′ BS 10 Remarks. 1. The solution implies that there are three suitable paths. (Why?) 2. An analogous problem on the plane is known: A beetle crawls inside a triangle with sides a, b, c. What is the shortest length of a path that visits each side and returns to the initial point? page 124 It the case of an acute triangle the answer is the path joining the bases of the altitudes; this is known as Fagnano’s problem. (See Coxeter, Chapter 4, CITE Coxeter § 5.) For a right or obtuse triangle the path degenerates into an altitude traveled twice; see Problem 70 in ShklyarskiChentsovYaglom. CITE ShklyarskiChentsovYaglom

Year 1994 Olympiad Level A Problem 1. First solution. A bottle of mix used to require 16 of a jug of 1 of a jug of grape juice. Hence, the volume of a bottle is apple juice and 10 1 1 4 equal to 6 + 10 = 15 of the volume of a jug. After the change in the recipe, a bottle of drink requires 51 of a jug of a apple juice and x1 of a jug of grape juice. Hence, the volume of the bottle is equal to 1 1 4 + = 5 x 15 4 1 1 1 − = . Thus, of the volume of the jug. So we have an equation = x 15 5 15 x = 15. Second solution. To make 30 bottles of the drink according to the old recipe, one needed 5 jugs of apple juice and 3 of grape juice: a total of 8 jugs. After the change, 30 bottles require 6 jugs of apple juice, and hence 8 − 6 = 2 jugs of grape juice. Now one jug of grape juice serves 30/2 = 15 bottles. Problem 2. Let x and y be the three-digit numbers to be found. If we write three zeros after x, we get 1000x. If instead we write the digits of y, we get 1000x + y. See Fact 11. Thus we have the equation 7xy = 1000x + y.

(1)

First solution. Divide both sides of (1) by x: y 7y = 1000 + . x The number y/x is positive and smaller than 10, since y ≤ 999 and x ≥ 100. Therefore 1000 < 7y < 1010. Dividing this inequality by 7 we get 142 67 < y < 144 72 .

page 125

82

SOLUTIONS

Since y is an integer, y is equal to either 143 or to 144. Let y = 143. Substituting this value of y into (1) we deduce that 7x · 143 = 1000x + 143.

Solving this equation we find that x = 143. If y = 144, a similar equation yields x = 18, which has too few digits. Second solution. Rewrite (1) in the form 1000x = (7x − 1)y.

It is not difficult to see that x and 7x − 1 have no common divisors except for 1 and −1. Indeed, if d is a common divisor of x and 7x − 1, then d divides 7x, and hence it divides 1 = 7x − (7x − 1); see Fact 5. But 1 is only divisible by 1 and −1. Thus 7x − 1 divides the product 1000x and is relatively prime to x. This implies that 7x − 1 divides 1000 (see Fact 9). Since 1000 , 7x − 1 ≥ 7 · 100 − 1 = 699 > 2 the only possibility is 7x − 1 = 1000. Hence x = 143. Substituting back in (1), we find y = 143. Remark. Compare with Problem 57.10.1.

Problem 3. Extend the segments BQ and BP so they intersect the line AC at A′ and C ′ , respectively. In the triangle ABC ′ , the segment AP is both an altitude and a bisector (see figure). Hence, this triangle is isosceles Q with AB = AC ′ , and AP is a median too: BP = C ′P . Similarly, the segment CQ is a median of △CBA′ , that is, BQ = QA′ . Hence P Q is a midline of △A′ BC ′ , i.e., P Q k A′ C ′ . But then P Q k AC. A′ A

REF 57.10.1 page 126

B

P

C

C′

Problem 4. Let the initial configuration be a 1×1 square on a piece of graph paper. Whenever a grasshopper jumps, it must land on a crossing of the grid, because the jump is symmetric with respect to a crossing and starts from a crossing (see figure; we use the fact that a square grid is centrally symmetric with respect to any of its crossing). Another way to see this is to choose coordinates so that one of the corners of the initial square is the origin; then the x- and y-coordinates of each grasshopper must remain whole numbers, because the difference in each coordinate between the two grasshoppers involved in a jump is replaced by its negative. In particular, since the distance between two grasshoppers cannot be less than 1, at no time can their positions coincide with the corners of a smaller square than the original one. Now suppose the grasshoppers could land on the corners of a larger square than the original one. The reverse of a legal jump is also a legal

YEAR 1994 OLYMPIAD

83

jump, so by performing the jumps backwards, we’d be in a situation where grasshoppers starting at the vertices of a (larger) square ended up occupying the vertices of a smaller square. But we know this is impossible. Remark. If the grasshoppers had started at the vertices of some other parallelogram, they would be destined to always jump to crossings of the slanted grid made of copies of the initial parallelogram. The set of crossings of such a grid is called a lattice.

page 127

Problem 5. We start with some obvious remarks. • Any two hands of the clock define an unfavorable sector between their extensions, in the sense that if the third hand is in that sector, the moment is unfavorable. The size of this sector does not exceed 180◦ (see figure). • After a whole number of hours, the minute and second hands will return to the exact same position they occupy now. • After six hours, any unfavorable moment is replaced by a favorable one (the hour hand rotates through 180◦ and moves from an unfavorable sector to a favorable one). • The converse is not true: there are favorable moments, six hours after which the time is again favorable. For example, any time between 3:00:00 and 3:00:05 is favorable, but so are the times six hours later, from 9:00:00 to 9:00:05. Now imagine the day divided into favorable and unfavorable intervals of time. The operation of adding six hours maps all unfavorable intervals to favorable ones of the same length (though the latter might be part of larger favorable intervals). So the total length of favorable intervals is at least as much as that of unfavorable intervals. The same operation of adding six hours also maps some favorable intervals to favorable intervals (consider the 5-second interval discussed in the previous paragraph). Thus the favorable intervals add up to strictly more time than the unfavorable ones. Remarks. 1. A similar situation is considered in Problem 56.8.4. REF 56.8.4 2. The original question asked “what is there more of, favorable time or unfavorable time?” Obviously what we’re interested in is the total length of the intervals of favorable and unfavorable time, not whether there are more moments of each kind of time. There are infinitely many moments of each of the two kinds, and they can be placed in one-to-one correspondence, just as the intervals [0, 1] and [0, 2] can be placed in one-to-one correspondence through the map x 7→ 2x. The question of when it is possible to establish a one-to-one correspondence between two infinite sets is interesting and deep. (For finite sets, the answer is simple: a one-to-one correspondence exists if and only if the sets have the same number of elements.) Here are some example results; for details, see Settheory: CITE Settheory (a) There is a one-to-one correspondence between the set of all integers and the set of positive integers. (b) There is a one-to-one correspondence between the set of all rational numbers and that of positive integers.

84

SOLUTIONS (c) There is no one-to-one correspondence between the set of all rational numbers and that of all irrational numbers. (d) There is a one-to-one correspondence between the points of a line segment and the points of a square.

Problem 6. The first player colors the 18×18 square adjacent to the longer side of the rectangle so that it is equidistant from the right and left edges of the rectangle (see figure for an analogous situation with a 7×14 board). The rest of the rectangle is imagined to be divided into congruent halves. Now the second player’s every move can be countered by the first with a symmetric move; the second player would only be able to prevent this by coloring a square that straddles the symmetry axis, which is impossible. Level B Problem 1. Let D and E be points inside a triangle ABC. Draw all 10 segments between pairs of points; they can be divided into two non-self-intersecting polygonal lines of five segments each (see figure). Either of these two polygonal lines can be taken as the desired pentagon; the other polygonal line is made of the diagonals of A the pentagon.

B

E D C

Problem 2. First case. If k > l, Sue wins, by securing a part longer than the sum of all the other parts. For example, she can cut k into parts of length l + 32 (k − l),

1 6 (k

− l),

1 6 (k

− l).

(See figure below.) The largest of these, l + 32 (k − l), cannot serve as a side of any triangle: by the triangle inequality the sum of the lengths of the other two sides should be greater than this, but the sum of the lengths of all other segments is only l + 31 (k − l). k l + 32 (k−l)

1 (k−l) 6

1 (k−l) 6

l

Second case. If k ≤ l, Leo wins. Let Sue divide her segment into parts of length k1 , k2 , k3 , where k1 ≥ k2 ≥ k3 . Then Leo divides his segment l so that his largest part is of the same length as Sue’s largest part, and the remaining two are of the same size: k1 , 12 (l − k1 ) and 21 (l − k1 ). (See figure below.) Since k1 ≥ k2 , it is possible to construct two isosceles triangles from

page 129

YEAR 1994 OLYMPIAD

85

the resulting segments, of lengths l − k1 l − k1 , , k3 . 2 2 Indeed, from segments of lengths a, a, b it is possible to construct an isosceles triangle if and only if b < 2a. Obviously k2 < 2k1 . On the other hand, l − k1 2· = l − k1 > k3 , 2 since k1 + k3 < k ≤ l. k1 , k1 , k2 ,

k

k1

k2

k3

k1

k1

1 (l−k1 ) 2

1 (l−k1 ) 2

l k2 k1

1 (l−k1 ) 2

k3

1 (l−k1 ) 2

Problem 3. Here is an infinite set of solutions: 2

2

2

x = k(2k + 1), y = 2k + 1, z = −k(2k + 1),

page 130

for k = 0, 1, 2, . . . .

Verification is straightforward. How can this answer be guessed? Since we only have to find infinitely many solutions, not all of them, we can get rid of one variable and two cubes by taking z = −x. Then we get the equation 2x2 + y 2 = y 3 , which we can rewrite as x2 = 12 (y−1)y 2 or r y−1 x=± y. 2 So if (y−1)/2 is a perfect square, x is an integer. Thus we take (y−1)/2 = k2 , which gives y = 2k2 + 1, x = k(2k2 + 1). Remark. There certainly exist other solutions, e.g., x = 12 k(k 2 − 1) + 1, y = − 21 k(k 2 − 1) + 1, z = −k 2 + 1 for any k ∈ Z.

Problem 4. First solution. Let the circles be disposed Q M as in the figure, that is, with the center of each outside the other. Other configurations are treated similarly; B see the remark after the second solution. It suffices to prove that the triangles AQN and AM P are congruent. For this, we prove that they are similar and that the correspond- N A ing sides AQ and AM are equal. By the theorem on inscribed angles, \P = \N and \Q = \M , so △AQN is similar to P △AM P . To prove that the segments AQ and AM are congruent, we prove that the corresponding arcs, measured counterclockwise, are the same. The first

86

SOLUTIONS

of these arcs has measure twice the inscribed angle ABQ; see Fact 15. Similarly, the arc AM equals twice the angle MAN . Thus, it suffices to prove that \ABQ = \MAN . Now, the angle ABQ is an exterior angle of the triangle ABN , and hence \ABQ = \ANB + \BAN . We see that \MAN = \BAM + \BAN ; but \ANB = \BAM , since each of them equals half the arc AB on the circle through A, B and N . Hence \MAN = \ANB + \BAN = \ABQ. The problem is solved. Second solution. By the tangent-secant theorem (power of a point), we have MP =

AM 2 , MB

NQ =

AN 2 . NB

Thus it suffices to prove that AM 2

page 131

MB . (1) NB Let’s show first that the triangles AMB and NAB are similar. We have \AMB = \BAN , since they both have half the measure of the arc AB on the circle ABM (see figure). By the same token, \M AB = \NAB. This proves the similarity N of AMB and NAB, which in turn implies that AN 2

=

AB AM MB AM = , and = . AN NB AN AB Multiplying these equalities we get (1).

Q

M B

A

P

Remark. If the center of one of the circles lies inside the other circle, the picture will be different. If the center of each of the circles is inside the other circle, we get still another picture. In these cases, in the first solution, instead of proving that \ABQ = \MAN , we have to prove that \ABQ + \MAN = 180◦ . The second solution remains valid in either case. The reader is encouraged to work out the details.

Problem 5. Let x be the dropped digit, a the part of the number to the left of x, and c the part of the number to the right of x. Let c contain n digits. Then the original number equals a · 10n+1 + x · 10n + c; see Fact 11 on page 104. After x is struck out it becomes a · 10n + c. (These formulas encompass the case n = 0, with the convention c = 0, but we can dispose of this case quickly. Indeed, in this case 10a + x is divisible by a. It follows that x is divisible by a, and hence a ≤ x ≤ 9. This would mean the original number has only two digits. We are going to construct a six-digit number below, so the case c = 0 is irrelevant.) Now consider the ratio between the two numbers: a · 10n+1 + x · 10n + c , where c < 10n . r= a · 10n + c

YEAR 1994 OLYMPIAD

87

Subtracting 10 from both sides of this equation and simplifying we reach x 9 x · 10n − 9c ≤ ≤ ≤ 9. (1) r − 10 = a · 10n + c a a Set l = r − 10. Multiplying both sides of the equality in (1) by a · 10n + c and manipulating we get (x − la) · 10n = (l + 9)c.

(2)

If l ≤ 0, the left-hand side is positive, so l + 9 > 0. Thus,

page 132

−8 ≤ l ≤ 9.

Moreover, l 6= 0; otherwise the original number ends in 0. Lemma. a < 10. Proof. Consider the two cases: l > 0 and l < 0. If l > 0, equality (2) implies that x − la > 0, and hence x 9 a < ≤ ≤ 9. l l If l < 0, then (l + 9)c 9 · 10n x − la = < = 9, 10n 10n implying −la < 9, so a < 9.

˜

The lemma implies that the original number has n + 2 digits. The next step is to find the greatest possible value of n. By assumption c does not end in 0. Thus c cannot be divisible by both 2 and 5 (see Fact 10 on page 104). Suppose first that c is not divisible by 2. Consider the right-hand side of (2). Since 1 ≤ l + 9 ≤ 18, the number l + 9 can be divisible by 24 = 16, but it cannot be divisible by 25 . Hence n ≤ 4. If n = 4, then l + 9 = 16, and (2) becomes (x − 7a) · 54 = c. Since x is a one-digit number, it follows that a = 1, and hence x = 8 or x = 9. If x = 9, then c ends with a 0 and hence does not fit. If x = 8, then c = 625 and the final answer is 180625 = 10625 · 17.

If, on the other hand, c is not a multiple of 5, then l + 9 can be divisible by 5 but not by 52 , so n ≤ 1, and the number will certainly not be the largest possible. Problem 6. For part (b), it is easy to give an example where the first few ships are arranged so that there is no room for the last one; see figure on the right. In part (a), there is a hidden difficulty: many students assumed it’s enough to prove that one can place the last 1 × 1 ship, but in fact one must show that each ship in sequence can be placed.

page 133

88

SOLUTIONS

The 1 × 4 ship can be placed. To prove that each of the two 1 × 3 ships can be placed, too, we mark eight “phantom” 1 × 3 ships, all in the same direction and leaving two cells in between them (see first diagram below). If a 1 × 4 and at most one 1 × 3 ship have been placed, each of them can touch or overlap with at most two phantom ships; therefore some phantom ship will be untouched and can be replaced by an actual 1 × 3 ship.

The reasoning for the 1 × 2 ships is similar; it involves twelve parallel 1×2 phantom ships, again with two cells between them (see middle diagram). Suppose the 1 × 4 ship, two 1 × 3 ships and no more than two 1 × 2 ships have been laid down. Each touches at most two phantom ships, so at least one phantom ship will be untouched and can be replaced by an actual ship. For the final step of the proof we consider sixteen 1×1 phantom ships (see rightmost diagram above). Each of the six long ships already placed touches at most two phantom ships, while each of the (at most three) 1 × 1 ships already placed touches one phantom ship. Thus at most fifteen phantom ships are affected, and one is left over, showing that the last 1 × 1 ship can be placed. Remark. It is interesting to find the maximal number of identical ships of, for example, size 1 × 4, that one can place.

Level C Problem 1. We offered two detailed solutions to the very similar problem 57.8.2. Here we give the corresponding arguments in slightly abbreviated form. First solution. Let x and y be the numbers to be found. Then 3xy = 7 10 x + y, or 3y = 107 + y/x. Since 0 ≤ y/x ≤ 10, we have 107 < 3y < 107 + 10, and therefore 3333333 13 < y < 3333336 23 . Now one can either consider the three cases y = 3333334, y = 3333335, and y = 3333336, or observe that y 3333336 ≤ < 4, x 1000000 and hence 107 + 1 ≤ 3y < 107 + 4. Only one number in this interval is divisible by 3, namely, 107 + 2. So 3y = 107 + 2, and hence y = 3333334 and x = 1666667. Second solution. Since 107 x = (3x − 1)y

(1)

page 134 REF 57.8.2

YEAR 1994 OLYMPIAD

89

and since x and 3x − 1 have no common divisors, 107 is a multiple of 3x − 1. But 3x − 1 ≥ 3 · 106 − 1, so either 3x − 1 = 5 · 106 or 3x − 1 = 107 . Only 3x − 1 = 5 · 106 fits, so x = 1666667. The relation (1) implies y = 3333334. Problem 2. We first show that all the xn lie in the interval [0, 1]. Indeed, 0 ≤ xn ≤ 1 =⇒ −1 ≤ 1 − 2xn ≤ 1 =⇒ 0 ≤ |1 − 2xn | ≤ 1 =⇒ 0 ≤ xn+1 ≤ 1.

Rationality implies periodicity. If xn is rational, then so is xn+1 and its denominator is no greater than that of xn (assuming both are irreducible, of course). Indeed, let xn = pn /qn be irreducible. Then qn −2pn qn − |qn −2pn | = xn+1 = 1 − . qn qn

If this fraction is irreducible, its denominator is the same as that of xn ; if it is reducible, its denominator decreases after simplification. Thus, all the terms of the sequence are rational numbers between 0 and 1. But there are only finitely many such irreducible fractions with denominators not greater than a given q. Therefore some term of the sequence will reappear and at that point the sequence will become periodic.

Periodicity implies rationality. First solution. Eliminating the absolute value sign in the equation xn+1 = 1−|1−2xn |, we see that either xn+1 = 2xn or xn+1 = 2 − 2xn . Therefore xn+1 = a1 + 2b1 xn , where a1 is an integer and b1 = ±1. Similarly, xn+2 = a2 + 4b2 xn . Continuing this process we get xn+k = ak + 2k bk xn ,

where ak is an integer and bk = ±1. Suppose that xn+k = xn . Then xn is a solution of a linear equation with integer coefficients: ak + 2k bk xn = xn . This equation has a unique solution, since 2k bk = ±2k 6= 1. The solution is rational, so xn is rational. But then xn−1 is rational, xn−2 is rational, and so on. Finally, x1 is a rational number. Second solution. Write x1 in binary notation (Fact 12 on page 104): xn = 0.a1 a2 a3 . . . . If a1 = 0, the binary representation of xn+1 is obtained from the binary representation of xn by a shift (verify this): xn+1 = 0.a2 a3 a4 . . . If a1 = 1, the binary representation of xn+1 is obtained from the binary representation of xn by a shift and the simultaneous interchange of 0 with 1, a process we will call inversion. Let us prove that if xn is periodic, then an is also periodic. If xn+k = xn and an even number of inversions took place between them, then an+k = an . If an odd number of inversions occurred between xn and xn+k , we have an+2k = 1 − an+k = an . Either way the

page 135

90

SOLUTIONS

sequence an is periodic. But periodic binary representations, like decimal ones, correspond to rational numbers (see Fact 13 on page 104). Remarks. function y = f (x) = 1 − |1 − 2x| is linear on each of the segments  1  1. The   0, 2 and 12 , 1 :  2x for 0 ≤ x ≤ 21 , y= 2 − 2x for 12 ≤ x ≤ 1. Similarly, the function page 136 y = fn (x) = f (f . . . f (x) . . . ) | {z } n times

sending x1 to xn is linear on each segment [k/2n , (k+1)/2n ]. The graphs of the first three of these functions are as follows: y

y

y = f (x)

x

y

y = f1 (x)

x

y = f2 (x)

x

Because of the shape of its graph, the function f is called the tent map. It is an important object in symbolic dynamics. 2. For every T = 2, 3, . . . , there exists at least one point with period T : for instance, the x-coordinate of the last intersection point of the segment {(x, y) | y = x, 0 ≤ x ≤ 1} with the graph of the function y = fT (x). Explicitly, this value is x1 = 2T /(2T + 1). Contemplate the question: How many periodic trajectories are there for each period T ? (Or, which is almost the same, how many points x are there for which x = fT (x) and x 6= fk (x) if k < T ?)

Problem 3. First solution. Two members of Parliament will be said to be enemies if one of them has slapped the other. The statement of the problem is the particular case n = 665 of the following result: Lemma. If there are M ≥ 3n − 2 members of Parliament and each has slapped exactly one colleague, it is possible to make up a parliamentary committee with n members, none of whom is enemies with another. Base of the induction. For n = 1, the statement is obvious. Induction step. Since there are as many members as there were slaps, the pigeonhole principle (Fact 1) implies that there is a member who was slapped at most once. This member, therefore, has at most two enemies (one slapee and perhaps one slapper). We draft this member into the committee, and set aside his or her enemies, reducing the number of available members by at most 3. Thus there are at least M −3 ≥ 3(n−1)−3 members. Therefore the conditions of the lemma are satisfied with n replaced by n − 1, and by the inductive hypothesis one can select a committee of n− 1 members of this

page 137

YEAR 1994 OLYMPIAD

91

reduced parliament containing no enemies. With the one previously chosen we get n committee members none of which are enemies. Second solution. Consider a directed graph Γ (see fact 3 on page 101), whose vertices are members and whose edges correspond to slaps, the source being the slapper and the target the slapee. Define a directed tree as any directed graph whose vertices can be partitioned into levels satisfying the following properties: there is one vertex of level 1, called the root, with no edges emanating from it; and every vertex of level j > 1 has one outgoing edge, which goes into a vertex of level j − 1. Lemma. Each connected component of Γ consists of disjoint directed trees whose roots are linked together in an oriented cycle. 3

1 4

3

2

2

1

3

1 1

1

Proof. Starting from any vertex, we follow directed edges until we return to a vertex already visited, which must happen since the graph is finite. Hence, each connected component has a cycle. We assign the vertices in these cycles level 1. Among the remaining vertices, we assign level 2 to those from which there is an edge leading to a vertex of level 1. Vertices of level 3 will be those from which there is an edge leading to a vertex of level 2, and so on. More formally, the level of a vertex not in a cycle is 1 plus the number of edges that separate that vertex from a cycle. (This number is well-defined since the path leading from a vertex to a cycle is unique.) ˜ Next we assign one of three colors to each vertex, so that no edge connects vertices of the same color. We start with the cycles. In cycles of even length, we color the vertices alternatingly red (R) and green (G). In cycles of odd length, we paint one vertex blue (B), then alternatingly paint the other vertices R and G. R

R R

G

R

G

G

B R

R

G

We next color the trees according to their roots. If the root is R or B, even levels are colored G and odd levels starting from 3 are colored R. If the root is G, even levels are colored R and odd levels are colored G.

page 138

92

SOLUTIONS

Clearly, at least one of the colors is used in at least one third of vertices. The committee is then declared to consist of all members of that color. Remarks. 1. If each cycle contains an even number of nodes, two colors suffice. 2. Give an example where it is impossible to select a committee with 668 members (out of 1994). 3. The lemma in the first solution can be generalized: If there are M members of Parliament and each has slapped exactly k colleagues, it is possible to make up a parliamentary committee with [M/(2k + 1)] members containing no enemies. Supply a proof; you can follow closely the first solution given above.

Problem 4. Let’s try to guess the length of the segment AK. Consider the limit case as the point D tends to C. Then one of the circle shrinks to a point and the other becomes the circle inscribed in △ABC; the point K merges with P , the tangency point of this circle with side AC. Now, there is a formula for the length of the A segment between a vertex A of a triangle ABC and the tangency point of the incircle to a side M K containing A: it reads ′ B AB + AC − BC P N ′ . C 2 Q Let us prove that AK is given by this formula, D F B E C not only in the limit case but in general. ′ Denote the tangency points as in the figure. Observe that AB = AP , ′ AC = AQ, KM = KP , KN = KQ, since these are pairs of tangent segments drawn from a single point. Moreover, MN = EF since these are segments of common tangents to two circles (see Fact 17). We can now write AB + AC − BC = AB ′ + AC ′ − EF = AB ′ + AC ′ − MN

= AP + AQ − MN = AP + AQ − (MK + KN ) = (AP − KP ) + (AQ − KQ) = 2AK.

Problem 5. (a) The answer is negative. For example, take the polygon in the figure, consisting of three identical square rooms connected by thin bent corridors. We will prove that no 0.3S chord splits the polygon into two pieces of equal area. Let the area of the polygon be S, and let the area of each room be 0.3S, so the total area of the corri0.3S dors is 0.1S. If the chord intersects only a corridor, then two rooms are on one side of it, and their area 0.3S is greater than 0.5S. If the chord intersects one of the rooms, it cannot intersect the “hub” of the corridors, and so again there are two rooms on one side of the chord. (b) First solution. The idea is to move the ends of a chord through the polygon in such a way that the area of the smallest piece is maximized. For notational simplicity we assume the total area is 1.

page 139

YEAR 1994 OLYMPIAD

93

Select the direction of the chord so it is not parallel to any side or diagonal of the polygon; let this be the vertical direction. Then the chord cannot ever pass through more than one vertex of the polygon. The chord divides the polygon into exactly two pieces if no vertex lies in the interior of the chord, and into exactly three pieces otherwise. Choose any vertical chord with no vertex in its interior. It divides the polygong into two pieces. If the smallest piece has area at least 13 , the problem is solved. Otherwise, we start moving the chord either right or left, so as to make the area of the smallest piece grow. As long as the chord does not touch a concave vertex, the area grows continuously (see Remark to the solution of Problem 63.11.4). However, when the chord reaches a concave vertex, one of its endpoints may jump suddenly to another part of the perimeter. What happens when the chord encounters such a troublesome vertex? (Note that the chord cannot meet two vertices simultaneously.) Consider the situation at the exact moment when three regions exist. Let a be the area of the region where the chord had been so far, and b, c, with b ≥ c, the areas of the other two regions. Note that b > 31 , since a < 31 (otherwise we’d be done). There are three possible configurations, shown below, and in all of them we push the chord slightly into the b region, thus combining the two regions of smallest area. In one case, shown in the rightmost diagram, this involves a reversal in the direction of movement of the chord.

c

c

a

b c

b b

a

a

If b < 23 we are done, because the combined piece has area just above a + c = 1 − b ≥ 13 , while the other has area just below b, which as we know exceeds 13 . If b ≥ 23 , we continue moving the chord into the area b; the area of the smallest piece has jumped from a to a + c, but remains less than 13 , and we repeat the process. Now, a troublesome vertex can only be crossed once, because the recently swept region, of area a, keeps growing throughout — that is, at each time it contains all earlier versions of itself — whereas the “recently swept regions” available for a given troublesome vertex are all disjoint. (These are the regions of area a, b, c in the figure.) The number of vertices being finite, we eventually run out of troublesome vertices, meaning that no obstacle remains for the smaller piece to attain area 31 . (Compare Fact 2 on page 100.)

page 140

REF 63.11.4

94

SOLUTIONS

Second solution. Any polygon can be partitioned into triangles whose sides are sides or internal diagonals of the polygon. (We will not prove this fact, so this solution is only a sketch.) Consider such a triangulation. Each of the internal edges of the triangulation splits the polygon into two pieces. Let AB be an edge for which the area of the smaller piece is maximal. Let ABC be a triangle of the triangulation adjacent to AB on the side of the larger piece. Denote by SAB , SBC and SAC the areas of the pieces adjacent to the corresponding sides of the triangle ABC and not containing it. (If a side is an edge of the polygon, the corresponding area is defined to be 0.) Then

page 141

SAB + SBC + SAC + SABC = 1, where, for simplicity, we have defined the area of the polygon to be 1. We may assume that SBC ≥ SAC . Further, from the maximality condition imposed on the chord one can deduce that SAB ≥ SBC ≥ SAC .

These inequalities imply that SAB + SABC ≥ 31 . If SAB ≥ 31 , then the problem is solved; otherwise let X be a point on side AC such that SAB + SABX = 13 . The chord passing through B and X satisfies the conditions of the problem. (Check it!) Remarks. 1. A variation of part (a) is “Can a chord always be found that divides the polygon into some number of pieces of equal area?” The answer is still no; it is not difficult to adapt the solution given above to this case. For example, let the areas of the rooms in the figure on page 92 be 0.3S, 0.33S and 0.36S. Prove that such a polygon cannot be cut by a single chord into any number of pieces of equal area. 2. A polygonal annulus is a region whose boundary consists of two polygonal curves, instead of one. There are polygonal annuli that cannot be divided by a chord into two polygons whose areas are more than a third the total area. Can you come up with an example? Also, explain what breaks down in the proof given in the first solution to (b) above. 3. The second solution of part (b) shows that it is always possible to select the desired chord so that it contains a vertex.

Problem 6. We call a polynomial an xn +an−1 xn−1 +· · ·+a1 x+a0 positive if all its coefficients are positive (ai > 0 for i = 0, . . . , n). Clearly, the product of positive polynomials is positive. Therefore if the square and cube of a polynomial are positive, so is any higher power, because any higher power is a product of squares and cubes. We first find a nonpositive polynomial whose square and cube are positive. Experimenting with 0 and 1 as coefficients — a process made especially easy using the shorthand discussed in Remark 2 below — we find that the polynomial f (x) = x4 + x3 + x + 1, although it has a 0 coefficient, has a positive square and cube. Now we need a polynomial with a negative coefficient. For this, we

page 142

YEAR 1994 OLYMPIAD

95

“jiggle” the polynomial f a bit (mathematicians say “perturb”). That is, we replace the 0 coefficient by a small negative coefficient, defining g(x) = f (x) − εx2 ,

(1)

where ε > 0 is very small. Then the coefficients of g 2 and g 3 are close to those of f 2 and f 3 , and so must still be positive. So the answer to the question is affirmative, and an example is given by x4 + x3 − εx2 + x + 1 for any ε positive and small enough. One can check 1 explicitly that ε = 10 will work. Remarks. 1. To be rigorous, the closeness argument must use the fact that the coefficients of the square of a polynomial F (and also those of the cube) depend continuously on the coefficients of F . (For this notion, see for instance the remark to the solution of Problem 63.11.4.) Continuity follows from the fact REF 63.11.4 that the coefficients of F 2 are themselves given by polynomial expressions in the coefficients of F : if F (x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0 , the coefficients of F 2 are a24 , 2a3 a4 , a23 + 2a2 a4 , and so on. Therefore a small change in the ai causes only a small change in the coefficients of F 2 . 2. The multiplication of polynomials can be sped up by writing just the coefficients, particularly if they’re simple. For instance, the multiplication of x2 + 1 by x + 1 can be represented by the schematic calculation 101 +1 0 1 1111 3

where we interpret the result as x + x2 + x + 1. Also note that the “hole” in x2 +1 — that single 0 between 1’s — gets filled when we multiply this polynomial by x + 1. Using this idea it’s not hard to come up with the polynomial in the solution, which “fills its own hole” when multiplied by itself: 11011 +1 1 0 1 1 + 11011 + 11011 121242121 Incidentally, the square of the integer 11011 is 121242121. How does this relate to the previous calculation? Something similar happens with the cube of 11, which is 1331 — these are the coefficients of (x + 1)3 . How far can you push this observation? What happens if you square the polynomial x4 + 2x3 + 2x + 1 and the number 12021?

Level D Problem 1. The figure shows an example with 6 faces: two triangles, two quadrilaterals, and two pentagons. Such a polyhedron can be obtained from a tetrahedron by clipping off two corners. page 143

96

SOLUTIONS

Remarks. 1. This is essentially the simplest example. Indeed, let the polyhedron have n faces. Since  than two faces have the same number of edges,  no more there are at least 21 (n+1) types of faces (triangles, quadrilaterals, and so on). Since n ≥ 4, this means there are faces with 4 or more edges. But such a face has at least 4 neighbors, so n ≥ 5. Therefore there are faces with 5 or more edges. But then n ≥ 6. Another solution, with more than the minimal number of faces, can be obtained from a cube by clipping off two neighboring corners. 2. Solve the following related problem from the 1973 Moscow Mathematical Olympiad: Prove that any convex polyhedron has two faces with the same number of edges. (Hint: Euler’s formula.) 3. Prove or disprove: Any polyhedron with 10n faces has n faces with the same number of edges.

Problem 3. We first solve an auxiliary problem: to find the smallest circle with center on the y axis, tangent to the x axis, and intersecting the graph of the function y = x4 somewhere beside the origin y y = x4 (see figure). In other words, what is the least r for which the system x2 + (y − r)2 = r2 ,

y = x4 .

(1) (2)

has a nonzero solution? It is intuitively clear that this problem is equivalent to the initial one, and we shall x prove this equivalence rigorously later. Substituting (2) into (1), simplifying and dividing by x2 (assuming x > 0) we get x6 − 2rx2 + 1 = 0 whence we deduce that

  1 1 4 x + 2 . 2 x The smallest value r0 is the minimum value of r(x) for x > 0. We have r(x) =

r′ (x) = 2x3 −

1 . x3

√ On the semiaxis x > 0 we see that r′ (x) < 0 for x < x0 = 1/ 6 2, r′ (x0√ ) = 0, and r′ (x) > 0 for x > x0 . Hence r′ (x) decreases for x < x0 = 1/ 6 2, it attains its minimum at x0 , and it increases for x > x0 . Thus, the smallest value of r for which the circle has a common point with the curve y = x4 (away from the origin) is √ 332 . r0 = r(x0 ) = 4 It remains to show that the same r0 solves the initial problem as well. First, let’s check that the circle of radius r0 with center at (0, r0 ) lies above

page 144

YEAR 1994 OLYMPIAD

97

the graph of y = x4 . Indeed, r(x) ≥ r0

for any x 6= 0.

(3)

Substituting the expression for r(x) in (3) we get x6 − 2r0 x2 + 1 ≥ 0

for any x 6= 0.

Multiplying both sides of this inequality by x2 and replacing x4 by y we get x2 + (y − r0 )2 ≥ r02

for any x and y = x4 .

But this means that the cherry of this radius can lie in the goblet. Finally we verify that if r > r0 the cherry will not touch the bottom of the goblet. Indeed, in this case, so for y0 = x40 , we have

x60 − 2rx20 + 1 < 0,

x20 + (y0 − r)2 < r2 . This means that the circle of radius r tangent to the x-axis at the origin intersects the graph of the function y = x4 ; so the cherry will not touch the bottom. Remarks. 1. It is not difficult to see that, in cross-section, the biggest cherry is tangent to the goblet at the bottom and at two more points. In space, the set of points where the cherry is tangent to the goblet is the union of a point and a circle. 2. A similar problem can be solved for the goblet whose vertical cross-section is of the form y = |x|a , for any real a > 0. For the x-coordinate of the tangency point of the thickest cherry with the goblet we have an equation a−2 . x2(a−1) = a (a) For a > 2 there is a nonzero solution; the situation is similar to that of a = 4. (b) For a = 2, we get x = 0 and r = 21 , which is the curvature radius of the graph at the origin; so the biggest cherry touches the goblet only once, at the bottom. (c) For 0 < a < 2 no cherry can touch the bottom of the goblet.

Problem 4. The homothety with center A and scale factor 2 sends the initial polyhedron M into a polyhedron M ′ whose volume is 8 times greater than that of M . Let us prove that all 8 translated polyhedra B Y lie inside M ′ . Let B be the translate of the vertex A, let X be an arbitrary point of M , and let Y be the image of the point K X under the same translation (see figure). We must prove A X that Y belongs to M ′ . Since M is convex, the segment BX is entirely contained in M ; in particular, its midpoint K belongs to M . The quadrilateral ABY X is a parallelogram, so Y is obtained from K under a homothety with the center at point A and factor 2; hence Y belongs to M ′ .

page 145

98

SOLUTIONS

Next we observe that points close to A do not belong to any of the translated polyhedra. Indeed, let us turn M so that A lies above all the other vertices of M . Then there exists a plane that passes below A but above all the other vertices of M . This plane cuts off of M a small polyhedron N containing A. Clearly, N does not intersect any of the 8 translated polyhedra. If all the translated polyhedra had disjoint interiors, the volume of M ′ would be at least the sum of volumes of the translated polyhedra plus the volume of N . But the sum of volumes of the translated polyhedra is equal to the volume of M ′ . This is a contradiction. Remarks. 1. It is easy to generalize the result to n-dimensional space. 2. At the Thirteenth International Mathematical Olympiad, held in the town of ˇ Zilina, Czechoslovakia, in 1971, a weaker formulation was offered: Prove that at least two of the nine (not eight) polyhedra intersect.

Problem 5. It is difficult to illustrate the problem (the intersection points lie far from each other), so let’s think. First, recall that the points of a bisector are equidistant from the sides of the angle it bisects. For each of the four lines containing a side of the quadrilateral ABCD, define a function fi , where i = 1, 2, 3, 4, equal to the directed distance to this line: if the point lies on the same side of the line as the quadrilateral, we take the usual distance, otherwise we take minus the distance. The functions fi depend linearly on the coordinates of the points (see Fact 25), i.e., fi (x, y) = ai x + bi y + ci . A point lies on the bisector of an outer angle of the quadrilateral if and only if the sum of the values of the two functions corresponding to the legs of the angle vanishes. For each of the two intersection points of the bisectors spoken about in the formulation of the problem, we see that the sum of all four functions vanishes. But the sum of linear functions is a linear function, and the set of points where a nonconstant linear function vanishes is a straight line. The sum of the fi does not vanish identically since it is positive inside the quadrilateral. Hence the intersection points of the bisectors lie on one line. Remarks. 1. A spatial generalization of the result of the problem can be obtained using the same idea. 2. Here is another problem that can be solved using a similar idea: in a triangle, find the locus of the points the sum of whose distances to two sides equals the distance to the third. The intersection of a bisector with the opposite side satisfies this condition; therefore so does any point on a line segment connecting two of these intersections. (Why? Be careful.) 3. In school, a linear function is any polynomial of degree at most 1. In linear algebra, a linear function is any polynomial of degree at most 1 that does not have a constant term. Functions with a constant term are said to be affine.

page 146

YEAR 1994 OLYMPIAD

99

Problem 6. When does a power of 2 have a string of 9s among its last few digits? When it’s a bit less than a number divisible by a high power of 10. For example, 212 + 4 is divisible by 100, 253 + 8 is divisible by 1000. Let’s first try to find numbers of the form 2n +1 divisible by a high power of 5. Then we will multiply those numbers by the corresponding power of 2 and obtain numbers of the form 2k (2n + 1) divisible by a high power of 10. Simplifying and subtracting the smaller summand we get the required power of 2. 2·5k−1

Lemma 1. For all k ≥ 1, the number 2

+ 1 is divisible by 5k .

Proof. We use induction on k (see Fact 24 on page 24). The base of induction (k = 1) is obvious. To prove the induction step, we write 22·5 Let a =

k−1 45 .

k−1

+ 1 = 45

k−1

+ 1.

By the induction hypothesis a + 1 is divisible by 5k . Then k

45 + 1 = a5 + 1 = (a + 1)(a4 − a3 + a2 − a + 1).

Since a + 1 is divisible by 5k , it suffices to prove that the second factor is divisible by 5. Indeed, a is of the form 5m − 1, so all the summands of the second factor give residue 1 after division by 5 (see Fact 7), and their sum is divisible by 5. The lemma is proved. ˜ k−1

Thus, the number 2k (22·5 + 1) ends with at least k zeros. It is easy to see that 2k has at most k/2 digits, if k > 1. Hence, among the last k digits of the number k−1 22·5 +k , no more than k/2 digits can differ from 9. Remarks. 1. Similar arguments prove that the numbers 40 ,

41 ,

42 ,

...,

45

k

−1

have different residues after division by 5k . Try to prove a more general fact often encountered in number theory, called Hensel’s lemma: If x − 1 is divisible by pk , where p > 2 is a prime and k > 0, but is not divisible by pk+1, then xn − 1 is divisible by pk+r if and only if n is divisible by pr . 2. At the Leningrad Mathematical Olympiad in 1981, a similar problem was posed: Is there a positive integer power of 5, whose last 100 digits contain at least 30 zeros in a row?

page 147

Year 1995 Olympiad Level A Problem 1. Let a and b be the initial prices (in groats) of ale and bread, respectively. Then 1 = a + b. After the first rise, the prices become 1.2a for ale and 1.2b for bread, so 1 = 1.2a +

1 2

· 1.2b = 1.2a + 0.6b.

Subtracting this equation from 1 = a+b gives 0 = 0.2a−0.4b, which simplifies to a = 2b. Substituting into 1 = a + b we get 3b = 1, so b = 31 and a = 32 . The price of ale after the second rise is (1.2a) · 1.2 = 0.8 · 1.2 = 0.96. Thus, Isaac can still buy his ale with a groat. Problem 2. We work by induction (see Fact 24 on page 109).

∗ ∗ ∗

page 148

∗ ∗

Base of the induction. Since 10017 = 53 · 189, the claim is true for the first element of the sequence. Induction step. Suppose a number in the sequence is divisible by 53, and look at the difference between this number and the next: 1001 . . 1}7 − 1001 . . 1}7 | .{z | .{z k times

∗

(1)

k−1 times

This difference equals (1001 − 100) · 10k = 901 · 10k , because the last k digits cancel out; it is therefore divisible by 53, since 901 = 53 · 17. Now, if the number on the right in (1) is a multiple of 53 and the difference is too, so is the number on the left (Fact 5, page 102). This proves the induction step.

∗ ∗ ∗

Remarks. 1. One can show that the quotients upon division by 53 are of the form 18 . . . 89. A 2. See problem 58.9.1 for a similar question. REF 58.9.1 K

Problem 3. (a) The segment KL is a midline of the triangle ABC, so its length is half that of AC. Similarly, LM has half the length of BD. But AC is taken to BD under a 120◦ rotation about the point O (clockwise in the case of the figure). Hence AC = BD, so KL = 12 AC = 12 BD = LM . D

O

B L

◦ ◦

120 120◦◦◦◦◦◦ 120 120 120

∗

C M

∗

102

SOLUTIONS

−→ −→ (b) Again by the midline theorem, the vectors AC and KL are parallel, as −→ −−→ −−→ −−→ are LM and BD. The angle between AC and BD is 120◦ , since a 120◦ −→ −−→ rotation takes one to the other. Therefore the angle between KL and LM − → − − → is 120◦ , and the angle LK and LM is the complement of that, or 60◦ . Using part (a), we deduce that the triangle KLM is equilateral.

∗ ∗ page 149 ∗∗

Problem 4. Among parallelipipeds of fixed volume, the one with the least area is the cube; see Fact 26 on page 111. Therefore the minimum amount of material needed is √ 3 (1) 6 · ( 1995)2 . This is a messy number, so let’s try to find a solution in integers. The nearest cubes to 1995 are 1728 = 123 and 2197 = 133 ; moreover, a 12 × 12 × 13 box doesn’t have enough volume (1872). A 12 × 13 × 13 box does (2028), but its area, 2 · (12 · 13 + 12 · 13 + 13 · 13) = 962, only gives an answer to part (a). To do better, we can try adding 1 to one dimension while subtracting 1 from another (see also Remark 3 below). Indeed, the 11 × 13 × 14 box has volume 2002, and area 958, so it works for (a), (b) and (c).

∗

Remarks. 1. One can check that this is the best possible answer in integers: 957 units are not enough. Allowing real numbers, the minimum area is 950.853, the value of (1). 2. The two-dimensional counterpart of the result we used is that the rectangle of least perimenter for a given area is a square. Derive this from the inequality between the geometric and arithmetic means. 3. Because the dimensions are roughly the same, adding 1 to one dimension while subtracting 1 from another changes the volume and area very little (recall that (x − 1)(x + 1) is “almost equal” to x2 ). In the case of 12 × 13 × 13 we have a 1.5% excess margin in the volume — we can go down from 2028 to 1995 — while the area only needs to go down by about 0.5%, from 962 to 958. So the strategy works.

∗ ∗ ∗

Problem 5. First solution. It suffices to prove that the fuel cost does not change under an interchange of two consecutive deliveries. Indeed, by applying such transpositions we can reorder the deliveries in any way we want. (This statement, which is easily proved by induction on the number of villages, is an important theorem in permutation theory; see Chapter 2 in Transpositions.) CITE Transpositions Suppose that at some point the bus visits village M and then visits N page 150 next, delivering goods of respective weights m and n. By assumption, the one-way distances to M and N are m and n as well. If we change the route just by interchanging N and M , this obviously does not affect the amount of fuel spent on other legs of the trip. More subtly, it does not affect the portion of the fuel bill due to the goods delivered to any other villages. This is a consequence of the fact that the cost per mile is proportional to the total removed “ This is weight, and so can be written as a sum of costs for each package separately. ∗ clear ... that come Apart from the packages delivered to M and N , all packages on the truck later” travel the same miles under either route (before or after the interchange).

YEAR 1995 OLYMPIAD

103

Thus we just have to compare the cost of carrying goods of weights m and n to the villages M and N . In the first case, a load of m + n will be carried over a distance m (to village M ), then a load n will be carried over a distance m + n — back to the city and out to N . Hence, the cost is proportional to (m + n)m + n(m + n) = (m + n)2 .

∗

In the second case the cost is proportional to (m + n)n + m(m + n), which is the same.

∗

∗

Second solution. Let the weights of the goods be a1 , a2 , . . . , an in the order they are delivered. The fuel cost is proportional to a1 (a1 + 2a2 + 2a3 + · · · + 2an ) + a2 (a2 + 2a3 + · · · + 2an )

+ · · · + an−1 (an−1 + 2an ) + a2n .

Expanding the products and reordering we get a21 + a22 + · · · + a2n + 2a1 a2 + 2a1 a3 + · · · + 2a1 an + 2a2 a3 + · · · + 2an−1 an ,

which equals (a1 + a2 + a3 + · · · + an )2 . This expression does not depend on the order of enumeration. Problem 6. Let N lie on AB and K on AF , as in the figure. Observe that F K = AN . Select points P on BC, R on CD, S on DE and T on EF so that F K = AN = BP = CR = DS = ET . C P Then \KBN = \T AK, \KCN = \SAT , \KDN = R B D \RAS, \KEN = \P AR, \KF N = \NAP . Thus \KAN +\KBN +\KCN +\KDN +\KEN +\KF N = \KAN +\T AK+\SAT +\RAS+\P AR+\NAP = \KAN +\KAN = 120◦ +120◦ = 240◦ .

N S A

E

K F

T

Level B Problem 1. We work by induction (see Fact 24 on page 109). Base of the induction. Since 12008 = 19 · 632, the claim is true for the first element of the sequence. Induction step. We need to show that if one of our numbers is divisible by 19, so is the next. To this end it is enough to show that the difference between two consecutive numbers is divisible by 19 (see Fact 5 on page 102). But 1203 . . 3}08 − 1203 . . 3}08 = 1083 · 10n = 19 · 57 · 10n . | .{z | .{z n times

page 151

∗

n−1 times

Remarks. 1. It is not difficult to show that the quotients upon division by 19 are of the form 63 . . . 32. 2. See problem 58.8.2 for a similar question. REF 58.8.2

104

SOLUTIONS B

Problem 2. At first let A′ be any point on BC, and draw the segment AA′ . We claim that, among the segments C2 from C to points on the side AB, there are at most two whose length equals AA′ : namely, the segments CC1 C1 and CC2 such that P1 ′ ′ \C1 CA = \A AC and \C2 CB = \A AC. A

A′ P2 C

(The two coincide if AC1 = AC2 is the bisector.) Indeed, CC1 = AA′ because △ACC1 and △CAA′ are congruent (case ASA). Similarly, CC2 = CC1 by the congruence of △ACC1 and △BCC2 . That these are the only two points on AB whose distance to C equals AA′ follows from the fact that a circle intersects a line in at most two points. Now we observe that the desired locus is made up of all intersections of AA′ with CC1 and CC2 , as A′ varies. Now recall that an isosceles triangle has a symmetry axis that contains the altitude, bisector, and median from the corresponding vertex, and is also the same as the perpendicular bisector of the opposite side. Let P1 be the intersection of AA′ with CC1 . It lies on the altitude of the triangle ABC dropped from B: indeed, the triangle AP1 C is isosceles, since the angles at its base are equal. Therefore AP1 = CP1 , that is, P1 lies on the perpendicular bisector of AC, which is also the altitude in question. Conversely, all points on this altitude satisfy the condition of the problem, by symmetry. The locus of P2 , the intersection of AA′ and CC2 , is less obvious. We know that the triangles A′AC and C2 CB are congruent, and rotating the latter by 120◦ about C (clockwise in the figure above) aligns it with the former. Thus CC2 rotated by 120◦ becomes parallel to AA′ ; that is, the angle AP2 C is 120◦ . This can also be seen as follows:

∗

page 152

∗

\AP2 C = 180◦ − \A′AC − \C2 CA

= 180◦ − \A′AC − (60◦ − \A′AC) = 120◦ .

The locus of points (inside the triangle ABC) that see the segment AC under an angle of 120◦ is an arc with endpoints A and C and measure 2 · (180◦ − 120◦ ) = 120◦ . (Since the sides of an equilateral triangle are seen from its center at B 120◦ , this arc intersects the other part of the locus — the altitude — at the center of the triangle. This corresponds to C1 = C2 being the midpoint of AB and A′ being the midpoint of BC.) Reversing the argument, we see that all points on this arc satisfy the condition of the problem. A (Check it!)

∗

C

Problem 3. Suppose first that n ≤ 1995, so no strip can be longer than 1995. Since all strips have different, integer lengths, the most area they can cover is 1+2+· · ·+1995 = 21 ·1995·(1995+1), or 1995·998 (see Remark after solution). So the maximum possible value of n (not exceeding 1995) is 998.

∗

page 153

YEAR 1995 OLYMPIAD

105

Now, one can show that n can be 998, by exhibiting a decomposition of the 1995×998 rectangle into strips of different lengths. We follow the pattern of the figure on the right (which was drawn with longest side 9 instead of 1995). That is, one row of the rectangle is occupied by a single strip of length m = 1995, the next by strips of lengths 1 and m − 1, and so on until the last row, which is occupied by strips of lengths (m − 1)/2 and (m + 1)/2. Note that m = 1995 is odd. It is also obvious that any value of n less than 998 will work; starting from the 1995 × 998 rectangle, delete some rows until only n are left. Inspired by this particular case, we prove:

∗

Lemma. An M × N rectangle, with M ≥ N , can be divided into strips all of different lengths if and only if M ≥ 2N − 1. Proof. We follow the technique used above for M = 1995. The maximum area that can be covered is 12 M (M +1) (the sum of integers from 1 to M ), so MN ≤ 21 M (M +1), or N ≤ 21 (M +1). There remains to show that for all such values of N the rectangle can be covered. The case of M odd is verbatim the same as for M = 1995. Now take the case that M is even. Then N ≤ (M +1)/2 means that N ≤ M/2. If N = M/2 we can cover the M ×N rectangle by strips of lengths M , 1, M −1, 2, M −2, . . . , M/2−1, M/2+1. Any rectange with fewer rows (that is, N < M/2) can be covered by deleting some rows from the previous example. ˜

∗

We now solve the problem in the case where 1995 is the shorter dideleted spurious mension (n > 1995). We simply take N = 1995 in the lemma and obtain ∗ “are” M ≥ 2 · 1995 − 1 = 3989. Thus the final answer is “n ≤ 998 or n ≥ 3989”. page 154

Remarks. 1. We used the equality 1 + 2 + · · · + n = 21 n(n+1). This formula can easily be proved by induction. We take this occasion to give another elegant proof, whose underlying idea is often used to find the sum of any arithmetic sequence. Set X = 1 + 2 + · · · + n and compute the sum of all numbers in the table 1

2

n n−1

3

...

n−2

...

n−2 n−1 3

2

n 1

Since the sum of the elements in each column is equal to n + 1 and there are n columns, the sum is equal to n(n + 1). On the other hand, the sum of elements in each row is X, so 2X = n(n + 1), implying our statement. 2. One can show a priori that, in the situation of the problem, the sides of the strips should be parallel to the sides of the given rectangle. Indeed, assume otherwise, and let A be a vertex of a slanted strip closest to the boundary of the original rectangle. Consider the sides of all the strips containing A on their boundary. The angle between such consecutive sides is right, which means that all rectangles having A on their boundary are slanted. But this contradicts minimality.

∗

106

SOLUTIONS

Problem 4. First solution. The hypothesis implies that the number (a + c)(b + c) ab = a+b+c+d=a+b+c+ c c is an integer. Hence the fraction is reducible. Since both factors in the numerator are greater than in the denominator, after division by c each of the factors turns into a number greater than 1. Thus, a + b + c + d is the product of two factors, each greater than 1, so it cannot be prime. Remark. We used a nontrivial statement: If the fraction (xy)/z is an integer, then z can be represented as a product of integers ts so that x is divisible by t and y is divisible by s. If you master the second solution, you will see how to prove this statement rigorously.

Second solution. We know that a, b, c, d are positive integers and that ab = cd. We claim that in this situation there exist positive integers u, v, w, z such that a = uv, b = wz, c = uw, d = vz. Indeed, consider the prime factorizations of a, b and c. Since c divides ab, all its prime factors are present among the prime factors of a and b combined; more precisely, a factor appearing m times in the prime factorization of c must be accounted for at least m times in the prime factorizations of a and b taken together. Let p1 , . . . , pn be all the prime factors of a and b, and write a = pk11 pk22 . . . pknn ,

b = pl11 pl22 . . . plnn ,

mn 1 m2 c = pm 1 p 2 . . . pn ,

where some of the numbers ki , li , mi can equal zero. As we observed above, mi ≤ ki + li for i = 1, . . . , n. Therefore we can write mi = ti + si , where ti ≤ ki and si ≤ li . Set u = pt11 pt22 . . . ptnn

∗

and w = ps11 ps22 . . . psnn .

Then c = uw. Besides, u divides a and w divides b. Hence a = uv and b = wz, where v and z are integer. Finally, ab uvwz d= = = vz, c uw proving our claim. Now note that a + b + c + d = uv + wz + uw + vz = (u + z)(v + w). Both factors are greater than 1, so a + b + c + d is not a prime. Problem 5. First solution. The general idea is this: At some point, three at least of the starting triangles must be cut. After that there will be two groups of three identical triangles each. Again, from each of these two trios, two triangles must be cut. Two children coming from one group will be identical with one another and with two children from the other group (see shaded triangles in the figure). We’re back to having four identical triangles!

∗

page 155

YEAR 1995 OLYMPIAD

107

But we must be careful about the argument: what if the cuts are made first in one triangle and its children, leaving the other starting triangles for later? Perhaps then we might not get the trios of the previous paragraph? To help organize our thoughts, let’s call a cut in an initial triangle a level-1 cut. Any cut of a triangle obtained from a level-1 cut will be called a level-2 cut, and so on. Suppose we have a recipe — a sequence of cuts — that takes us from four equal triangles to all different triangles. The key observation is that the order of the cuts doesn’t matter, so long as the recipe still makes sense. In particular, the end result of a recipe can be obtained via a different recipe consisting of a reordering of the same cuts: all level-one cuts made first, then all level-two cuts, etc. Now start with a recipe having as few steps (cuts) as possible. After the reordering, which leaves the number of steps unchanged, there must be three level-one cuts right at the beginning; otherwise two starting triangles will never be touched. After all level-one cuts are made, there must be four level-two cuts; otherwise there will be two first-generation children left untouched at the end. These four level-two cuts produce four identical triangles, as discussed (see figure on previous page). We discard all other triangles apart from these four. Now, if we apply to these triangles all the cuts in the recipe that still make sense (skipping steps that involve discarded triangles or their descendents), we end up with a subset of the triangles obtained from the full recipe, and of course these must be all different. But then we’ve found a shorter recipe to go from four identical triangles to all different triangles! (At least 7 steps shorter.) But we had assumed our starting recipe was as short as possible, so we have reached a contradiction. Second solution. We will look for an invariant of the problem (see Fact 2 on page 100) that we can use to our advantage. An obvious choice is the area. Let each starting triangle have area 1 and let p and q = 1−p, be the areas of the triangles resulting from the first cut. We can assume p < q, the case of an isoceles triangle being trivial. By induction, the area of any triangle is pm q n , where m and n indicate respectively how many times the smaller or the larger of the two children was chosen after a cut. Triangles with the same pair (m, n) have the same area and, since they also have the same shape, must be congruent. Obviously, the total area of the triangles is 4 throughout the process. Now suppose that after some sequence of cuts, no congruent triangles are left; we will try to obtain a contradiction. Each final triangle corresponds to a different pair (m, n). If we sum the areas of all (m, n)-triangles for 0 ≤ m < ∞ and 0 ≤ n < ∞, whether or not they occur, we should obtain a number strictly greater than 4, since, by assumption, the final triangles are among this set, but are finite in number.

∗∗

∗

SL: make sure figure position remains the same

page 156

∗

∗∗

108

SOLUTIONS

Using the formula for the sum of a geometric series, we have 4 0 (see the remark below). So we can eliminate the factor a2 + b2 from both sides to obtain 1/a = 1/b. Hence a = b. Remark. The following obvious statements are often used in olympiad problem solving (and in mathematics generally): (1) the square of a nonzero number is positive; (2) the sum of the squares of several numbers is nonnegative. If such a sum is zero, then each of the numbers is zero.

Problem 2. Denote by mi the masses of the weights and by xi the masses of the balls. The sum (m1 − m2 )+ (m2 − m3 )+ · · · + (m9 − m10 )+ (m10 − m1 ) equals 0, because each mi occurs exactly twice, with opposite signs, leading to complete cancellation. The mass xi of the i-th ball is the absolute value of the difference mi − mi+1 , so the preceding condition can be rewritten as ±x1 ± x2 ± · · · ± x9 ± x10 = 0,

where some signs of the xi are pluses and the others are minuses. Choose the balls whose masses enter into this sum with a plus sign and put them on the left pan of the balance. Put the other balls on the right pan. The balance will then be in equilibrium. Problem 3. Consider a grid square with a flower in it. Divide it into four equal quadrants (small squares). We can assume without loss of generality that the sides of the quadrants have unit length. Suppose our flower is in the upper left quadrant, as in the diagram; the other cases are all similar. Denote the gardeners at the vertices of the square by A, B, C, D.

A A A111

A A A222

B B B111

A A A

B B B

C C

D D

B B B222

page 171

122

SOLUTIONS

We will prove that the flower will be looked after by gardeners A, B, and C. We divide all the gardeners into four groups, each group being located in one of the four shaded sectors in the figure. Gardener A is closer to the flower than B. This follows from the Pythagorean Theorem, because the vertical separation from the flower is equal for both gardeners and the horizontal separation is less for A than for B (being less than 1 for the former and greater than 1 for the latter). Similarly, gardener A is closer to the flower than C, and gardeners B and C are closer than D. Therefore, D does not look after the flower. Neither does any other gardener from the D sector: they are all even farther from the flower than D (consider the vertical and horizontal distances again). Now consider the B sector. Clearly, all its gardeners (apart from B) are further from the flower than B, and so even further than A. We will show they are also further than C, and so do not look after the flower. Take, for instance, gardener B1 . The vertical separation from B1 to the flower is greater than 2, and the horizontal separation is greater than 1. So this gardener is farther from the flower than C. It is exactly analogous to show that no gardener in the C sector, apart from C, takes care of the flower. Finally, consider the A sector. We must show that A1 and A2 , the nearest gardeners apart from A, are further from the flower than B and C are. Consider A1 (the proof for A2 is similar). To see that A1 is further than B, just look at horizontal separations. To prove that A1 is further than C, consider the perpendicular bisector to the segment A1 C. Its points are equidistant from A1 and C. Points below the bisector are closer to C than to A1 . Clearly, the flower is below the bisector; therefore, it is closer to C. Thus, the gardeners that look after the flower are A, B, and C. Now consider the gardener at the origin, say X. We already know that X does not look after flowers unless they’re in one of the four squares (of the original grid) touching the origin. And we need only find the intersection of X’s area of responsibility with one of these squares: the pieces in the other three squares are found by symmetry. For instance, consider the square XY ZT in the figure. Again, we divide it into quadrants. As we have seen, gardener X is one of those who looks after the three shaded squares. Flowers in the fourth quadrant Y Z are looked after by Y , Z, and T . Hence, gardener X looks after the flowers growing in these three small squares and in the nine other squares obtained from them by reflection. X T A

Problem 4. Without loss of generality, assume K is closer to B than L, as in the figure. Then MKC is an equilateral triangle (since M C = KC and \M CK = 60◦ ). Hence, AB is parallel to MK (since \MKC = \ABC = 60◦ ) and the angles AKM and BAK are equal (being alternate interior angles). B K

M

L

C

page 172

page 173

YEAR 1996 OLYMPIAD

123

Notice that the angles BAK and CAL are equal by symmetry (or by the congruence of △BAK and △CAL, which follows from the SAS property). Hence \AKM = \CAL and \AKM + \ALM = \CAL + \ALM = \LM C. The last equation follows from the fact that \LM C, being an exterior angle of triangle AM L, equals the sum of its two opposite interior angles. Thus, it remains to prove that \LM C = 30◦ . But M L is a median in the equilateral triangle MKC; hence M L is an angle bisector of this triangle. Remark. Compare with the solution to Problem 3 on page 130.

Problem 5. We can assume that the rook starts from the upper left corner. A solution for even n is illustrated on the right. The rook visits first all the squares on the top two rows, then the squares on the third and fourth rows, and so on. We now show that the task is impossible for odd n. Consider any row apart from the top one. When the rook hits this row, its next move must be to another square of the same row, and then to another row. Thus, the squares of the chosen row are paired up by the moves within that row. (Note that the rook visits each square exactly once, since the number of moves is n2 .) Therefore any row contains an even number of squares. In other words, after its first visit to a row, the rook will have visited two of its squares. After the second visit, there will be four such squares. If n is odd, a moment would come when the rook must visit the last unvisited square on a row, and then it would be unable to move. (See also Fact 23 on page 109.) Problem 6. (a) Let N be the greatest number of problems solved by a single student. So, for instance, if N = 8 one student solved all problems and we’re done; if N = 7 we take a student who solved 7 problems and one who solved the remaining problem, and again we’re done. Consider the student/problem pairs that resulted in solutions. There are at least 40 such pairs (exactly 40 if we interpret “solved by five students” more strictly; either interpretation works). By the pigeonhole principle (page 100), at least one student solved 5 or more problems. So we’re left with two cases for N : N = 6, so there is a student who solved exactly six problems. Each of the remaining two problems was solved by 5 among the remaining 7 students, and since 7 is less than 5 × 2, there must be some student who solved both problems (pigeonhole principle again). N = 5, that is, each student solved at most five problems. Since there were at least 40 solutions, each student must have solved exactly five problems, and each problem must have been solved by exactly five students.

page 174

124

SOLUTIONS

So suppose student 1 solved problems 1–5. We claim there is a student among the other seven who solved problems 6, 7, and 8. If this were not true, each of the seven would have solved at least three of the first five problems. Together with student 1, this would give 5 + 7 × 3 = 26 solutions of problems 1–5. But this cannot be, because each problem was solved by exactly five students. Therefore our claim is true. (Another way to think of this last paragraph is as follows. Each problem was left unsolved by three students; in the case of problems 6, 7, and 8, these are student 1 and two more. Therefore, all in all, there are at most seven students — including student 1 — who failed to solve one or more problems among the last three. That leaves at least one student who solved all three.)

problems

(b) We need to fill an 8 × 8 table (rows representing students, columns representing problems) so that each column has four or more entries and no two rows complete each other to form the full set {1, 2, . . . , 8}. We may as well assume that each column has exactly four entries. Any solution is likely to involve some trial and error, but we use what we learned in part (a) to direct the search. In particular, we keep the definition of N . The cases N = 8, N = 7 and N = 6 won’t help us, because our earlier arguments show that in these cases, there are two students who solved all problems between themselves. (In the case N = 6 the relevant inequality is now 7 < 4 × 2.) Next we try for a counterexample assuming N = 5. Naturally, we start by filling row 1 in positions 1–5. By assumption, in the subtable consisting of the last three columns and seven rows, there are 3 × 4 = 12 entries. Each of these rows has at most two entries in columns 6–8; otherwise it would complete row 1. It is reasonable to place exactly two entries on each of rows 3–8, leaving row 2 with nothing in columns 6–8; this way we can make row 2 a duplicate of row 1 without 1 2 losing any freedom. 3 The simplest way to choose two of the numbers 6, 4 7, 8 for each of the last six rows so that each number 5 6 is chosen four times is to group the rows in pairs, as in 7 the figure: rows 3 and 4 skip column 6 and get entries 8 in columns 7 and 8; the next two rows skip column 7 1 2 3 4 5 6 7 8 students instead; and the last two rows skip column 8. We have now filled everything in, except for the subtable consiting of columns 1–5, rows 3–7. This is similar to the original task, but with two entries per column (out of six), so it’s easy to solve. Also, we must take care that no two rows that complete each other in the first five columns should also complete each other in the last three. This is best achieved by keeping the pairing of rows already introduced; we divide the first five columns into three chunks (of lengths 2,2,1 or 3,1,1) and fill one chunk per pair of rows,

page 175

YEAR 1996 OLYMPIAD

125

like this:

or

Level B Problem 1. First solution. Suppose the statement is not true. Then a certain n-gon has at least 36 angles less than 170◦ , and the remaining n − 36 angles less than 180◦ (convexity). Therefore, the sum of all angles of this polygon is less than 36 · 170◦ + (n − 36) · 180◦ . But it is well known that this sum is equal to (n − 2)180◦ . This yields the inequality (n − 2) · 180 < 36 · 170 + (n − 36) · 180. After removing the parentheses and an obvious simplification, we obtain the inequality 180 · 34 < 170 · 36, which is false. Second solution. If an interior angle of a polygon is less than 170◦ , the corresponding exterior angle is greater than 10◦ . If we had 36 or more such angles, their sum would be greater than 360◦ . But the sum of exterior angles of a convex polygon is 360◦ . This contradiction completes the proof. Problem 2. First solution. First assume that one of the numbers is zero. For instance, let a = 0 (the other cases are similar). Then we have the inequalities |b| ≥ |c| and |c| ≥ |b|, whence |b| = |c|, that is, b = c or b = −c. In the first case, we have b = a + c; in the second case, a = b + c, and we’re done. Now suppose that none of the numbers a, b, and c is zero. Without loss of generality we can assume that a is the greatest of the three numbers in absolute value (that is, |a| ≥ |b| and |a| ≥ |c|). We can also assume that a > 0 (if it isn’t, we replace all three variables by their negatives, an operation that affects neither the problem’s hypotheses nor the conclusion). Then |a| = a, |a−b| = a−b, |a−c| = a−c. Under these assumptions, the inequality |b − c| ≥ |a| implies that b and c have different signs. (Why?) Consider the two possible cases. (1) b > 0, c < 0. Then |b| = b, |c| = −c, |b − c| = b − c, and the given inequalities take the form a − b ≥ −c, b − c ≥ a, a − c ≥ b. The first inequality implies that b ≤ a + c; the second one, that b ≥ a + c; therefore, b = a + c. (2) b < 0, c > 0. Then, as in the previous case, we obtain the inequalities a−b ≥ c, c−b ≥ a, a−c ≥ −b. Therefore, in this case we have simultaneous

page 176

126

SOLUTIONS

inequalities c ≥ a + b and c ≤ a + b, that is, c = a + b. Thus, the statement is proved in both cases. Second solution. Square both sides of the inequality |a − b| ≥ |c| and carry all terms to the left-hand side. We get (a − b)2 − c2 ≥ 0. The left-hand side can be factored as a difference of squares: (a − b − c)(a − b + c) ≥ 0, or, which is the same, (a − b − c)(b − c − a) ≤ 0. By a similar argument, the products (b−c−a)(c−a−b) and (c−a−b)(a−b−c) are also nonpositive. Multiplying the three products, we find that

page 177

(a − b − c)2 (b − c − a)2 (c − a − b)2 ≤ 0.

But if a product of nonnegative numbers is not positive, it must be zero. Hence at least one of the numbers is zero and we’re done. B

Problem 3. The angles BNM and N CA N M are equal, because the lines MN and AC are parallel. The inscribed angle N CA is equal⌢to half the measure of its intercepted arc AB . The angle BAM between a tangent and a chord⌢is also equal to half the measure A of the arc AB which it intercepts (Fact 15 on page 105), so \N CA = \BAM . Hence the angles BNM and BAM are equal, and the quadrilateral AMBN is cyclic. We have the following chain of equations for angles:

C

\N CA = \BAM = \MBA = \MNA = \NAC. The second equation follows from the fact that AMB is an isosceles triangle (the tangent segments drawn to a circle from an external point are equal), the third one from the fact that AMBN is a cyclic quadrilateral, and the last one from the equality of alternate interior angles of parallel lines. It follows that \N CA = \NAC; hence AN C is an isosceles triangle, i.e., AN = N C. Remark. The attentive reader may have noticed that we used the fact that \BCA is acute. Check that otherwise there is no point N on the side BC such that MN k AC.

Problem 4. (a) The sum of the number 9 and the number beneath it lies between 10 and 18. Since there is only one perfect square in this interval, the number under 9 must be 7. Similarly, the number 7 must be written above the number 9 on the second row. In the same manner, it can be shown that the numbers under 4, 5, and 6 must be 5, 4, and 3, respectively. Now it is not difficult to complete the answer: 1 2 3 4 5 6 7 8 9 8 2 6 5 4 3 9 1 7

page 178

YEAR 1996 OLYMPIAD

127

(b) Using the same reasoning we see that the only number that can be written under 11 is 5. But the number under 4 must also be 5, which is impossible. (c) The idea is to reduce the problem to a similar one for smaller n. The first square past 1996 is 2025 = 452 , which equals 1996 + 29. So if under each number k from 29 through 1996 we write the number 2025 − k, the sum-is-a-square condition is satisfied for all columns after the first 28. This reduces the problem to the case n = 28. Repeating the same strategy we write under each of the numbers k = 21, 22, . . . , 28 its counterpart 49−k, reducing the problem to the case n = 20. Then we write 36 − k under each k = 16, 17, 18, 19, 20, reducing the problem to n = 15. Finally, under each k = 1, 2, . . . , 15 we write 16 − k.

Remark. When do you run into trouble with this strategy? Try to prove that the problem can be solved whenever n > 11.

Problem 5. We begin with two simple statements: (1) The segment joining the midpoint of a chord to the circle’s center is perpendicular to the chord. (2) The condition that the endpoints of a chord lie on different arcs AB is equivalent to the chord intersecting the segment AB (in an interior point). Thus the problem can be reworded as follows: Given three points A, B, and O with AO = BO, find the locus of all points M such that the perpendicular to MO going through M crosses the open segment AB. We now prove that the perpendicular to MO at M intersects the open segment AB if and only if exactly one of the two angles OMA and OMB is obtuse. Indeed, the perpendicular intersects AB if and only if A and B lie in opposite half-planes with respect to this B perpendicular; this condition can also be expressed by saying that exactly one of the two points (A or B) lies in the same half-plane as the point O, and the other lies in the half-plane opposite O. But a point P lies in the A half-plane opposite to O if and only if the angle P MO is obtuse (as illustrated by point A in the figure on the M O right), and P lies in same half-plane as O if it is acute (point B in the figure). The locus of points M such that \AMO is the interior of the circle with diameter AO, and the locus of points M such that \BMO A B exceeds 90◦ is the interior of the circle with diameter BO (see Fact 14 on page 105). This means that the locus in question consists of the points lying inside one, but not O both, circles with respective diameters AO and BO. In other words, it is the union of the interior of these circles minus their intersection — the shaded region in the figure.

page 179

128

SOLUTIONS

Problem 6. Let us show that Ali Baba can always arrange for seven piles to end up with at most four coins, and the thief can ensure that no pile ever contains less than four coins. That will mean that Ali Baba can walk away with 100 − 7 · 4 = 72 coins. We start by proving that the thief can ensure there are at least four coins in each pile after each exchange. This is certainly true at the beginning. Suppose it is still true at a certain point and some coins are then placed in the cups. If there are two cups with the same number of coins, the thief can simply swap them, leaving the situation unchanged. If the number of coins is different for each cup, then the two cups with most coins will contain at least three and four coins, respectively, and the thief can swap these cups. Then the new piles will again consist of at least four coins each. Now we show that Ali Baba can always reduce seven piles to four coins or less. Suppose there are four piles with more than four coins each. Let x1 ≥ x2 ≥ x3 ≥ x4 ≥ 5 be the numbers of coins in these piles. We’ll show that Ali Baba can work with these piles alone until one of them has four coins or less. He chooses the partitions x1 = y1 + 1,

x2 = y2 + 2,

x3 = y3 + 3,

x4 = y4 + 4,

and puts 1, 2, 3, and 4 coins from these piles into the respective cups. After the thief rearranges the cups, the new piles will consist of y1 + z1 ,

y2 + z2 ,

y3 + z3 ,

y4 + z4

(1)

coins, where z1 , z2 , z3 , z4 is a certain permutation of the numbers 1, 2, 3, 4. Let x′1 , . . . , x′4 be the four numbers in (1), reordered if necessary to preserve their nonincreasing order. There are three possibilities: (1) If the first cup was moved, the tallest pile got even taller (since z1 > 1). In other words, x′1 > x1 . (2) If the first cup stayed put and the second was moved, the second tallest pile grew; it either surpassed the first and we have x′1 > x1 (since we renumbered the piles), or this did not happen and we have x′1 = x1 and x′2 > x2 . (3) If the first two cups stayed put, the last two were interchanged; in this case x′1 = x1 , x′2 = x2 , and x′3 > x3 . Ali Baba then repeats this process so long as these same four piles still have more than four coins each. We must show that this cannot go on forever. Conclusions (1)–(3) can be rephrased as follows: on each application of the process, x1 can only go up, never down; if x1 doesn’t go up, x2 can only go up, never down; if neither x1 and x2 go up, x3 must go up. This means that no triple (x1 , x2 , x3 ) can ever occur more than once. (To convince yourself of this, consider the parallel situation where x1 , x2 , x3 represent the digits of a decimal number and obey the same rules: the number as a whole must increase at each step.) But there are only finitely many triples, so

page 180

YEAR 1996 OLYMPIAD

129

eventually it must be impossible to apply the process again, indicating that some pile now has four coins or less. Considering now all the piles, the net effect so far is that the number of piles with four coins or less has increased. As long as Ali Baba can repeat the whole procedure, this number will keep increasing. It only stops when there are at least seven piles with four coins or less. Level C

page 181

Problem 1. First solution. We can assume that a ≥ b (the case a ≤ b is treated similarly). Then b2 ≤ ab, a2 ≥ ab, and therefore, a2 ≥ a2 + b2 − ab ≥ b2 , whence a ≥ c ≥ b. It follows that the first factor in the product (a − c)(b − c) is nonnegative and the second one is nonpositive. But then the product is nonpositive, completing the proof. Second solution. Consider a triangle with a 60◦ angle between two sides of lengths a and b. By the law of cosines, the length of the third side is p a2 + b2 − 2ab cos 60◦ = c.

60 ◦ b

a

c

Since the greatest angle of any triangle is at least 60◦ and the smallest angle is at most 60◦ , the angle opposite side c is of intermediate value in our triangle. Since a greater angle in a triangle lies opposite a longer side, either a ≤ c ≤ b or b ≤ c ≤ a. Hence one of the two factors a − c and b − c is nonnegative and the other is nonpositive. Therefore, their product is nonpositive. Problem 2. Draw all 17 lines parallel to a diagonal of the square and passing through at least two of the marked points (see figure). This takes care of all points except two corners of the array, which require an 18th line. To see that fewer lines will not suffice, consider the centers of the 36 unit squares around the edges of the array. Clearly, any line not parallel to the edges of the array crosses at most two of these points, so we need at least 18 lines to account for all of them (pigeonhole principle, see page 100). Problem 3. Since an exterior angle of a triangle is equal to the sum of its two opposite interior angles, the triangle APk M yields the equation \Pk M C = \Pk AC + \APk M , or \APk M = \Pk M C − \Pk AC. Adding together all these equations, we see that the sum in question is equal to (\P1 M C + · · · + \Pn−1 M C) − (\P1 AC + · · · + \Pn−1 AC).

page 182

130

SOLUTIONS

Suppose that n is odd; the complementary case is similar. Reflection about the altitude of the equilateral triangle P1 M C shows that the angles Pk M C and Pn+1−k M C add up to 60◦ (see figure on the next page). A Then the terms in the first sum fall into (n − 3)/2 pairs, M each of which adds up to 60◦ , and two unpaired terms: \P1 M C = 60◦ and \P(n+1)/2 M C = 30◦ . Hence the first sum is equal to 30◦ · n. The terms of the second sum fall into (n − 1)/2 pairs with a total of 60◦ each. Hence the second sum equals 30◦ · (n − 1). The statement of the problem follows immediately. B P1 Pk P n + 1 −k C Remark. Compare with Problem 4 on page 15.

Problem 4. The case m = n = 1 is obvious: there’s no room for even a single move. For definiteness, assume that the board consists of n rows and m columns, with m ≥ n, and that the rook starts from the upper left corner. It turns out that the winning strategy for the first player is to make each move the longest possible. To prove this, we’ll assume that there are boards on which this strategy does not win and refute this conjecture by deriving a contradiction. Among all such boards, we can take one with smallest possible area. Since on an 1 × n board the first player obviously wins, we can assume that both dimensions are greater than 1: m ≥ n ≥ 2. The first player moves the rook along the entire longer side, the top row. The second one is forced to move down. Consider three cases:

page 183

(a) The second player makes a one-square move. We’re now in the starting situation of the same game played on an m × (n−1) board (see figure). Since m ≥ 2, this is not the 1 × 1 board.

(b) The second player moves all the way down. Then the first moves all the way across, per the strategy. If m = n = 2, the first player wins immediately. If not, we’re in the situation we’d be in after the initial move if the game were being played on an (m−1) × (n−1) board.

(c) The second player moves by k squares, where 1 < k < n−1. The first player moves all the way across, per the strategy. At this point, whether the next move by the second player is up or down, we’re in the situation we’d be in after the first player’s initial move if the game were being played on a smaller board: either an (m−1) × (n−1) board (thick solid line in the figure; note that this cannot be the 1 × 1 board since n ≥ 4) or an m × (n−k) board (double thin line).

In every case, we have reduced the situation to one that is encountered, if the strategy is followed, during a game played on a board of lesser area (and distinct from 1 × 1). But, by our least-area

in the original, the black-edged rectangle is too long.

YEAR 1996 OLYMPIAD

131

assumption, on such a board the first-player strategy wins. Hence it wins on the m × n board as well, and we arrive at a contradiction. Problem 5. We answer the question with an example. Suppose there are 10 inhabitants in the country, and that their houses are placed along a straight line in ascending order of their owners’ heights. Suppose the intervals between the houses, in kilometers, are 1, 2, 3, 4, 5, 4, 3, 2, and 1:

page 184

Then everyone except the tallest person can travel free of charge. Indeed, the five shortest persons can choose circles of enormous radius, so each will be shorter than 5 of their 9 neighbors. The rest of them must choose a circle enclosing just one neighbor. On the other hand, everyone except the shortest person will be allowed to play basketball, if the five tallest choose a large circle and all the rest choose a circle enclosing just one neighbor. Problem 6. It is readily seen that P (N ) > P (M ) for N > M > 0, because the coefficients are positive. We also have P (N ) > 1 for N > 0. Next we claim that if k divides x − y, then k divides P (x) − P (y). This follows by induction on the number of monomials in P . For one monomial it is obvious since xn − y n = (x − y)(xn−1 + xn−2 y + · · · + y n−1 ). But if the statement is true for two polynomials, it’s also true for their sum. Now let’s set A = P (1)P (2) . . . P (1996). Since P (k) divides A, it also divides P (A + k) − P (k), for k = 1, . . . , 1996. Hence P (k) divides P (A + k). But P (k) > 1 and P (A + k) > P (k). It follows that P (A + k) is a composite number for k = 1, . . . , 1996, completing the proof. Level D p 5

√

p 5

√ 2 + 3 and b = 2 − 3, so Problem 2. Introduce the symbols a = the desired root is x = a + b. By direct substitution we see that a5 + b5 = 4 and ab = 1, so 1 1 a + = x and a5 + 5 = 4. a a

page 185

132

SOLUTIONS

Our goal is to relate the expression on the right to powers of x. By expanding the binomial powers we see that       1 1 1 1 5 , = a5 + 5 + 5 a3 + 3 + 10 a + x5 = a + a a a a     1 3 1 1 x3 = a + . = a3 + 3 + 3 a + a a a 1 Substituting the value of a3 + 3 from the second equation into the first, we a obtain           1 3 1 1 1 1 5 5 . + 10 a + = a + 5 +5 a+ −3 a+ a+ a a a a a Hence or

x5 = 4 + 5(x3 − 3x) + 10x x5 − 5x3 + 5x − 4 = 0.

1 1 Remark. It is possible to express an + n can be expressed in terms of a + for a a any n:   1 1 . an + n = P n a + a a The polynomials Pn are related to the so-called Chebyshev polynomials Cn by the relation Cn (x) = 21 Pn (2x). The Chebyshev polynomials are defined by the formula cos(nx) = Cn (cos x). The link between these formulas follows from the relation cos x = 12 (eix +e−ix ), √ CITE Zorich where i = −1; see Zorich. See also Remark 2 to Problem 2 (page 110).

Problem 3. Any set of evenly spaced parallel planes is equivalent to any other under a similarity, and any cube is also similar to any other. So the question can be rephrased thus: Can the vertices of the unit cube [0, 1]3 lie on evenly spaced parallel planes? Now, a family of parallel planes is made up of the constant sets of a linear function ax + by + cz; and the spacing between two such planes is proportional to the difference between the values of the function on each. We wish to find a, b, c so that the values of the function on the vertices of the unit cube are evenly spaced. These values are 0, a, b, c, a + b, b + c, a+c, and a+b+c. A moment’s thought will show that our condition will be satisfied if, for example, a = 1, b = 2 and c = 4. In other words, the planes defined by x + 2y + 4z = k, for k = 0, . . . , 7, contain each one vertex of the unit cube. Problem 4. This problem and many similar ones can be solved using modular arithmetic, which is the consideration of remainders upon division by various integers. Choosing the best integer by which to divide in order to draw useful conclusions (the modulus) usually involves some trial and error, but where squares are involved, it’s often useful to work modulo 4. The remainder of a perfect square upon division by 4 is either 0 or 1 (see Remark

page 186

YEAR 1996 OLYMPIAD

133

1 after the solution). So sums of two squares can only give 0 + 0, 0 + 1 or 1 + 1 modulo 4; numbers that give a remainder of 3 modulo 4 cannot be sums of two squares. Similarly, the remainders of a square modulo 9 are 0, 1, 4, or 7 (see Remark 1); considering the possible sums of these remainders we find that sums of two squares can only give 0, 1, 2, 4, 5, 7, 8 modulo 9. (Note that 4 + 7 ≡ 2 modulo 9, since 11 = 9 + 2.) Here are two of many possible solutions based on these facts: First solution. The sum n = (36k + 2)2 + 42 , where k is an integer, gives the same as 22 + 42 modulo 4 and also modulo 9, since 4 and 9 both divide 36. Therefore: • n =≡ 0 modulo 4, which means that n−1 ≡ 3 modulo 4. So n − 1

cannot be a sum of two squares.

• n =≡ 0 modulo 9, which means that n+1 ≡ 3 modulo 9. So n + 1

cannot be a sum of two squares.

Since k can take any value and n grows with k if k ≥ 0, we have found infinitely many values of n as desired. Second solution. The sum n = 9k + 1, where k > 0 is an integer, gives 2 modulo 4 and gives 1 modulo 9. Therefore: • n + 1 cannot be the sum of two squares since n+1 ≡ 3 modulo 4. • n − 1 gives 1 modulo 4 and 0 modulo 9, which are both allowed. So this

path doesn’t look promising, but now we look at arithmetic modulo 3. If n − 1 = 9k can be written in the form 9k = a2 + b2 , there are two possibilities: either a and b are both non-multiples of 3, so a2 and b2 each have remainder 1 modulo 3, which disagrees with their sum being a multiple of 3; or a and b are both multiples of 3, and we can divide by 9 to obtain a similar equation, 9k−1 = (a/3)2 +(b/3)2 . In the latter case we just apply the argument again to the new equation, and conclude the proof by reverse induction (see Fact 24). Note that if k − 1 = 0 we cannot apply the argument, but then there is obviously no solution. page 187

Remarks. 1. It is easy to find what remainders are allowed for a square, for any modulus q. For suppose that n gives r modulo q; that is, we have an equation n = kp + r, with k an integer. Then n2 = k 2 q 2 + 2kqr + r2 ; that is, n2 has the same remainder as r2 , modulo q. So all we need to do is look at all values or r and take their squares modulo q. • q = 3: We have 02 = 0, 12 = 1, 22 = 4 ≡ 1 modulo 3. So the possibilities are 0 and 1. The square of any non-multiple of 3 leaves a remainder of 1 modulo 3. • q = 4: We have 02 = 0, 12 = 1, 22 = 4 ≡ 0, 32 = 9 ≡ 1 modulo 4. So the possibilities are again 0 and 1. • q = 9: We can save time by noting that the remainders 5 to 8 can be related to their “complements”; that is, 9 − t is the same (modulo 9) as −t, so (9−t)2 is also the same (modulo 9) as (−t)2 = t2 . Thus we only need

134

SOLUTIONS

to consider the squares of 0, 1, 2, 3, 4. They give 02 = 0, 12 = 1, 22 = 4, 32 ≡ 0 and 42 ≡ 7 modulo 9. • Also useful: the square of an odd number always yields a remainder of 1 when divided by 8. (Check it!) 2. A famous criterion for a positive integer n to be representable as the sum of two squares requires that any prime factor of n of the form 4k + 3 must appear in the prime factorization of n with an even power [§ 4–5]NumbRepresSumSquar. CITE [ See also Remark 2 on page 57.

Problem 5. Denote by O1 and O2 the centers of the circles and by r1 and r2 their radii, respectively. Consider the point D at which a common interior tangent to the circles meets the segment O1 O2 . Then (see Fact 17) r1 DO1 = . DO2 r2 Let AC be a diagonal of the quadrilateral in question and let S be the point at which it meets O1 O2 (see figure). X B A D=S O1

O2 C

By the law of sines, we have SO1 r1 SO2 r2 = , = . (1) sin \O1 AS sin \O1 SA sin \O2 CS sin \O2 SC Consider the circle with center X passing through A and C. It is not hard to see that the lines AO1 and CO2 are tangent to this circle. Expressing the angles between these tangents and the chord AC in terms of intercepted arcs (see Fact 15), we obtain the equation ⌢ ⌢ \O1 AS + \O2 CS = 12 (AC + CA) = π (where the second arc contains the point B in the figure and the first doesn’t); hence sin \O1 AS = sin \O2 CS. The angles O1 SA and O2 SC, being vertical angles, are equal. Therefore, by (1), we have SO1 r1 = . SO2 r2 Since there is only one point dividing a line segment in a given ratio, the points D and S coincide. A similar reasoning is true for the second diagonal.

page 188

YEAR 1996 OLYMPIAD

135

This means that the diagonals of the quadrilateral and the common internal tangents cross the line of centers at the same point, and we’re done. Remark. This problem implies the following remarkable statement: If a quadrilateral ABCD is circumscribed around a circle, the intersection point of the lines joining the opposite points of contact coincides with the intersection point of its diagonals. Indeed, take as X the incenter of the quadrilateral. Let ω1 be the circle centered at A and passing through the points at which the incircle of the quadrilateral touches AB and AD. Denote by ω2 a similar circle with center C. Applying the statement of the problem to the configuration thus obtained, we see that the lines joining the opposite points of contact of the incircle meet on AC and, for similar reasons, on BD at the same time.

Problem 6. First solution. We will choose certain rows of the modified array one by one, and keep track of the sum S = S(m) of the m rows chosen so far. The first row we take is the modified version of the row (1, 1, . . . , 1) of the original array. The string S(1) simply coincides with this row and, obviously, consists of zeros and ones. If it has only zeros, this row alone yields the desired set. Otherwise, we find in the original array the row obtained from S(1) by replacing 1 by −1 and 0 by 1. The corresponding rwo in the modified array row is chosen to be our second string. Then S(2), the sum of S(1) and the chosen row, consists of 0s and 1s again. Suppose that we have already chosen k rows. If the sum S(k) coincides with one of the previous sums S(m), where m < k, then the sum of the k−m rows indexed from m+1 to k is the zero string and we’re done. If the sum S(k) does not coincide with any of the previous sums, then we take up next the row in the modified array corresponding to the row in the original array obtained from S(k) by replacing 1 by −1 and 0 by 1. This row could not have been chosen before, because different sums S(m) specify different rows to be chosen in the array, and the sum S(k) has never occurred before. If at a certain step we obtain a sum that has occurred before, then, as we have already seen, the problem will be solved; if not, we will eventually have exhausted all the rows of the array. That is, we’ll have 2n different sums S(k). Since the number of different strings of 0s and 1s of length n is also equal to 2n , each of these strings will coincide with one of the sums S(k). In particular, for a certain k, the string S(k) will consist of zeros, and the desired set is the set of the first k chosen rows. Second solution. Denote by ai the rows of the initial array and by bi the rows of the modified array, where i = 1, 2, . . . , 2n . We will construct yet a third array, with the rows ci = ai − 2bi . In other words, any two corresponding elements of ci and ai either coincide (if the element of ai was replaced by zero) or are opposite (in all the other places). In particular, the new array consists of ±1. Therefore, for any i there exists a j(i) such that ci = aj(i) . Now let us consider the sequence ik specified by the recurrence relation ik+1 = j(ik ). (The first term of the sequence is chosen at will.) Since this sequence can take only finitely many values, it must contain equal terms.

page 189

136

SOLUTIONS

Suppose that ik = il for certain k and l, k < l, and all the terms of the sequence with numbers less than l are different. Then bik + bik+1 + · · · + bil−1 = 12 (aik − cik ) + 21 (aik+1 − cik+1 ) + · · · + 12 (ail−1 − cil−1 ) = 12 (aik − aik+1 + aik+1 − aik+2 + · · · + ail−1 − ail )

= 12 (aik − ail ) = 0,

which completes the proof.

Year 1997 Olympiad page 190

Level A Problem 1. The numbers of pieces on each row ranges from one to eight. Since these numbers are different for different rows, there is a row with exactly one piece, another with two, and so on. We number the rows according to the number of pieces in them. On row 1, we mark its single piece. On row 2 there are two pieces, so we can choose one that’s not in the same column as the first marked piece. We mark this second piece. Similarly, we can mark at least one of the three pieces on row 3 in a column not already occupied by a marked piece, and so on. See also Fact 1. Problem 2. The round trip along the road and the path takes 16 hours. Therefore, no trouble will come from the first crater if the walk starts right after that crater has finished an eruption. The round trip along the path takes 8 hours. This means that the walk on the path will be safe from the second crater if it starts immediately after that crater has finished an eruption. Therefore the whole trip will be safe if the first crater has just finished an eruption at the time Professor Garibaldi sets out, and the second crater has just settled down 4 hours later, by the time the volcanologist reaches the path. Let us find such a moment. The first crater erupts during the 1st, 19th, 37th hours. The second erupts during the 1st, 11th, 21st, 31st, 41st hours. Therefore, if Professor Garibaldi sets out at the beginning of the 38th hour, reaching the path at the beginning of the 42nd hour, everything will be fine. Remark. We have solved the problem by listing possibilities. However, we could set up a Diophantine equation. The first crater erupts during hours of the form 18x + 1, where x is a positive integer, and the second crater during hours of the form 10y + 1. We want to have a 4-hour interval between eruptions; this leads to the equation 10y − 18x = 4. The smallest positive integer solution to this equation is y = 4, x = 2.

page 191

138

SOLUTIONS

Problem 3. Let L and K be the reflections of M in the lines OX and OY , respectively (see diagram below and to the left). Then K, P , and N lie on the same line, and NK = NP + P K = NP + P M . Indeed, segment MK is perpendicular to the line OY , and if A is their intersection point, then MA = AK (by the definition of reflection). The right triangles MAP and KAP have equal legs; therefore, they are congruent and \KP A = \M P Y = \NP O. In addition, we have P K = P M . X

Y

M

X

Y

M

A N L

Q

K

N

P L

P

K

Q

O

O

Similarly, points N , Q, and L lie on the same line, and NL = N Q+QL = N Q + QM . It remains to prove that NL = NK. We do this by showing that the triangles KON and LON are congruent. Indeed, since K and M are symmetric about the line OY , we have KO = MO (see diagram above and to the right). Likewise, MO = LO, and hence LO = KO. Further, the triangles KON and LON have a common side, and \KON = \KOP + \P ON = \P OM + \P ON = \QON + \P ON = \XOY. One shows similarly that \LON = \XOY . Thus, \KON = \LON, and the triangles KON and LON are congruent by the SAS property, which completes the solution. Remark. Given two points M and N on the same side of a line OY , the shortest path from M to N touching the line is the path MP N considered in our problem. This property, which the ancient Greeks knew and could demonstrate — it appears as a theorem in Heron’s Catoptrics — is one way to explain the fact that a ray of light reflected in a mirror makes the same angle coming in and going out (because of the physical principle that light “likes” to take the shortest path). Or, put in a whimsical way: A fireman at point N has to put out a fire at point M having filled his pail in a river, represented by the straight line page 192 OY . To what point on the river should he go in order to make as short a run as possible? Now the initial problem can be interpreted as follows: if the river is the angle XOY and the angles N QO and M QX are equal, it doesn’t matter to which side of the angle the fireman runs.

YEAR 1997 OLYMPIAD

139

Problem 4. Any number ending in 0 is composite. Therefore if we choose a number ending in 0, we only have to worry about changing the last triple of digits, because changing any other triple will leave it a composite number. Next, if we choose our number N to end in 000, the condition “replacing the last triple by anything gives a composite number” is the same as the condition that N +1, N +2, . . . , N +999 are all composite. Now, if N is a multiple of all numbers from 1 to 999 — for instance, N = 999!, it is clear that N +k is a multiple of k for k = 1, . . . , 999. Thus N +2, . . . , N +999 are all composite. But what to do about N +1? One idea is to make N even bigger, and look at the second hundred numbers after N , instead of the first hundred. For instance, if N = 2000!, then N +1000, N +1001, N +1002, . . . , N +1999 are all composite, and so N +1000 satisfies the conditions of the “Prove that” part of the problem. But 2000! has more than 1997 digits, because it’s greater than 10001000 . To find an N with 1997 digits, we note that many factors are superfluous. We can start by dropping all the factors from 1 to 999. We keep the factor 1000 since we want N to end in 000, but we can drop all other even factors, because, N being even, N +1002, N +1004, . . . , N +1998 are also even and therefore composite! So now N is the product of 501 numbers under 2000, and we see that it has less than 1997 digits from an easy estimate:

Rewritten to give reader a chance to understand how one might come up with such an idea.

2000501 = 2501 · 1000501 < 10501/3 · 10501×3 = 101670 .

(Here the inequality comes from the fact that 23 < 10.) To conclude we extend N with zeros to the right so it acquires exactly 1997 digits, and we finally add 1000 according to the earlier plan; this gives a 1997-digit number satisfying all the required conditions. Remark. You may have encountered the idea in the second paragraph when solving a simpler problem: Prove that there exist 1000 consecutive composite numbers.

Problem 5. First solution. Let the lines DE and AB B A G intersect at G. The triangles DEC and BEG are congruent, by ASA (see figure). Thus BG = E CD = BA, which means that A, G, and C F D C lie on a circle with center B; moreover AG is a diameter of this circle. Since \AF G = 90◦ , the point F must lie on the same circle, by the 90◦ case of the inscribed angle theorem (Fact 14). Again by the inscribed angle theorem, we have \GF C = 12 \GBC = 1 ◦ ◦ ◦ ◦ ◦ 2 (180 − 40 ) = 70 . It follows that \DF C = 180 − \GF C = 110 . Second solution. We can do without the inscribed angle theorem. To see that F lies on the circle, note that BF = BA = BG, because BF is the median of the right triangle AF G, and hence equals half the hypotenuse AG. Now consider the sum of the angles of the quadrilateral ABCF . We have \BCF + \CF A + \F AB + \ABC = 360◦ .

page 193

140

SOLUTIONS

But \BCF + \F AB = \CF A because the triangles CBF and F BA are isosceles. Therefore we get 2\CF A + 40◦ = 360◦ , or \CF A = 160◦ . It follows that \CF D = 360◦ − \AF C − \AF D = 360◦ − 160◦ − 90◦ = 110◦ . Problem 6. First, let’s solve a simpler problem: suppose each coin can be used in only one weighing, rather than two. What is the greatest number of coins such that the expert can be sure of being able to determine the counterfeit coin after k weighings? If two coins are placed on the same pan of the balance in one of the weighings, it will be impossible to identify which of them is counterfeit (assuming one of them is), because they cannot take part in a second weighing. Therefore, each weighing involves one coin on each pan. If the balance is not in equilibrium, the counterfeit coin is obviously the lighter of the two. If it is in equilibrium, the number of questionable coins decreases by two. It follows that k weighings allow us to identify the counterfeit coin from among 2k + 1 coins. Returning to the original problem, let’s denote by f (n) the unknown number of coins. Suppose that in the first weighing, there were s coins on either pan. If the balance did not come to equilibrium, then we must look for the counterfeit coin among s coins each of which can be used only once, and we are permitted to make n − 1 weighings. As we have seen above, s ≤ 2(n − 1) + 1 = 2n − 1. If the balance is in equilibrium, then we obtain the original problem for the f (n)−2s coins that were not weighed the first time, with n−1 weighings allowed; thus, f (n) − 2s ≤ f (n − 1). Hence f (n) ≤ f (n − 1) + 2s ≤ f (n − 1) + 2(2n − 1). It follows that

page 194

f (n) ≤ 2(2n − 1) + 2(2n − 3) + · · · + 2 · 3 + f (1).

It is easy to see that f (1) = 3; therefore, by the formula for the sum of arithmetic progression, we have f (n) ≤ 2n2 + 1. On the other hand, if we have 2n2 + 1 coins and each time we take the largest possible s, that is, s = 2n − 1 at the first step, s = 2n − 3 at the second step, and so on, then the expert will be able to find the counterfeit coin. Thus, f (n) = 2n2 + 1. Level B Problem 1. Let a, b, and c be the lengths of the sides of the triangle, with a = 31 (b + c). Since the smallest angle of a triangle is opposite the shortest side, it suffices to prove that a is the smallest of the side lengths. That is, we want to show that b+c a= c; therefore, 31 (b + c) + b > c, that is, 2b > c, which proves the inequality a < b. page 195

Problem 2. Denote by m1 , m2 , . . . , m9 the weights of the pieces, all in increasing order. Put the pieces of weights m1 , m3 , m5 , and m7 on the left, and those of weights m2 , m4 , m6 , and m8 on the right. Then m 1 + m 3 + m 5 + m 7 < m2 + m 4 + m 6 + m 8 . But if we add the heaviest piece on the left, the balance tips the other way: m 1 + m 3 + m 5 + m 7 + m 9 > m2 + m 4 + m 6 + m 8 . So moving some part of the heaviest piece to the right will achieve equality. Problem 3. The sum of the angles of any hexagon is 720◦ ; this can be shown, for instance, by cutting the hexagon into two quadrilaterals. Hence \A + \B + \C = \A1 + \B1 + \C1 = 360◦ .

(1)

The area of the hexagon is the sum of the areas of triangles AB1 C, BCA1 , AC1 B, and ABC (see figure). Therefore we need to show that the area of △ABC is equal to the sum of the areas of triangles AB1 C, BCA1 , and AC1 B. A′ It turns out that these three triangles can be put B together to form a copy of triangle ABC. To see A1 this, we first turn the triangle AB1 C about point C C1 so that its vertex B1 goes to point A1 ; this is possible since CA1 = CB1 . Suppose that the point A is taken under this rotation to A′ . We have \A′A1 B = 360◦ − \A′A1 C − \BA1 C = A C ◦ 360 −\CB1 A−\CA1 B = \AC1 B (we have used (1) and the fact that rotations preserve angles). Now it is clear B1 that triangles A′A1 B and AC1 B are congruent by the SAS property. It remains to prove that triangles A′ BC and ABC are congruent. It follows from the congruence of triangles A′A1 B and AC1 B that A′B = AB. So triangles A′BC and ABC are congruent by the SSS property. Problem 4. For simplicity, we think of the trains and stations as points (or we can consider corresponding points in each train and each station). Clearly, if n is divisible by 3, Ira and Lyosha always leave simultaneously, so there are no underground portions. Suppose that n is not divisible by 3. Then, no matter when they come to their stations, either Ira or Lyosha will take the train first. Denote by l the length of the interval between trains. If a certain point X is underground, any point Y whose distance from X is a multiple of l is

Translation said “see Fact “ref{fact:G5}”, but no such label exists. Original has no such reference.

page 196

142

SOLUTIONS

underground as well. Indeed, if Ira enters the station when Roma is at X, then Ira will catch a train first. But when Roma is at Y , the arrangement of trains is the same as when Roma is at X, so in this case Ira will also be the first to catch a train, hence point Y is also underground. Thus, the underground portions occur in a periodical pattern of period l, and it’s enough to study the situation within one interval of length l. (See figure; “hairy” portions of the line lie underground.) x Consider the moment when a train leaves station B. A Suppose the train Ira will take — the nearest train upline C from A — is x units away from A at this moment. Then the entire interval between this train and point A lies underground. Indeed, if the train Roma is conducting is in this interval, it will pick up Ira first, because Lyosha B has just missed a train. Let’s show that the interval of length l − x just down the line from A is not underground. Indeed, when a train arrives at station A, the nearest train upline from B will be at a distance l − x from B. It follows that if Roma’s train is in the interval of length l − x mentioned above, then Lyosha will get on a train first, because Ira has just missed a train. Thus, in an interval of length l, a portion of length x lies underground and the rest, of length l − x, above ground. Since the pattern is periodic, the ratio of underground to above-ground portions along the entire road is the same, x to l − x. We still have to find x. Clearly, the value of x is equal to the remainder of the division of the distance BA by l. If D is the total length of the track, the distance BA is 32 D and l is equal to D/n, so x = {2n/3} D, where the braces denote the fractional part of a number. Thus, if n yields the remainder 1 upon division by 3, then x = 2D/3, and if the remainder is 2, then x = D/3. Problem 5. We divide the participants of the tournament into two groups: those whose second score was higher than the first and those whose first score was higher. At least one of these groups has n players or more. We can assume, for instance, that the first group does so, and that it consists of x players, where x ≥ n. Suppose that the total score of the first group in the second tournament is D points more than in the first. Then, by assumption, D ≥ xn.

(1)

These x players increased their score at the expense of the other 2n − x players, since matches inside one group always contribute the same total number of points, 21 x(x − 1), to the intra-group total score. Each match with a player from the other group can add at most one point to the total score; hence D ≤ x(2n − x).

(2)

page 197

YEAR 1997 OLYMPIAD

143

Comparing inequalities (1) and (2), we obtain 2n − x ≥ n. If inequality (1) and/or (2) were strict, we would have 2n−x > n, contrary to the assumption x ≥ n. Therefore, x = n and D = n · n, and so the players from the first group increased their personal scores by exactly n points each. The same argument applies to the second group, because it also consists of n players.

page 198

Remark. It follows from the solution that the situation in the problem can only occur if the participants fall into into two groups of n players each, such that everyone in the first group beat everyone in the second group in the first tournament, and the second group beat everyone in the first group in the second tournament.

Problem 6. Suppose that F (x) = a0 + a1 x + a2 x2 + · · ·

By assumption,

and G(x) = b0 + b1 x + b2 x2 + · · · .

(a0 + a1 x + a2 x2 + · · · )(b0 + b1 x + b2 x2 + · · · ) = 1 + x + x2 + · · · + xn .

The constant term of a product of polynomials is equal to the product of their constant terms, so a0 ·b0 = 1. Hence a0 = 1 and b0 = 1. The coefficient of the first power of x in the product is a0 b1 +a1 b0 . Since it equals 1, we have either a1 = 1, b1 = 0 or a1 = 0, b1 = 1. Assume without loss of generality that a1 = 1 and b1 = 0. If all the coefficients ai of F (x) are equal to 1, the statement of the problem is true. If at least one of them is zero, there is some m for which a0 = a1 = · · · = am−1 = 1,

am = 0.

We want to prove that the polynomial F (x) can be represented in the form (1 + x + x2 + · · · + xm−1 )T (x), where T is also a 0-1 polynomial. If some coefficient bl is equal to one, where 1 ≤ l < m, we immediately see that the coefficient of xl greater than 1, because the term xl of the product can be obtained by multiplying both a0 by bl xl or al by b0 xl . Hence bl = 0 for 1 ≤ l < m. Now it is clear that bm = 1: otherwise the coefficient of xm would be zero. The sequence of coefficients a0 , a1 , . . . can be viewed as a sequence of alternating intervals: first a interval of ones, then a interval of zeros, then again ones and so on. Consider one of these intervals: ar = · · · = ar+s−1 = 1 with ar−1 = 0, ar+s = 0. The length s of this interval cannot be greater than m: otherwise the term xr+m of the product could be obtained both by multiplying ar xr by bm xm and by multiplying ar+m xr+m by b0 . We prove that the length of such a interval cannot be less than m. Suppose that this is not true and consider the leftmost interval ar , . . . , ar+s−1 of ones surrounded by zeros whose length s is less than m. Let’s examine the computation of the term xr+s in the product F (x) · G(x). Since ar+s = 0, it must appear as the product of some term of the form au xu and some term bv xv . (Of course, u + v = r + s.)

page 199

144

SOLUTIONS

It is not difficult to see that if u 6= 0, then au−1 = 0; otherwise we could obtain xr+s−1 in two ways: au−1 xu−1 · bv xv = ar+s−1 xr+s−1 · 1. The case u = 0 is similar. Thus there must be a interval of ones of the form au , . . . , au+m−1 . (Its length is m, because it lies to the left of the interval ar , . . . , ar+s−1 .) But the number r + m − v = r + m − (r + s − u) = u + (m − s) lies on the interval u, . . . , u + m − 1, whence ar+m−v = 1, and we arrive at a contradiction: ar xr · bm xm = ar+m−v xr+m−v · bv xv . Consequently, all intervals of ones have the same length m. Long division (see Fact 19) convinces us that the polynomial F can be represented as the product of the polynomial 1 + x + · · · + xm−1 by a polynomial all of whose coefficients are zeros or ones. Remark. It is easier to understand this solution — and especially to come up with it — by drawing intervals on a line and studying what happens when they are shifted. Geometrically speaking, the problem can be formulated as follows: Suppose a interval of length L can be covered by translates of a certain union S of disjoint little intervals contained in L (so each point of the interval is covered by one of the little intervals and these little intervals do not overlap, but only touch one another at their endpoints). Then all the intervals in S have the same length l, and, of course, the quotient L/l is an integer. See how simple and beautiful the solution becomes when translated into geometric language!

Level C Problem 1. First solution. We take cartesian coordinates in space and call α, β, and γ the coordinate planes, with equations x = 0, y = 0, and y = 0, respectively. Consider the ball B defined by

page 200

x2 + y 2 + z 2 ≤ 1. Its projection on the plane α is a disk of radius 1 with center at the origin. The set of points whose projections hit this disk is a cylinder, denoted by C1 ; it is defined by the inequality y 2 + z 2 ≤ 1. In a similar way we define the cylinders C2 and C3 as the sets of points whose projections onto the planes β and γ fall inside the respective unit disks centered at the origin. Let C be the intersection of C1 , C2 , and C3 . We will show that C satisfies the conditions of the problem. We know that C is convex, being the intersection of convex sets. Let’s show that the projections of C on the planes α, β, and γ are disks. For instance, consider the plane α. The projection of C on α lies in the unit disk of α, by construction. But the unit disk in α is itself contained in C, because it projects onto the two other planes inside the respective unit

Original said x2 + y 2 ≤ 1, in conflict with the definition of α.

YEAR 1997 OLYMPIAD

145

disks (the projections are diameters). Thus the projection of the solid C on α both contains and is contained in the unit disk; therefore, they coincide. Next, C is not a ball. Indeed, the point (a, a, a) lies in C1 , C2 , and C3 if 2a2 < 1, and it lies outside of the unit ball B centered at the origin if 3a2 > 1; clearly there are values of a satisfying both conditions. One last thing needs to be checked, though: could C be some other ball? The answer is no, because a ball having a different size or center than the unit ball B centered at the origin would have a different projection than B (and hence than C) in at least two coordinate planes.

page 201

Second solution. Consider a ball and its projections on the three planes, which are circles. Take any point A on the sphere (the boundary of the ball) that does not project to the circumference of any of the three circles; in effect A can be any point of the sphere not lying on a coordinate plane. Now take any plane intersecting the ball but so close to A that the intersection is entirely contained in the same octant of space as A. (For instance, if the three coordinates of A are positive, the octant it defines is the set of points whose three coordinates are positive.) If we lop off the little spherical cap on one side of this plane, what’s left of the ball is no longer a ball, but it has the same projections on the coordinate planes as the ball! This set is convex, being the intersection of a ball and a half-space. A

B Problem 2. Translate the quadrilateral ABCD by −→ the vector AC (see figure). The image is a quadrilatB0 A0 eral A′B ′ C ′D′ , where A′ = C, and the quadrilateral B′ BB ′D′D is a parallelogram, because the segments D C BD and B ′D′ are parallel and congruent. Let A0 , D0 C0 B0 , C0 , and D0 be the midpoints of segments BD, ′ ′ ′ ′ ′ BB , B D , and D D, respectively. D C′ We claim that A0 B0 C0 D0 is a parallelogram whose diagonals form the same angle and have the same lengths as the diagonals of the quadrilateral ABCD. The statement about the lengths follows from the fact that the segments A0 B0 and C0 D0 are the midlines of triangles B ′BD and B ′D′D, respectively. The statement about the angles is true because the segments B0 D0 and BD are parallel and congruent, and so are A0 C0 and AC. Therefore, it remains to prove that the perimeter of quadrilateral ABCD is not less than the perimeter of parallelogram A0 B0 C0 D0 . But the perimeter of the parallelogram is equal to B ′D + BD′ (by the property of triangle’s midlines). By the triangle inequality, BC + CD′ ≥ BD′ and B ′C + CD ≥ B ′D. The proof is completed by adding these inequalities.

Problem 3. We go directly to part (b), since (a) is a particular case of it. Obviously, our construction applied to a regular polygon A1 A2 . . . An yields a regular polygon B1 B2 . . . Bn . (To justify this rigorously, notice that the polygon A1 A2 . . . An maps to itself under a 2π/n rotation; hence the polygon B1 B2 . . . Bn also maps to itself under the same rotation.)

page 202

146

SOLUTIONS

Since all regular n-gons are similar, any of them can be obtained by the extension procedure from some other regular n-gon. It remains to prove that the polygon A1 A2 . . . An is uniquely determined by the polygon B1 B2 . . . Bn . First solution. We prove by induction (see Fact 24) a more general statement: Let A1 A2 . . . An be a polygon. Let B1 be the point on the ray A1 A2 beyond the point A2 such that A1 B1 /A1 A2 = α1 ; let B2 be the point on the ray A2 A3 beyond the point A3 such that A2 B2 /A2 A3 = α2 ; and so on. Then the points A1 A2 . . . An are uniquely determined by the polygon B1 B2 . . . Bn and the coefficients α1 , α2 , . . . , αn . Base of the induction (n = 3). Let B ′ be the intersection point of the line A1 A2 with the segment B2 B3 (see figure below and to the left). We can find the ratio B2 B ′ /B ′ B3 using Menelaus’ theorem (see remark below), because we know the ratios A2 B2 /A2 A3 and A3 B3 /A3 A1 . Therefore, the point B ′ , and hence the line A1 A2 , are uniquely determined. Similarly, we can determine the lines A2 A3 and A3 A1 ; therefore the triangle A1 A2 A3 is unique. B1 B′

A2

B3

B′

A3

B2

A2 Bn

A3

A1

B1

A1 B2

Bn−1

B3 An

Induction step. Let the line An A2 intersect the segment B1 Bn at B ′ (see figure above and to the right). By Menelaus’ theorem, ifwe know the ratios An Bn /An A1 and A1 B1 /A1 A2 , we can find the ratio B1 B ′ /B ′ Bn . This determines B ′ uniquely. Another application of Menelaus’ theorem allows us to find the ratio An B ′ /An A2 . The induction conjecture applied to the polygons A2 A3 . . . An and B ′ B2 . . . Bn−1 shows that the points A2 , A3 , . . . , An are determined uniquely. Now it is not difficult to determine the point A1 . Remark. Menelaus’ Theorem. Let A′ , B ′ , C ′ be points on the sides BC, AC, AB of a triangle ABC or on their extensions. The points A′ , B ′ and C ′ are collinear if and only if BA′ CB ′ AC ′ · · =1 (1) CA′ AB ′ BC ′ and either all the points lie on the side extensions or only one of them does. We prove the implication we actually used above: collinearity implies (1). Suppose that the points A′ , B ′ , and C ′ lie on the same line. Projection to a perpendicular line maps these points to the same point. Denote this point by P and the projections of A, B, and C by A0 , B0 , and C0 . Then BA′ CB ′ AC ′ B0 P C0 P A0 P · · = · · = 1. ′ ′ ′ CA AB BC C0 P A0 P B0 P

page 203

YEAR 1997 OLYMPIAD

147

Second solution. Suppose the polygons B1 B2 . . . Bn and A1 A2 . . . An are related as in the problem. Then A2 is the midpoint of A1 and B1 , a fact that we can express by the equation A2 = 21 A1 + 12 B1 . Similarly we have A3 = 12 A2 + 21 B2 = 14 A1 + 14 B1 + 21 B2 . We interpret these sums as “weighted averages” of points on the plane, which make sense as long as the weights add up to 1. (Such averages are also called affine combinations.) Continuing in this way we eventually get An = 21 An−1 + 12 Bn−1 = 41 An−2 + 14 Bn−2 + 21 Bn−1 = · · · 1 1 1 1 1 = n−1 A1 + n−1 B1 + n−2 B2 + · · · + Bn−2 + Bn−1 2 4 2 2 2 and finally, since A1 is also the midopint of An and Bn , 1 1 1 1 1 1 1 A1 = An + Bn = · · · = n A1 + n B1 + n−1 B2 + · · · + Bn−1 + Bn . 2 2 2 4 2 2 2 In this last equation we have A1 on one side and a weighted average of A1 , B1 , . . . , Bn on the other. This implies that A1 is a weighted average of B1 , . . . , Bn alone, with known coefficients. (Formally we can treat the equation A1 = 2−n A1 +2−n B1 +· · ·+ 21 Bn as if A1 and the Bi were numbers, subtracting 2−n A1 from both sides and then multiplying by the right number to adjust the weights back to 1.) And since now A1 is uniquely determined by the Bi , so are A2 , A3 , . . . , An , as we wished to show. Remark. A slick way of phrasing the same argument in a different language is this: Place weights of masses 1, 2, . . . , 2n at the points B1 , B2 , . . . , Bn . Then A1 is the (uniquely determined) center of mass of this system, for the following reason: page 204 Let’s modify the system by placing an additional weight of mass 1 at A1 . Now the center of mass of the weights at A1 and B1 is at A2 , so we can replace the unit weights at A1 and B1 by a double weight at A2 . Next we replace the weights at A2 and B2 by a weight of mass 4 at A3 . We continue in this way until we end up with a single weight, located at A1 , which is therefore the center of mass of the modify system. Going back to the original system involves subtracting the auxiliary weight at A1 ; but adding or removing weights at the center of mass cannot change the center of mass.

Third solution. The homothety (scaling) with center B1 and scale factor 1 2 takes the point A1 to A2 . Similarly, the homothety with center B2 and factor 21 takes A2 to A3 , and so on. Finally, the homothety with center Bn and factor 12 takes An to A1 . Thus the composition of these n homotheties fixes the point A1 . But a composition of homotheties is also a homothety, and a homothety leaves exactly one point fixed (unless it fixes every point; but this cannot be the case because our composition has scale factor 12 · 21 · 12 · · · = 21n ). Hence the point A1 is defined uniquely. Problem 4. The expressions in the problem occur in the relations between a cubic polynomial and its roots. Let’s recall these relations, which are

148

SOLUTIONS

the degree-three case of Vieta’s formulas (see also Fact 20). A polynomial P (x) = (x−a1 )(x−a2 )(x−a3 ) becomes, upon expanding the product, P (x) = x3 − (a1 +a2 +a3 )x2 + (a1 a2 +a1 a3 +a2 a3 )x − a1 a2 a3 .

A similar expression can be written for Q(x) = (x−b1 )(x−b2 )(x−b3 ). The equalities in the problem mean that these polynomials differ only in the constant term. Therefore, the graph of one polynomial is obtained from the graph of the other by a vertical translation. For x ≤ b1 , we have Q(x) ≤ 0, because each of the three factors in the expression for Q(x) is negative or zero, and there are an odd number of factors. Thus, Q(a1 ) ≤ 0 and P (a1 ) = 0. Therefore the graph of y = Q(x) is obtained from the graph of y = P (x) by a downward translation, or the two coincide. In particular, Q(a3 ) ≤ P (a3 ) = 0. But for x > b3 we have Q(x) > 0. Consequently, a3 ≤ b3 .

Problem 5. Suppose that the scores of competitors are not the same. Let M be the highest score and m the lowest score. We call players whose score is M leaders, and those whose score is m trailers. The superscore of a leader is at least M m, being the sum of M numbers, each of which is at least m. For an analogous reason, the superscore of a trailer is at most M m. Therefore, if the superscores are all the same, leaders and trailers all have a superscore of M m. Several stunning conclusions follow: • All the wins of a leader were against a trailer. Otherwise, the leader’s superscore would be strictly more than M m. • There is only one leader. For if there were more, one would have beat the other, contradicting the previous item. • All the wins of a trailer were against a leader. Otherwise, the trailer’s superscore would be strictly less than M m. • There is only one trailer. Otherwise, contradiction with the previous item. • M = 1 and m = 0, since there was only one match between a leader and a trailer, and only such matches can contribute to M and m, as a consequence of the “All the wins” statements. • There were only two players in the tournament. For a third player would have beat both the leader and the trailer (again because of the “All the wins” statements). Therefore there would be a score greater than M , contrary to assumption. Hence, if there are more than two players and the superscores are all the same, the scores must also be all the same. Remark. For the solution to another problem involving tournament scores, see Problem 5 on page 142.

Problem 6. It is enough to show that any initial segment of the sequence of first digits of powers of 5 occurs in reverse in the sequence of first digits of powers of 2.

page 205

YEAR 1997 OLYMPIAD

149

This would be very easy if the sequence of powers of 2 were extended to the left with the first nonzero digit of each negative power. That’s because the first nonzero digit of 1/2n is the same as the leftmost digit of 5n . So the problem is to find among positive powers of 2 the same sequence of starting digits as among negative powers of 2. If some power of 2 were equal to some power of 10 this would be easy, because the sequence of first digits would be periodic (if 2i = 10j with i, j > 0, the decimal representation of 2n+i would be the same as that of 2n followed by j zeros). This is too much to hope for (why?), but it points us in the right direction: Since we only need to find matches for sequences of finite length, it suffices to show that for any k there exists a power of two, x = 2n , whose decimal representation is of the form 100 . . . 0}∗, | {z

(1)

k zeros

where ∗ stands for an arbitrary strings of digits. If this is so, we will have 2n−1 = 2n/2 = 500 . . . 0∗, then 2n−2 = 250 . . . 0∗, then 2n−3 = 1250 . . . 0∗, etc. To be more precise, write x = 10N + y, with y < 10N −k ; then 2n−l =

page 206

x 10N + y y = = 10N −l 5 l + l ; l l 2 2 2

so the first digit of 2n−l coincides with the first digit of 5 l for l < k. The existence of a power of two of the form (1) can be proved using a standard fact about irrational numbers (see Remark 1 below). But we give here an alternate, explicit construction. Since there are finitely many k+1-tuples of digits and infinitely many powers of 2, we can certainly find two powers of 2 sharing the same first k + 1 digits, say 2a and 2b , with a > b. We take the quotient 2a−b and apply the following observation: Lemma. Suppose A and B, with A > B, are positive integers that have the same first k+1 digits. Consider the first place, counting from the left, where they have different digits. If A has the larger digit there, or if B only has k+1 digits, the fraction A/B has an infinite decimal representation beginning with a one and k zeros. If B has the larger digit, the decimal representation of A/B begins with k nines. Proof. Denote by D the number formed by the first k +1 digits of A and B. Suppose that A is a p-digit number and B is a q-digit number, with p ≥ q. Then A = 10p−k−1 D + α and B = 10q−k−1 D + β, where α < 10p−k−1 , β < 10q−k−1 . We have 10p−k−1 D + α |α − 10p−q β| A p−q p−q − 10 = q−k−1 − 10 = q−k−1 B 10 D+β 10 D+β
b, has to be an integer, and it cannot be a power of 10 (and if it could, this would only help, as already seen). So the k zeros or nines are all in the integer decimal representation of 2a−b . So far we have shown that for a given k we can find a power of 2 starting either with a 1 and k zeros or with k nines. In the first case, the problem is solved. In the second case, let N be the number of digits of a power of 2 starting with k nines; note that N > k, since a power of 2 cannot end in 9. Applying the italicized statement with N instead of k we obtain a new power of 2 that either begins with a 1 and N zeros (and then the problem is solved) or begins with N nines, which is more nines than the earlier power of two. In the latter case, by the lemma, the quotient of this power of 2 by the earlier one is a power of 2 beginning with a 1 and k−1 zeros. So again in this case the problem is solved (for k−1 rather than k, but this does not matter since we could just start with a larger k). Remarks. 1. Finding a power of 2 of the form (1) is the same as finding i, j > 0 such that 0 ≤ 2i − 10j < 10j−k , or equivalently 1≤

1 2i < 1+ k. j 10 10

(3)

Taking base-10 logarithms, these inequalities become 0 ≤ i log10 2 − j < ε, where we denote by ε the number log10 (1 + 10−k ). Finally, this inequality can be rewritten in the form {i log10 2} < ε, where the braces denote the fractional part of a number. page 208 So we must look at the sequence of fractional parts ai = {i log10 2}. It is convenient to think of the fractional part of a number as a point on a circle of unit length. (This circle can also be seen as the interval [0, 1] with endpoints identified. See the solution to Problem 4 on page 69 for a similar situation.) On the circle, each point ai is separated from the next one, ai+1 , by an arc of the same length α, always in the same direction. Here α = log10 2, but the discussion below holds for any α. If α is a rational number, written in lowest terms as p/q, the points ai form a periodic sequence of period q. (Why?) On the circle they occupy the vertices of a regular q-gon. We already know that this cannot happen in our case, since the ≤ in (3) cannot be an equality. Thus log10 2 is irrational.

YEAR 1997 OLYMPIAD

151

Now, for α irrational it is an easily proved theorem (see below) that the sequence ai = {iα} will eventually land arbitrarily close to any point on the circle. In particular, for α = log10 2, there is some i such that ai = {iα} < ε.

2. Theorem (Density of irrational orbits). If α > 0 is irrational, the points ai = {iα}, for i ≥ 1, form an everywhere dense sequence in the circle of unit length or in the interval [0, 1]. That is, given a point x ∈ [0, 1] and a positive number ε, no matter how small, there exists some i such that |iα − x| ≤ ε. Proof. If α is irrational, all the ai are distinct. (Why?) Take an integer N with 1/N < ε, and divide the circle into intervals [0, 1/N ), [1/N, 2/N ), . . . , [(N −1)/N, 1). Since the ai are all distinct and there are infinitely many of them, we can find two in the same interval, say am and an , with n > m. Set k = n − m. Assume first that an > am . Then ak = {kα} = {nα − mα} = {an − am } = an − am < 1/N . From the sequence of numbers 0, ak , 2ak , . . . , take the largest that does not exceed x, say jak . Then x − jak < (j+1)ak − jak = 1/N < ε. Thus we have found a point in the sequence ai that’s within the required distance from x. (Note that jak = ajk since jak ≤ x < 1.) The case an < am can be turned into the previous case by replacing α by 1 − α and x by {1 − x}; this replaces every ai by 1 − ai . (Why?) ˜

I’ve turned the “proof sketch” into a proof since it does not take that much space and it’s a good example of rigorous argumentation with fractional parts.

Level D Problem 1. One formula for the area of a triangle ABC ia AB·BC·CA/4R, where R is the circumradius. So what we must show is that the ratio of the areas of triangles A′ B ′ C ′ and ABC is equal to AB ′ · BC ′ · CA′ + AC ′ · CB ′ · BA′ . AB · BC · CA Define x = AB ′/AB, y = BC ′/BC, z = CA′/CA. Then it is easy to compute the ratios of the areas of the triangles AB ′C ′ , A′ BC ′ , and A′B ′ C to the area of the triangle ABC. They are equal to x(1 − y), y(1 − z), and z(1 − x). Therefore, the solution reduces to checking the identity 1 − x(1 − y) − y(1 − z) − z(1 − x) = xyz + (1 − x)(1 − y)(1 − z). Multiply out and see! Problem 2. We have Z π/2 Z 2 2 cos (cos x) + sin (sin x) dx = 0

π/2

2

cos (cos x) dx + 0

Z

π/2

sin2 (sin x) dx.

0

π Perform the substitution (see Fact 28) y = − x in the second integral; 2 then dy = −dx, and Z π/2 Z π/2 Z 0    π 2 2 − y (−dy) = sin2 (cos y) dy. sin (sin x) dx = sin sin 2 0 0 π/2

page 209

152

SOLUTIONS

Since cos2 (cos x) + sin2 (cos x) = 1, we have Z π/2 Z π/2 Z cos2 (cos x) + sin2 (sin x) dx = cos2 (cos x) dx + 0

=

Z

0

π/2

dx = 0

π/2

sin2 (cos x) dx

0

π . 2

Problem 3. Notice that f2 (x) − f3 (x) = 2x − 1; we can add 1 and multiply the expression 2x thus obtained by f rac12. This yields the function x. Subtracting it from f1 (x), we obtain 1 = f1 (x) − 21 (f2 (x) − f3 (x) + 1). x We turn to the proofs of impossibility. Since we are not allowed to divide, it is impossible to express 1/x only in terms of the functions f2 and f3 : we cannot obtain x in the denominator. In other words, the addition and multiplication of polynomials yields only polynomials, whereas the function 1/x is not a polynomial. The proof that we cannot do without the function f2 is more interesting. Let’s compute the derivatives of f1 and f3 at the point x = 1. Both these derivatives are equal to zero. If two functions have zero derivatives at the point 1, then the derivatives of both their sum and product is also zero. The latter follows from the computation (f g)′ (1) = f ′ (1)g(1) + f (1)g ′ (1) = 0. Thus, all the functions that can be obtained by our operations from the functions f1 and f3 have zero derivatives at point 1. And the derivative of 1/x at this point is nonzero. Only one Olympiad competitor managed to solve this part of the problem. He considered the series expansions of f1 and f3 in terms of x−1 and pointed out that both expansions start with (x − 1)2 , while that of 1/x has a linear term. This is equivalent to the argument above. It remains to prove that we cannot do without the function f3 . We will give three proofs. First solution. Any function obtained from f1 and f2 will clearly be of the form f (x)/xn , where n ≥ 0 and f (x) is a polynomial. Such functions are called Laurent polynomials. (Warning: a Laurent polynomials may not be a polynomial!) By multiplying top and bottom by x if necessary, any Laurent polynomial can be written in the form f (x)/xn with n even. Let A be the set of Laurent polynomials f (x)/x2k such that the polynomial f (x) − f (−x) is divisible by x2 + 1. We know that f1 ∈ A, because f1 (x) = f (x)/x2 , where f (x) = x(x2 +1), and so f (x) − f (−x) = 2x(x2 + 1) is divisible by x2 + 1. It is even easier to check that f2 ∈ A: in this case f (x) = x and f (x)−f (−x) = 0. On the other

page 210

YEAR 1997 OLYMPIAD

153

hand, 1/x is not in A, because for f (x) = x the function f (x) − f (−x) = 2x is not divisible by x2 + 1. It remains to show that the sum and the product of Laurent polynomials in A belong to A as well. The statement about the sum is left as an exercise to the reader. Consider two Laurent polynomials f (x)/x2k and g(x)/x2l in the set A. Then f (x)g(x) f (x) g(x) · 2l = 2(k+l) , 2k x x x

and so f (x)g(x)−f (−x)g(−x) = f (x) g(x)−g(−x) +g(−x) f (x)−f (−x) is divisible by x2 + 1. Remark. An alert reader may have noticed that the definition of A is a bit fuzzy, in that membership in A might depend on how an element is expressed. That is, could we have f1 (x)/x2k1 = f2 (x)/x2k2 , yet f1 (x)−f1 (−x) is divisible by x2 +1, while f2 (x) − f2 (−x) is not? The answer is no, because f1 and f2 are obtained from one another only by multiplication by a power of x, and therefore the same is true about f1 (x) − f1 (−x) and f2 (x) − f2 (−x). But the factorization of a polynomial is unique (see Fact 21), so xn f (x) is divisible by x2 + 1 if and only if f (x) is.

Second solution. (Uses complex numbers — see Remark 1 below.) The functions f1 , f2 and 1/x are defined for all Let’s look √ nonzero complex numbers. 1 at their values at the point x = i = −1. We have f1 (i) = i + i = i − i = 0 and f2 (i) = i2 = −1, which are real numbers. Therefore, the value of any function obtained from f1 and f2 at the point i will be real. But the value of 1/x at the same point is 1/i = −i, which cannot be expressed in terms of real numbers via multiplication and addition. Remarks. 1. Complex numbers are extremely important in mathematics, physics and most other sciences. They are a generalization of the ordinary (“real”) numbers, just as real numbers are a generalization of the integers. To work with complex numbers √ we include among our numbers one whose square is −1. It is denoted i, or −1. Then, for addition and multiplication to make sense, we need to include also all the numbers of the form a + bi, where a and b are real. (For b = 0 this formula yields ordinary real numbers a, showing that real numbers are a particular case of complex numbers.) Complex numbers can be added, subtracted, multiplied, and divided (if the divisor is not zero). These operations satisfy the same properties — commutativity, associativity, distributivity — as the corresponding ones on real numbers. The sum of two complex numbers a + bi and c + di is the complex number (a+c) + (b+d)i. The product can be found using the properties listed above, including the defining property i2 = −1: (a+bi)(c+di) = ac + adi + bci + bdi2 = (ac−bd) + (ad+bc)i.

Every polynomial has a root among the complex numbers. This is known as the fundamental theorem of algebra. Complex numbers have many other beautiful properties and applications. Googling the words “complex numbers introduction” brings up several good, well-explained sites.

page 211

154

SOLUTIONS

2. Our second solution works if we interpret “arbitrary number” in the statement of the problem to mean “arbitrary real number”. But if we’re using complex numbers, it’s natural to ask: what if we allow multiplication by, and addition of, complex numbers? The answer is still that 1/x cannot be expressed in terms of f1 and f2 . Here’s one reason: f1 (i) = f1 (−i) and f2 (i) = f2 (−i). Therefore, any function obtained from f1 and f2 takes the same value at the points i and −i; but this is not true of the function 1/x. 3. There is a deep connection between this problem and the vast and beautiful area of mathematics known as algebraic geometry. We will say some words about it, even though it goes far beyond the high-school curriculum. Consider all the functions that can be obtained from the single function x by the allowed operations, including multiplication by and addition of a complex number. We can get only polynomials, and we get all polynomials in this way. They form a set denoted by C[x], where C stands for the complex numbers. This set is a ring — more exactly, a commutative and associative ring with identity — since it has an addition and a multiplication operation with these properties. If we start from the function x2 instead of x, we obtain the ring C[x2 ] of polynomials with even-degree terms only. This is a subring of C[x], and it is distinct from C[x] since x (for example) is in C[x] but not in C[x2 ]. Next, consider all functions that can be obtained from x and 1/x by the allowed operations. We get another ring, the ring of all Laurent polynomials; we denote it by C[x, x−1 ]. Any function in this ring is nicely behaved in C∗ , the complex plane minus the origin, just as any polynomial is nicely behaved in all of C. Note that C[x] is a subring of C[x, x−1 ], and the two do not coincide. More generally, take some finite number of Laurent polynomials, f1 , . . . , fn , to serve as seeds. The functions that can be obtained from these generators (seeds) by the allowed operations form a subring of the ring of all Laurent polynomials; this subring — let’s call it A — may or may not be all of C[x, x−1 ]. It can be shown that A is the ring of nicely behaved functions (algebraic functions) on some space X called an algebraic curve. The full ring of Laurent polynomials is the ring of algebraic functions on C∗ = C \ 0. The inclusion A ⊂ C[x, x−1 ] corresponds to a algebraic map (or morphism) of curves φ : C∗ → X. The ring A equals all of C[x, x−1 ] if and only if this map is an isomorphism, that is, it has an inverse and the inverse is also an algebraic map. There are three reasons why φ might not be an isomorphism: φ is not surjective, φ is not injective, or the derivative of φ at certain points is 0. If A = C[x] is the ring of functions generated by f2 and f3 in the problem, the first reason takes place: A is the set of algebraic functions on the space X = C, and φ : C∗ → C is the inclusion, which is not surjective because 0 is not in its image. If we start from f1 and f2 , the curve X can be taken as the set of (y, z) ∈ C2 such that y 2 z−(1+z)2 = 0, and the map φ takes x ∈ C∗ to the pair (y, z) defined by y = x + 1/x and z = x2 — that is, y = f1 (x) and z = f2 (x). (Note that the polynomial defining X vanishes identically when these values of y and z are substituted.) Since the points i and −i in C∗ have the same image in the curve X (check it out!), the map φ is not surjective and so is not an isomorphism. We say that X is a singular curve with an ordinary double point.

I expanded this a bit to give the reader some chance to understand what’s going on.

page 212

YEAR 1997 OLYMPIAD

155

Finally, suppose we start from f1 and f3 ; this time φ takes x ∈ C∗ to the pair (y, z) ∈ C2 defined by y = x + 1/x and z = (x − 1)2 . (The curve X has a complicated equation, y 2 z − (y + z)2 + 4y − 4 = 0.) The map φ is not an isomorphism because its derivative vanishes at the point 1. We say the algebraic curve X has a cusp, or pinch point.

Problem 4. Cutting the regular tetrahedron along the planes parallel to its faces and passing through the midpoints of its edges, we partition it into a regular octahedron and four regular tetrahedrons. Each small tetrahedron is obtained from the big one by a homothety with scale factor 21 and center at a vertex of the big tetrahedron (see figure on the left).

It is a little harder to visualize how to cut a regular octahedron. As in the case of the tetrahedron, let us shrink the octahedron in half fixing one of its vertices. Repeating this for all six vertices, we obtain six small octahedrons. After they are carved out of the big octahedron, we find that the remainder consists of eight regular tetrahedrons adjoining the faces of the big octahedron. One vertex of each of these tetrahedrons is the center of the initial octahedron, and all the others are midpoints of the octahedron’s edges (see figure above and to the right). After the first step (at which only the tetrahedron is sliced) we obtain an octahedron and tetrahedrons with edges of length 12 ; after the second step (at which four tetrahedrons and one octahedron are cut) we obtain octahedrons and tetrahedrons with edges of length 41 , and so on. After the seventh step the edge lengths of tetrahedrons and octahedrons has become 1 1 128 , which is less than 100 . page 213

√ Problem 5. Set a = x3 , b = y 3 , c = z 3 . Then xyz = 3 abc = 1. For any x and y we have the inequality x2 − xy + y 2 ≥ xy (expand the square (x−y)2 ). Multiplying both sides of this inequality by x + y, which is legal since because x > 0 and y > 0, we obtain x3 + y 3 ≥ (x + y)xy. It follows that xyz xyz z 1 = ≤ = . 1+a+b xyz + x3 + y 3 xyz + (x+y)xy x+y+z

156

SOLUTIONS

1 1 and . Now just add Similar inequalities can be obtained for 1+b+c 1+c+a together the three inequalities: 1 1 1 z x y + + ≤ + + = 1. 1+a+b 1+b+c 1+c+a x+y+z x+y+z x+y+z Problem 6. We start by combining parallel strips so no two directions are the same. We then fix a point O and assign to each strip either of the two vectors at O whose length is the strip’s width and whose direction is perpendicular to the strip’s boundary. Divide the plane into 12 sectors (angles) of measure 30◦ with vertex O. For each pair of opposite sectors, choose the vectors lying within the two sectors, including their boundaries, and find the sum of their lengths. This yields six sums, at least one of which must be 100 6 or more, since their total is 100. Choose a pair of opposite sectors whose corresponding sum is at least 100 6 . Replacing some of these vectors by their opposites, if needed, we can ensure that all of them fall within the same 30◦ sector. A sum of vectors does not depend on the order of summands. Take the vectors in the clockwise order of their directions — so the direction of each is obtained from the previous one’s by a clockwise On rotation through at most 30◦ — and draw them in this order so the head of each vector is the On−1 M tail of next. This gives a convex polygonal line O2 OO1 O2 . . . On , as in the figure. The length of O1 this broken line, as we have mentioned above, is 100 O at least 6 . The distance O0 On , where O0 is another name for O, cannot be less than 100 ◦ ◦ 6 · cos 30 . That’s because all the component vectors lie within 30 of the sum vector; that is, the length of the projection of each segment Oi−1 Oi , where i = 1, . . . , n, onto the line O0 On is at least Oi−1 Oi · cos 30◦ . Now translate each strip so its edges pass through the endpoints of the corresponding segment Oi−1 Oi , We claim that the polygon MOO1 O2 . . . On , where M is the intersection point of the perpendiculars to segments OO1 and On−1 On drawn through the points O and On , respectively, will be completely covered by the strips. Indeed, consider any point X in the polygon MOO1 O2 . . . On . A simple, but somewhat incomplete, argument is to consider the point Y on the broken line OO1 O2 . . . On closest to X. If Y lies on Oi Oi+1 , then XY ⊥ Oi Oi+1 , and hence X lies in the strip perpendicular to the segment Oi Oi+1 . Let’s give a longer, but more rigorous, reasoning. Suppose the strips do not cover the point X. The angle XOO1 is not obtuse. (Why?) Nor is the angle XO1 O acute; if it were, the strip X O2 perpendicular to OO1 would cover X. It follows that \XO1 O ≥ 90◦ , and, since \OO1 O2 < 180◦ , \XO1 O2 = \OO1 O2 − \XO1 O < 90◦ .

O1

O

page 214

YEAR 1997 OLYMPIAD

157

Similarly, one shows that the angle XO2 O3 is acute. Continuing by induction, we finally conclude that if point X is not covered by any of the strips perpendicular to segments OO1 , . . . , On−2 On−1 , then all the angles XO1 O2 , XO2 O3 , . . . , XOn−1 On are acute. But then X lies in the strip perpendicular to the segment On−1 On : a contradiction. Now recall that the angle between the vectors OO1 and On−1 On does not exceed 30◦ . Therefore, the angles MOOn and MOn O of the triangle MOOn are at least 60◦ . Therefore this triangle contains the On equilateral triangle constructed on the side OOn (see figure). The inradius of a regular triangle with 60◦◦◦◦◦◦ 60 60 √ side length a is equal to a/(2 3). It remains to M check the inequality 100 6

· cos 30◦ √ > 1, O 2 3 which is amply satisfied (any number above 24 would do in place of 100).

page 215

Year 1998 Olympiad Level A Problem 1. One solution is given by the number of months in a nonleap year that have 28, 30 and 31 days, respectively: 28 · 1 + 30 · 4 + 31 · 7 = 365. Another solution is x = 2, y = 1, z = 9. Remark. Compare with Problem 1 on page 24.

Problem 2. Let p1 , p2 , . . . , p8 be different prime numbers. We claim that the numbers n1 = p21 p2 . . . p8 , n2 = p1 p22 . . . p8 , . . . , n8 = p1 p2 . . . p28 satisfy the desired conditions. Indeed, each of the primes p1 , p2 , . . . , p8 appears in the prime factorization of n2i with a power of at least two. Therefore, n2i is divisible by each of the nj . To show that ni is not divisible by nj for i 6= j, just note that pj appears in the prime factorization of nj with power 2, but in that of ni with power 1 only. (Here of course we’re using the fact that the prime factorization of an integer is unique: Fact 10 on page 104). page 216

D

O

Problem 3. First, given the parallelogram ABCD, B there is at most one point M on the line AB with A M ′ D \AM O = \MAD, for the following reason. Since O and D are on the same side of the line AB, saying that \AM O = \MAD is equivalent to saying that M O is parallel to the line AD′ obtained from AD by reflection in AB (see figure above). But only one parallel to AD′ can go through O. Hence the line M O is unique, and so is M . Next, a point is equidistant from C and D if and only if it lies on the perpendicular bisector of the segment CD. We draw this line; it intersects CD at its midpoint P , and it intersects the line AB at P′ D P C some point N (see figure). We must show that N = M . The line OP is a midline of △BDC, because O is the midpoint of the diagonals. Hence OP k BC k AD. O Also, the line through O parallel to P N forms a rectangle P P ′N ′N , and O is the midpoint of the side P ′N ′ , being equidistant from opposite A B N′ N (=M ) sides of the parallelogram.

I reorganized this proof to avoid relying on the figures in the analysis of cases, and especiailly to deal cleanly with the exceptional cases.

160

SOLUTIONS

Then, if N is closer to B than to A, as in the figure, we have the following series of equalities: \NAD equals the exterior angle at vertex D of the parallelogram (corresponding angles), which equals \DP O (alternating internal angles), which equals \AN O (congruence of △OP P ′ and △ONN ′ , from the equalities P ′ O = ON ′ and P ′P = N ′N ). Thus \NAD = \AN O, and N satisfies the defining property of the point M . Since M is unique, we have N = M . If N is closer to A than to B, we have instead \NAD = \ADC = \CP O = \AN O, so again N must be equal to M . The same figure as before illustrates this situation, with the sole change we that switch A with B and C with D. There are two exceptional cases: When the initial parallelogram is a rectangle, the lines P N and P ′ N ′ coincide; then \NAD = \AN O because both are right angles. And when N = A, the angles NAD and AN O are undefined; there is, strictly speaking, no point M as in the statement, but still the point N is characterized by the condition in the first paragraph of this solution. Problem 4. Suppose that k among the numbers a1 , a2 , . . . , a100 are blue and 100 − k are red. Since the blue numbers are written in increasing order, these k numbers are the numbers from 1 through k. Similarly, the 100 − k red numbers are the numbers 100, 99, . . . , k + 1. Therefore, all the numbers from 1 to 100 occur among a1 , a2 , . . . , a100 . Problem 5. If a person X has acquaintances sitting next to each other — in particular, if X is acquainted with either of X’s neighbors — then X is acquainted with all the guests. Let us prove that there is such a guest X. Consider any two neighbors. If they are acquainted, we have found our X. If they are not, a common acquaintance of theirs can serve as X, since this person has two acquaintances who are neighbors. Having found X, we now observe that X’s neighbors are also acquainted with everybody, because they are acquainted with X, who is seated next to them. The neighbors of these neighbors are also acquainted with everybody, and so on around the table. Problem 6. Not necessarily. See the figure for the corresponding counterexample using eight squares B C instead of 100. The example can be described 9 explicitly as follows: Denote by A, B, C, D the vertices of the red 1 square. Divide the diagonal AC into 100 equal A D segments and denote their endpoints, in order, by the numbers 1, 2, . . . , 101, with point A denoted by 1 and point C by 101. For each pair of points k and k + 1 (k = 1, 2, . . . , 100), there exist exactly two squares of the given size whose sides are parallel to the sides of the red square and pass through points k and k + 1: one of these squares contains B and doesn’t contain D, while the

page 217

YEAR 1998 OLYMPIAD

161

other contains D but not B. If k is odd, we let the k-th white square be the square that contains B, and if k is even, the square that contains D. The 100 white squares thus selected completely cover the red square; but if we remove the square whose sides pass through points k and k + 1, the segment of the diagonal between points k and k + 1 will not be covered. (Compare with Problem 3 on page 24.) Level B Problem 1. We have 49 + 610 + 320 = (29 )2 + 2 · 29 · 310 + (310 )2 = (29 + 310 )2 .

page 218

Remark. One might be tempted to tackle this problem by calculating the remainders of the given number upon division by various primes, in search of a small divisor. But this will go nowhere, because 29 + 310 is a prime!

Problem 2. Draw the third altitude BS and extend it K to meet P Q at T (see figure). Then the rectangles AST Q and AEKL are of equal area. Indeed, L by the similarity of the right triangles ABS and A AEC, we have AE/AC = AS/AB, which implies AE · AB = AS · AC, which implies AE · AL = AS · AQ. We prove in the same way that CST P and CDMN have equal area. Q

M B E

D S

T

N C

P

Remark. The equation AE · AB = AS · AC can also be derived from the fact that its sides are both equal to the power of the point A with respect to the circle passing through the points B, E, S, and C.

Problem 3. Let x be the fraction of truthful villagers. Imagine that the truth tellers become liars and the liars become truth tellers. Then the traveler will receive exactly the same answers! Indeed, the honesty of each villager flips, but so does the honesty of the neighbor about whom the villager is informing; therefore, the answer remains the same. Now, after the change, the fraction of honest villagers is 1 − x rather than x. Thus, the traveler cannot distinguish between a circle in which the portion of truthful people is x and one in which this portion is 1 − x. This means the traveler can determine x only if x = 1 − x, or if x = 21 .

Remark. Number the villagers in clockwise order and set xi = 1 if the i-th villager is a liar, and xi = 0 otherwise. Then the answer of the i-th villager can be expressed as the value of xi + xi+1 , where the addition is performed modulo 2 (that is, 0 + 0 = 0, 0 + 1 = 1, 1 + 1 = 0). Therefore, the information received by the traveler can be viewed as a system of simultaneous linear equations over the field with two elements. See Field2elem,SystemLinearEq. CITE Field2elem,SystemLinearEq

Problem 4. By thinking of the bases as the nodes of a graph whose edges are the roads (see Fact 3), we see that the set of all bases can be divided

page 219

162

SOLUTIONS

into a certain number of connected components. Two bases belong to the same connected component if and only if it’s possible to go from one to the other using the available roads. By definition, a set of roads is essential if and only if after its removal there are at least two connected components. Initially, there is only connected component. Otherwise the empty set would be an essential set. It would therefore be the only strategic set. But the problem asserts the existence of two different strategic sets. Lemma. Closing all the roads in a strategic set causes the bases to break up into exactly two connected components. No road joining two bases within either of these components can belong to the strategic set. Proof. Let S be a strategic set and consider the connected components obtained by the removal of S. If a road b ∈ S joins two bases in the same connected component, restoring it will not change the number of connected components; that is, the set S \ b still breaks up the bases into more than one component, so it’s still essential, and S is not strategic. Now we show that the removal of S can only create two connected components. Let a be any road in S; as we have seen, it joins two different connected components, say X and Y . If we restore the road a, the components X and Y merge, but all other components are unaffected. So if there is a third component the set S \ a is still essential; that is, S is not strategic. ˜ Let the connected components arising from removing the strategic set V be called A and B, and let the ones arising from removing W be C and D. Set K = A ∩ C, L = A ∩ D, M = B ∩ C, and C N = B∩D (see figure). The sets K, L, M , N are pairK M wise disjoint and their union is the set of all bases. We claim that at most one of them is empty. Indeed, A B if K = L = ?, then A is empty, which is impossible. If K = N = ?, then A = D and B = C; but this conN L D tradicts the assumption that V and W are different strategic sets. The remaining cases are similar. By removing the strategic set V , we disconnect K from M , K from N , L from M , and L from N , but K remains connected to L and M to N . By removing the strategic set W , we disconnect K from L, K from N , L from M , M from N , while K remains connected to M and L to N . Thus, the set of roads that belong to exactly one of V and W are all the roads joining K and L, K and M , L and N , M and N . When this set of roads is removed, the bases split into at least two (nonempty) sets, K ∪ N and L ∪ M , which are not connected to each other. In other words, we have an essential set. Remark. This set is not necessarily strategic: for instance, sets K and N might be nonempty, with no roads from K to N .

page 220

YEAR 1998 OLYMPIAD

163 B

Problem 5. The locus of points O such that \AOD = 80◦ and O lies on the same side of AD as B is an arc of circle passing through A and D, and the locus of points O for O2 which \BOC = 100◦ with point O lying on the same C A side of BC as A is an arc of circle passing through O1 B and C (see (see the figure). The point O must lie on both these two circles. This means that there are at most two such points. Let’s locate the two possibilities for O. The first, O1 , D can be taken on the diagonal AC so that \BO1 C = 100◦ . ◦ Then, obviously, \AO1 B = 80 , and, by symmetry with respect to AC, we have \AO1 D = \AO1 B = 80◦ . Similarly, the second point, O2 , can be taken on the diagonal BD so that \BO2 C = 100◦ . In this case, \AO2 D = 80◦ and \AO2 B = 100◦ . The two points must be distinct, since they lie on different diagonals and do not coincide with the intersection point P of the diagonals, Also, both points lie inside the rhombus: for instance, O2 lies on the diagonal BD rather than on its extension. To see this, consider the triangle BP C. Since \BO1 P = 80◦ > 55◦ = \BCP , the point O1 lies inside the rhombus. Similarly, \CBP = 35◦ < 80◦ < \CO2 P ; hence O2 lies on the diagonal BD rather than on its extension. Therefore both values \AOB = 80◦ and \AOB = 100◦ , occur. Problem 6. Let x1 , . . . , xn be the coordinates of the marked points, which we will treat as unknowns. We assume x1 < x2 < · · · < xn . We also set x0 = 0 and xn+1 = 1. By assumption, for each i = 1, . . . , n, we have xi = 21 (ai + bi ),

(1)

where ai < xi < bi and ai , bi belong to the set {x0 , . . . , xn+1 }. Generalizing the notation, what we have are n equations of the form xi = β0 x0 + β1 x1 + · · · + βn xn + βn+1 xn+1 .

(2)

Each coefficient βj on the right-hand side is rational and nonnegative, since it equals 0 or 12 ; and at least one xj , with j > i, has a strictly positive coefficient, because bi in (1) equals xj for some j > i. Important: Though we always use the symbol βj for the coefficient of xj , the number it represents can of course vary across equations, and along the course of the solution. First solution. On the right-hand side of (2) we can omit the term in x0 (since x0 = 0) and the term in xi , whose coefficient is 0 by construction. We therefore write xi = β1 x1 + · · · + βd i xi + · · · + βn+1 xn+1 ,

where the hat indicates that the term underneath it is absent.

(3)

page 221

The draft translation of the first solution was different from the original. It was somewhat garbled, so I made adaptations for clarity. Also I’m factoring out the initial portion of the two solutions, which is common.

164

SOLUTIONS

We now simplify these n equations by substitution. The first equation gives x1 in terms of x2 , . . . , xn+1 ; we take this value of x1 and substitute it wherever x1 appears on the right-hand side of the other n − 1 equations. We thus get n − 1 equations involving only x2 , . . . , xn+1 . We claim that they have the same properties as the original set of equations, namely: the lefthand sides are respectively x2 , . . . , xn ; the right-hand sides have rational, nonnegative coefficients; and in the equation having xi on the left, at least one variable xj , with j > i, has a positive coefficient. That the new coefficients are rational and nonnegative is obvious, since they are products (or sums of products) of the old ones. To show the positivity condition, notice that a positive coefficient of xj , where j > i, either remains unchanged or is increased by a positive multiple of the coefficient of xj in the first equation. There is one way in which the new equations can differ from the old ones in (3): the i-th equation (for some values of i) might gain a xi term on the right, if it originally involved x1 and the value of x1 given by first equation involves xi . To make this i-th equation comply with the scheme in (3), we move the new term αxi to the left and divide by 1 − α. We must show that this is not a division by 0: but if 1 − α were 0, our equation at this stage would be 0 = β0 x0 + · · · + βn+1 xn+1 , where all the products on the right are nonnegative and at least one is positive, and this is an impossibility. For a similar reason the division by 1 − α preserves the nonnegativity of the coefficients: indeed, 1 − α cannot be negative, otherwise the negative product (1 − α)xi would equal a positive sum. And of course the coefficients remain rational, and at least one remains positive (for j > i). This reduces the system of equations to another one with the same structure as (3), but with one fewer equation and one fewer variable. We then repeat this procedure, eliminating x2 to get a system of n − 2 equations of the form (3) involving only the unknowns x3 , . . . , xn , together with the constants x0 and xn+1 . We continue eliminating variables one by one, until we finally reach a single equation xn = βn+1 xn+1 , which is a rational number (recall that xn+1 = 1). Then back-substitution gives the values of xn−1 , xn−2 , and so on back to x1 , always with rational numbers. Remarks. 1. What we did was in essence to use Gaussian elimination, a general method for solving systems of linear equations (see also Fact 25). I reinstated the remarks, 2. The fact that we could always divide by 1−α in applying the method is essential, as it ensures that our system has a unique solution. A system with rational coefficients admitting more than one solution has irrational solutions as well as rational ones; for example the system 

x1 = 2x2 − 2 x2 = 12 x1 + 1

√ √ has the solution x1 = 2 2, x2 = 2 + 1 among many others.

which were absent from the draft translation. (Did the authors decide to make changes between the original and the draft translation stage?) page 223

YEAR 1998 OLYMPIAD

165

On the other hand, if the solution is unique, it can be found by Gaussian elimination, which only involves addition, subtraction, multiplication, and division — operations that cannot turn rational numbers into irrational ones (see Fact 25 on page 110). This is the basis for the second approach to the problem, discussed next.

Second solution. (Above grade level.) The system of simultaneous linear equations (2) is rational ; that is, it has rational coefficients and rational constant terms. If a rational system of linear equations has a unique solution, the solution is rational (see last pargraph of Remark 2 above). Thus we will have solved the problem if we can show that the solution of (2) is unique. Let x1 , x2 , . . . , xn and y1 , y2 , . . . , yn be two different solutions. Then the numbers t1 = x1 − y1 , t2 = x2 − y2 , . . . , tn = xn − yn form a nonzero solution to the homogeneous system corresponding to (2), that is, the system in which all constant terms βn+1 xn+1 are replaced by 0. But then the numbers ti have the following property: each of them either lies halfway between two others or it lies halfway between 0 and another number tj . This is only possible if all the numbers are equal to zero. Indeed, choose among the ti the one with the greatest absolute value. If it isn’t zero, it can’t be the midpoint of an interval between two other numbers. Having shown that all the ti are 0, we see that our system has a unique solution, as needed. Level C Problem 1. Suppose a + b + c ≤ 11. Then 28a + 30b + 31c ≤ 31(a + b + c) ≤ 11 · 31 = 341, which is less than 365. So this cannot happen. Similarly, if a + b + c ≥ 14, we have 28a + 30b + 31c ≥ 28(a + b + c) ≥ 28 · 14 = 392

page 224

which is also impossible since 369 > 365. There remains to prove that a + b + c cannot be equal to 13. Suppose that a + b + c = 13. The possibility a = 13, b = c = 0 does not satisfy the assumption: 28 · 13 + 30 · 0 + 31 · 0 = 364 6= 365.

The last case to refute is a + b + c = 13, a < 13. In this case, we have b + c = 13 − a > 0 and 28a + 30b + 31c = 28(a + b + c) + 2b + 3c ≥ 28 · 13 + 2(b + c). The first term is equal to 364 and the second one is at least 2. It follows that the sum is at least 366 and can’t be equal to 365. Remark. Compare Problem 98.8.1.

REF 98.8.1

166

SOLUTIONS

Problem 2. First solution. Denote by ai the marked side of the i-th rectangle and by bi a perpendicular side. The P area of the initial square is equal to the sumPof the areas of the rectangles: ai bi = 1. But bi ≤ 1 for all i; therefore, ai ≥ 1.

Second solution. We project all the marked segments onto one side of the square. If it is completely covered by the projections, then their total length is at least 1. If there is a point on the side that is not covered by the projections, we draw the perpendicular to the side through that point. This perpendicular is covered by rectangles with a marked side parallel to it — otherwise the foot of the perpendicular would lie in the projection of the marked side. The total length of marked sides of the rectangles covering the perpendicular is at least 1.

Problem 3. We number the lamps in order along the road. If the segments lit by the n-th and (n+2)-nd lamps overlap, the (n+1)-st lamp can be turned off. Therefore, segments with different odd numbers are disjoint. It is impossible to place more than 999 disjoint (closed) segments of length 1 m on a segment of length 1000 m. Hence there are at most 1998 lamps, because if there were 1999 or more, there would be at least 1000 odd-numbered lamps. Next we show how to arrange 1998 lamps so none of them can be turned off. Let the centers of the segments they light form a uniform sequence whose endpoints are 21 m apart from the endpoints of the road. The interval 999 m. Hence the distance between the between two neighboring centers is 1997 1 1998 m between the n-th and (n+2)-nd lamps is 1997 m, leaving a gap of 1997 segments they light. This gap is lit only by the (n+1)-st lamp. Therefore, none of the lamps can be turned off. Remark. Compare Problem 98.8.6.

Problem 4. A number that is divisible by 1998 is also divisible by 999. We will show that the sum of digits of a positive number divisible by 999 is at least 27. As is well-known (or see Fact 6 on page 102), a number is divisible by 9 if and only if the sum of its digits in decimal notation is divisible by 9. The criterion for divisibility by 999 can be formulated similarly: Split the decimal representation of the number into groups of three digits from right to left. (We call these groups “triples”, even though the leftmost may have only one or two digits.) Add these triples together as numbers. The initial number is divisible by 999 if and only if the sum thus obtained is divisible by 999. For instance, 97902 is divisible by 999, since the number 97 + 902 = 999 is divisible by 999. One can view these triples as the “digits” of the number in 1000-ary notation. This criterion is proved just like the criterion for divisibility by 9; we leave this task to the reader.

at the end of 2nd sol., changed “is 1” to “is at least 1” because a point may presumably be on the boundary of two rectangles, leading to double counting. (Or does razrezali exclude that possibility?) Is it worth mentioning the possiblitity that with infinitely many rectangles, the sides needn’t be parallel to the square? omitted “by natural numbers” (the use of ordinals already implies that). page 225

REF 98.8.6

∗

YEAR 1998 OLYMPIAD

167

Now take a number divisible by 999, split it into triples of digits, and compute the sum of the triples. If the new number is greater than 1000, repeat the operation, and keep doit it until obtaining a number smaller than 1000. This will necessarily happen, because each operation decreases the number. (Proof: if a1 , . . . , ak are nonnegative integers, with ak 6= 0 and k ≥ 1, then a0 + 1000a1 + · · · + 1000k ak > a0 + a1 + · · · + ak .) The only positive integer smaller than 1000 and divisible by 999 is 999 itself, whose digits add up to 27. Thus, it remains to show that our operations never increase the sum of digits: this will mean that the sum of digits of the initial number could not be less than 27. Obviously, when a number is split into triples, the total sum of their digits is the same as that of the number. Let us show that when the triples are added, the sum of digits does not increase. Denote by S(X) the sum of digits of X. It follows from the algorithm of digitwise addition that

page 226

S(X + Y ) = S(X) + S(Y ) − 9P (X, Y ),

where P (X, Y ) is the number of carries performed when we add X and Y . This means that S(X + Y ) ≤ S(X) + S(Y ) and we’re done.

Problem 5. We start with the “problem of one nail.” Suppose that we have one nail hammered in the floor at a point M on the side AC. Fix a point O; it will be the center of the hypothetical rotation. Is it possible to rotate the triangle around the point O by a small angle, and if it is, then in what direction? B Suppose that the triangle is positioned as in the figure. Draw the perpendicular to AC at M . It divides the plane into two half-planes. If O lies in the same half-plane as C, then we can rotate the triangle around O counterO clockwise, and if it lies in the other half-plane, then a clockwise rotation is possible. Now imagine that O lies right on the perpendicular. C M A Then, if O and B are on the same side of the line AC (which is true whenever O lies inside the triangle), we’ll be unable to rotate the triangle in either direction. If O and B lie on different sides of the line AC, the triangle can be rotated in both directions. B Returning to the three-nail problem, draw the perpendiculars to the three sides of the triangle going through the nails. We claim that if they −++ −+− do not meet at the same point, the triangle can +++ −−+ ++− be rotated. Indeed, in this situation the per+−+ +−− pendiculars partition the plane into seven A regions. To each region we assign a string C of three plus or minus signs. The first sign for a region is + if the nail on AB does not prevent the triangle from rotating clockwise around points in this region, and so on; see figure on the right. Different regions will have different strings of signs. There are eight possible strings and seven regions. Hence,

page 227

168

SOLUTIONS

at least one of the strings will be either +++ or −−−. In the first case, we can rotate the triangle clockwise, and in the second, counterclockwise. If instead the perpendiculars meet at a single point, say C R, let D, E, and F be the positions of the nails on the sides AB, BC, and AC, respectively (see figure). Then the E nails at D and E block the rotation around any center K inside the angle DRE (and its vertical angle), so do HF R the nails at E and F for the angle ERF and F, D for the angle F RD. It follows that the only possible A D B position of the center we are left with is R, the intersection of the perpendiculars. But this point lies inside the triangle, and, as already seen, each of the nails blocks the rotation around it. Thus, rotation is impossible if and only if the perpendiculars to the sides drawn from the corresponding nails meet at one point. It remains to determine the position F of the third nail on the side AC. Denote by K the intersection of AC with the perpendicular to AB through D; similarly, let H be the intersection of AC with the perpendicular to BC through E. Then AD CE AK = = 21 AC and CH = = 32 AC. cos 60◦ cos 60◦ The triangle HRK is isosceles, since the angles HKR and KHR each measure 30◦ . Hence HF = FK and F divides the side AC in the ratio 5:7, measuring from A. Problem 6. Step 1. We prove that the numbers in a good arrangement can be colored with nine colors, denoted by 1, 2, . . . , 9, so that numbers of the same color form an increasing sequence. We do this by going through the numbers from left to right, assigning to each the lowest possible color such that the number most recently assigned this color is smaller than the current number. Suppose that nine colors are not enough to complete such a coloring. This means that, at a certain moment, we are unable to give the current number, a10 , any color from 1 to 9; in particular we must have encountered earlier some number a9 > a10 that received color 9. Next, a9 couldn’t get color 8 because of an earlier number a8 > a9 that got this color. Extending this reasoning down to color 1, we obtain 10 numbers in decreasing order. Step 2. An arrangement of the numbers from 1 to n colored as described above is completely determined by the color of each number from 1 to n and the color of each place in the string, because the numbers on the places of the same color increase. There are 9n ways to color the numbers from 1 to n using 9 colors. The number of colorings of the n places using 9 colors is the same. Thus, the number of good arrangements is at most 81n . Remark. There are exactly n! permutations of the numbers from 1 to n, and an ever smaller fraction of them is “good” as n increases. This follows from the bound in the problem together with Stirling’s formula, a result from calculus

page 228

YEAR 1998 OLYMPIAD

169

√ that says that n! is approximately equal to 2πn(n/e)n . More precisely, 81n 81n 81n = lim (81e/n)n = 0. = lim √ ≤ lim lim n→∞ n→∞ n→∞ (n/e)n n→∞ n! 2πn(n/e)n Here e ≈ 2.71828 is the base of natural logs. See also Fact 4 on page 101.

Level D Problem 1. We have x + y + z − 2(xy + yz + xz) + 4xyz − 21 = 21 (2x − 1)(2y − 1)(2z − 1).

(1)

If the left-hand side of the equation is zero, then at least one of the factors in the right-hand side is zero. Therefore, one of the numbers x, y, z is 21 . Remark. How would one come up with this factorization? We offer two suggestions. (a) The original polynomial involves the so-called symmetric functions of the variables x, y, z (x+y+z, xy+yz +xz, and xyz); these bring to mind Vieta’s formula (Fact??) in the cubic case. To be able to make the polynomial fit the formula we would like to get rid of the coefficients 2 and 4, which we can do by scaling the variables via the substitutions x = X/2, y = Y /2, and z = Z/2: x+y+z − 2(xy+yz+xz) + 4xyz

= 12 ((X +Y +Z)−(XY +Y Z +XZ)+XY Z). (2)

So let the symmetric functions on the right be the coefficients of a cubic polynomial in the dummy variable t: −t3 + (X + Y + Z)t − (XY + Y Z + XZ)t2 + XY Z = −(t − X)(t − Y )(t − Z), where the equality is exactly Vieta’s formula. Substituting t = 1 gives −(1 − X)(1 − Y )(1 − Z) = −1 + 2 × right-hand side of (2).

Going back to the original variables we get

−(1 − 2x)(1 − 2y)(1 − 2z) = −1 + 2 × left-hand side of (2),

which is the same as (1). (b) A general result (see Fact 19) says that if a polynomial in one or more variables takes the value 0 identically when x is replaced by a, this polynomial is divisible by x − a. Denoting the polynomial on the left-hand side of (1) by P (x, y, z), we see that P ( 12 , y, z) — the polynomial in y and z obtained by replacing x by 21 — is identically 0. Hence P (x, y, z) is divisible by x − 12 . Similarly, P (x, y, z) is divisible by y− 12 and z− 21 . Therefore, P (x, y, z) is equal to (x− 12 )(y− 21 )(z− 21 ) up to a constant factor (see Fact 21). f↼x↽

Problem 2. Here is a counterexample:  2 − x if x ≤ 1; f (x) = 1/x if x > 1. x The graph of this function is shown on the right. The first property is obvious, and the second is true because the functions 2 − x and 1/x have the same derivative at x = 1.

page 229

170

SOLUTIONS

Let us prove the third property. Obviously, if x is rational, then f (x) is also rational. Let y = f (x) be rational. Then either x = 2−y or x = 1/y. In any case, x is rational. Therefore, rational values of x correspond to rational values of f (x) and vice versa. P Problem 3. Denote by M the centroid of the tri30 angle ABC (the centroid is the intersection of the medians). Consider the equilateral triangle AP Q K C B 30 such that AK is its median and P lies on the line AB. Since the centroid divides the medians in M L the ratio 2:1, the point M is the centroid, and 30 60 thus, the center, of triangle AP Q. It follows Q A ◦ that \KP M = 30 = \KBM . Therefore M , K, P , and B lie on a circle. Since \MKP is a right angle, M P is a diameter of this circle. This circle meets AP at two points, one of which is P and the other one, B, is the midpoint of AP . √ Let the triangle AP Q have side length 2a. We have AB = a and AK = 3a. The triangle BKP is equilateral, BK = a = KC, and BC = 2a. Further, \ABK = 120◦ , whence by the law of cosines AC 2 = AB 2 + BC 2 − 2 2AB · BC cos 120◦ = a2 + 4a2 + 2a2 = 7a √ . Finally,√by the law of cosines we find cos \ACB = (4 √ + 7 − 1)/(2 · 2 7) = 5/(4 7) and cos \CAB = √ (7 + 1 − 4)/(2 7) = 2/ 7. ◦

◦

◦

◦

Problem 4. Both sides of the equation must yield the same remainder upon division by 3. The remainder of the left-hand side is 1 (see Fact 7); therefore, z is an even number (see Remark 1). Similarly, the left-hand side has remainder 1 upon division by 4; therefore, x is even, too. Thus, 4y = 5z − 3x = 52z0 − 32x0 , i.e., 22y = (5z0 − 3x0 )(5z0 + 3x0 ). Therefore, 5z0 − 3x0 = 2k and 5z0 + 3x0 = 2l , where k and l are nonnegative integers and k + l = 2y. It follows that 5z0 = 21 (2k + 2l ) and 3x0 = 21 (2l − 2k ). The number 2l−1 − 2k−1 = 3x0 is odd; hence k = 1. Then 2k = 2 and 3x0 = 2l−1 − 1. Consequently, l−1 is even: l−1 = 2s. It follows that 3x0 = (2s − 1)(2s + 1) is the product of two powers of 3 that differ by 2; the only possibility for these two numbers is 1 and 3. Then s = 1, l = 3, and 2y = 4. Thusj x = y = z = 2 is the unique solution to our equation.

page 230

page 231

Remark. It is not difficult to prove by induction that the remainder of 5z upon division by 3 is 1 if z is even and 2 if z is odd. In fact, the remainders of an upon division by b, for fixed a and b, always form a periodic sequence (Fact 4). This can be derived from the last statement in Fact 7. Also it should be mentioned that if b is prime, the period of this sequence is a divisor of p − 1. This statement is equivalent to Fermat’s Little Theorem; see the remark Problem 95.11.6. REF 95.11.6

Problem 5. On the face of it, it may seem that the desired construction is impossible: if the first cogwheel rotates clockwise, then the second one rotates counterclockwise, the third rotates clockwise, and so on. So the 61st

YEAR 1998 OLYMPIAD

171

cogwheel must rotate clockwise, which does not agree with the rotation of the first cogwheel. However we are in space, not on the plane. And the direction of rotation depends on the side from which we look at the cogwheel. Here is one possible solution: z

61

7

1 x 14

13

y 38

37

page 232

Let us describe the gear in the figure: cogwheels 13 through 61 and 1 lie in the xy-plane; all other cogwheels have their centers in the xz-plane, and lie in planes perpendicular to the xz-plane. Cogwheels 13, 15, 17, . . . , 61 rotate clockwise if we look at the xy-plane from above, while cogwheels 14, 16, . . . , 60, 1 rotate counterclockwise. The numbers of cogwheels undergo the “parity change” in the direction of motion occurs along the chain of cogwheels from 1 to 13; these two cogwheels at the endpoints of this chain rotate in opposite directions in the xy-plane. Remark. Experienced readers will have noticed that this solution in essence uses the M¨ obius strip. It would the draft translation was rather garbled here. I have been possible to draw the cogs on a M¨obius strip hope this is what you having the traditional shape shown in the figure, but mean. (Actually the it would have been harder to justify the correctness of drawing shown, with a the answer. ruled M. strip, cannot The phenomenon of parity change is connected with the fact that the contain any circles, but I guess the claim is M¨obius strip is a nonorientable, or one-sided, surface. substantially correct.)

Year 1999 Olympiad Level A Problem 1. Compare the fractions x = and arrange them in ascending order.

111110 111111 ,

y=

222221 222223 ,

and z =

333331 333334 ,

Problem 2. Show how to cut any quadrilateral into three trapezoids. (A parallelogram is also considered a trapezoid.) Problem 3. Find four pairwise distinct positive integers a, b, c, and d for which the numbers a2 + 2cd + b2 and c2 + 2ab + d2 are perfect squares. Problem 4. Annie has 500 dollars on her bank account. The bank allows only two kinds of operations: withdrawing $300 or adding $198. What is the largest sum that Annie can take from her account if she has no other money? Problem 5. In a right triangle ABC, the midpoint of the hypotenuse AC is labeled O. Points M and N are chosen on the legs AB and BC so that \M ON = 90◦ . Prove that AM 2 + CN 2 = M N 2 . Problem 6. Each participant in a chess tournament played each other twice: once as white and once as black. The final scores of all the players were the same. (A win is worth one point, a tie half a point, and a loss zero points.) Prove that at least two competitors won the same number of games playing white. Level B Problem 1. Two numbers are written on a blackboard in a laboratory. Every day at noon a researcher erases both numbers and writes their arithmetic and harmonic means instead. The numbers written on the blackboard on the morning of the first day were 1 and 2. Find the product of the numbers written on the blackboard at the end of the 1999th day. (The arithmetic and harmonic means of two numbers a and b are a+b 2 and 1 1 , 2 a + b

page 41

28

PROBLEMS

respectively.) Problem 2. Two players play the following game: the first writes one letter A or B per turn, from left to right, starting from nothing; the second player, after each play by the first, can either pass or interchange any two letters already written. When both players have had 1999 turns each, the game is over. Can the second player ensure that the final string is a palindrome no matter what the first player does? (A palindrome is a string that reads the same from left to right or from right to left.) Problem 3. The diagonals of a parallelogram ABCD meet at a point O. The circle passing through A, O, and B is tangent to the line BC. Prove that the circle passing through the points B, O, and C is tangent to the line CD. Problem 4. Find all positive integers k such that the number k

z }| { . . 2} 1 . . . 1 2 {z . . . 2} − |2 .{z | 2000

is a perfect square.

1001

Problem 5. The incircle of a triangle ABC, for which AB > BC, touches the sides AB and AC at P and Q, respectively. The midline parallel to AB is labeled RS, and it intersects P Q at T . Prove that T lies on the bisector of the angle B of the triangle. page 42

2n

Problem 6. A sports competition has n scored events and participants. The first event eliminates the bottom half the participants, according to the scores obtained; the second eliminates half of the remaining ones (a quarter of the total), and so on, until only an overall winner is left. Suppose a ranking of expected performance is made for each event separately, prior to the start of the competition. A contestant is called a potential winner if, for some ordering of the events and assuming the expected ranking holds true, that contestant will win the tournament. (a) Prove that it is possible for 2n−1 contestants to be potential winners. (b) Prove that it is not possible for more than 2n − n contestants to be potential winners. (c) ⋆ Prove that it is possible for exactly 2n − n contestants to be potential winners. Level C Problem 1. It is known that (a + b + c)c < 0. Prove that b2 > 4ac. Problem 2. Two circles meet at points P and Q. The third circle, centered at P , meets the first one at points A and B, and the second one at points

∗ ∗∗ ∗

∗

YEAR 1999 OLYMPIAD

B Q

C

C and D (see figure). Prove that the angles AQD and BQC are equal. Remark. To avoid having to consider numerous cases, at the Olympiad the problem was proposed only for the configuration shown in the figure. Nonetheless, the statement remains true for other cases.

29

D

A

P

Problem 3*. Find all the pairs of positive integers x, y such that x3 + y and y 3 + x are divisible by x2 + y 2 . Problem 4*. A disk is divided by 2n radii into 2n congruent sectors, n of them blue and n red. Two sectors, blue and red, are chosen at will. Starting with the chosen blue sector, the numbers from 1 to n are written in the blue sectors counterclockwise. Similarly, starting with the chosen red sector, the numbers from 1 to n are written in the red sectors clockwise. Prove that that there is a half-disk that contains all the numbers from 1 to n.

page 43

Problem 5. A grasshopper √ the interval √ [0, 1]. From a point x √ jumps along it can jump either to x/ 3 or to x/ 3 + (1 − 1/ 3). A point a is chosen on the interval [0, 1]. Prove that, starting from any point, the grasshopper can, after a number of jumps, reach a point at a distance of no more than 1 100 from a. Problem 6*. The numbers 1, . . . , 1999 are written around a circle in some order; then the sum of the products of all sets of 10 consecutive numbers is computed. Find the arrangement for which this sum is the greatest. Level D Remark. At the Olympiad, problems 5–7 were scored and the best two numbers were added to the scores of problems 1–4.

Problem 1. Let a, b, c be the sides of a triangle. Prove the inequality a2 + 2bc b2 + 2ac c2 + 2ab + 2 + 2 > 3. b2 + c2 c + a2 a + b2

A

B

Problem 2. A plane convex figure is bounded by two line segments, AB and AC, and an arc of circle BC (see figure). (a) Construct a line that bisects the perimeter of the figure. C (b) Construct a line that bisects the area of the figure. Problem 3*. The faces of a regular octahedron are colored white or black. Any two faces that share a common edge are of different colors. Prove that the sum of distances from any point inside the octahedron to the planes of the white faces is equal to the sum of its distances to the planes of the black faces.

∗

30

PROBLEMS

Problem 4. A square meadow contains a round clearing. A grasshopper is hopping around in the meadow. Before each jump it chooses a vertex of the square and jumps exactly halfway toward it. Can the grasshopper always arrange to land in the clearing after a number of jumps, no matter where it starts? Problem 5. A graph is a set of points, called nodes, some of which are connected by lines, called edges; each edge connects exactly two nodes. A coloring of the nodes is said to be regular if no two nodes of the same color are connected by an edge. A certain graph is regularly colored in k colors and cannot be regularly colored in fewer colors. Prove that there is a path in this graph that visits nodes of all k colors, each of them once. Problem 6*. Solve the equation (1 + nk )l = 1 + nm in positive integers, where l > 1. n

Problem 7. Prove that the first digits of the numbers of the form 22 form a nonperiodic sequence.

page 44

Year 1999 Olympiad Level A 1. x < z < y. Level B 1. 2. 2. Yes. Level C 3. x = 1, y = 1. and reflections.

3. For example, a = 1, b = 6, c = 2, d = 3.

4. $498.

4. k = 2. 4 6

2

1

3 5

6.

is the only solution, apart from rotations

∗

1998 1999 1997

Level D 4. Yes, it can.

6. Unique solution: n = 2, k = 1, l = 2, m = 3.

∗

Year 1999 Olympiad Level A 1. Consider the numbers 1 − x, 1 − y, and 1 − z. 3. If ab = cd, then a2 + 2cd + b2 and c2 + 2ab + d2 are perfect squares. 4. Both 300 and 198 are divisible by 6. 5. Consider a point symmetric to N with respect to O. 6. Otherwise there is a participant who won all the games as white and one who won no games as white. Level B 1. The product of the numbers on the blackboard does not change. 2. Try to arrange it so that the after 1001 moves the last three letters form a palindrome. 3. Use the theorem about the angle between a tangent and a chord. 4. A perfect square ends in an even number of zeros; the perfect square closest to n2 is n2 − 2n + 1. 5. Prove that ST = BS.

∗

Level C 1. If f (x) = ax2 + bx + c, then a + b + c = f (1) and c = f (0). 3. x(x2 + y 2 ) − (x3 + y) = y(xy − 1). 4. Consider two equal numbers with the smallest distance between them. 5. Both maps √ defined by the grasshopper’s jump contract the interval [0, 1] by a factor of 3. Level D 1. Apply the triangle inequality. 2. Consider the midpoint of the arc BC. 3. The planes containing white faces bound a regular tetrahedron. 4. Divide the square into 4n small squares and prove by induction that the grasshopper can hit any of them.

∗ removed rest of hint

page 87

62

HINTS

5. If, in a regular coloring, we reassign color 1 to the nodes of color 2 not connected with nodes of color 1, the entire coloring remains regular. 6. Use the binomial formula. 7. Consider the half-open intervals [0, lg 2), . . . , [lg 9, 0) on a circle of length 1. The first digit of 2k is determined by the interval that contains the fractional part of lg 2k = k · lg 2.

Year 1999 Olympiad Level A Problem 1. Consider the numbers 1 2 1−x= , 1−y = , 111111 222223 and their reciprocals 1 1 = 111111, = 111111 12 , 1−x 1−y

1−z =

3 , 333334

1 = 111111 13 . 1−z

Since 1/(1−x) < 1/(1−z) < 1/(1−y), and all three of these numbers are positive, we obtain 1 − x > 1 − z > 1 − y. Thus, x < z < y. Problem 2. The solution is shown in the figures on the next page. However, to make it rigorous, we must do some work. (At the Olympiad, even solutions without rigorous explanations were accepted by the graders.) Let ABCD be an arbitrary quadrilateral. It has two neighboring internal angles whose sum is at least 180◦ , because (\A + \B) + (\C + \D) = 360◦ . (This is true even for nonconvex quadrilaterals.) Hence one of the sums in parentheses is at least 180◦ . We may assume, by relabeling if necessary, that \A + \B ≥ 180◦ . If this sum equals 180◦ , then BC k AD and ABCD is a trapezoid. Any two cuts parallel to the bases then solve the problem. See the left diagram in the figure. C

B

C

N

C

N

ℓ

B α

M

L

ℓ

B α

A

D

A

K

D

A

M

L K

D

Next, suppose \A + \B > 180◦ . Consider the line ℓ passing through the point B parallel to AD, as in the middle and right diagrams above. (The rightmost diagram illustrates the case of a nonconvex quadrilateral.)

∗

page 233 I don’t understand why the original considered the nonconvex case separately.

182

SOLUTIONS

Let α be the angle formed by ℓ and the side AB, and situated opposite angle A relative to the parallel lines ℓ and AD. Then α = 180◦ − \A < \B, where the equality holds because interior opposite angles are supplementary. Therefore, the line ℓ goes inside the quadrilateral, and so, upon going out, it intersects the side CD at a certain point L. Now draw a line parallel to CD and intersecting the segments AD and removed “this ... BL. Suppose this line intersects the side AD at K, and the segment BL at done” M . If we now draw the line through M parallel to BC until it meets the ∗ segment CL at a point N , we will have cut our quadrilateral into trapezoids ∗ AKMB, KDNM , and CNMB. Problem 3. It suffices to choose the numbers so that the product of the first two is equal to that of the last two, ab = cd; indeed, in this case, we have a2 + 2cd + b2 = a2 + 2ab + b2 = (a + b)2 ,

page 234

c2 + 2ab + d2 = c2 + 2cd + d2 = (c + d)2 . Thus, we only need to find a number n that can be represented as the ∗ product of two different factors in two different ways: n = ab = cd. For removed “see also Fact 10”. example, we can take the number 6 = 1 · 6 = 2 · 3. Problem 4. Since 300 and 198 are divisible by 6, Annie can withdraw only multiples of 6 dollars (see Fact 5). The largest multiple of 6 not exceeding 500 is 498. Let’s see how to withdraw 498 dollars. After the following operations: 500−300 = 200, 200+198 = 398, 398−300 = 98, 98+198 = 296, 296+198 = 494, the sum on the account decreases by 6 dollars. Having repeated this procedure 16 times, Annie will have withdrawn 96 dollars. Then she can take out 300, deposit 198, and take out 300 again to end up with 498 dollars. Problem 5. Denote by N ′ the point symmetric to N with respect to O. The triangles ON C and ON ′ A are congruent by the SAS property. In addition, N ′ AM is a right angle, because \N ′ AM = \N ′ AO + \MAO = \ACB + \BAC = 90◦ .

N′

A

Then, by the Pythagorean Theorem, 2

2

AM 2 + CN 2 = AM 2 + AN ′ = MN ′ . Therefore, it remains to prove that MN ′ = MN . But this equality follows from the fact that the right triangles N ′ OM and N OM are congruent by the congruence of their legs. C

M

O

N

B

Problem 6. Suppose there were n participants in the tournament. The total number of games played was therefore n(n−1), and for each one of them, one point was awarded in total. Thus the final scores of the players, being all the same, must equal n − 1 points.

page 235

YEAR 1999 OLYMPIAD

183

Each participant played white in n−1 games, and the number of wins for a given player among his or her games played as white must be one of the numbers 0, . . . , n−1. Suppose that the statement of the problem is not true: each player won a different number of games while playing white. These different numbers then must take all the n available values 0, 1, . . . , n − 1. Let A be the player who won n − 1 games as white, and B the one who won no games as white. Who won the game between them in which A played black? A scored n−1 points playing white, and so won no games playing black. In particular, B won when playing white against A. On the other hand, we know that B won no games playing white, so we have a contradiction. Level B Problem 1. On the 1999th day the product of the numbers on the blackboard will still be the same as on the first day! This product does not change from day to day: 2 a + b 2ab a+b · = · = ab. 2 1/a + 1/b 2 a+b Problem 2. Here is the strategy of the second player. This player skips the first 1000 moves. Her k-th move after that is made so the last 2k+1 letters form a palindrome. Let’s prove by induction on k (see Fact 24) that this is always possible. For k = 0, this is obvious. Suppose that after 1000+(k−1) moves by the second player the last 2k−1 letters form a palindrome. If the next letter added by the first player, which is in position 1000+k, coincides with the one in position 1000−k, the second player has nothing to do. If the (1000+k)-th and (1000−k)-th letters are different, one of them also differs from the letter in the 1000th place. In this case, the second player exchanges it with the 1000th letter. This does not destroy the ealier (2k−1)-letter palindrome, because the middle letter has no match. Now the last 2k+1 letters form a palindrome. After 1999 moves — that is, when k = 999 — the entire word becomes a palindrome. B Problem 3. The angle between a tangent and a A chord drawn from the point of contact equals half the measure of the corresponding arc (see O Fact 15); hence \CBO = \BAC. At the same time we have \BAC = \ACD, because alterD C nate interior angles between two parallels are equal. Thus \CBO = \OCD. Applying the converse of the theorem about the angle between a tangent and a chord, we see that the line CD is tangent to the circle passing through the points B, O, and C.

∗

page 236

∗ moved from below ∗ for layout

184

SOLUTIONS

Remark. The parallelogram considered in the problem is such that √ the ratio of the lengths of the diagonal BD and the side CD is equal to 2. This property can be derived from the similarity of the triangles BCD and COD. Another consequence of this similarity is the fact that the midpoints of its sides are vertices of a parallelogram similar to the given one.

Problem 4. Set n = 1000. Consider two cases (see Fact 11): (a) k > n. Then k

z }| { 2 1 . . . 1 . . 0} . . . 2} |0 .{z . . 2} = |1 .{z . . 1} |2 .{z | {z . . . 2} − 2| .{z 2n

n+1

2n−k k−(n+1) n+1

This number ends in n + 1 = 1001 zeros. But the number of zeros at the end of the decimal representation of a perfect square must always be even! Thus this number is not a perfect square. (b) k ≤ n. Then k

z }| { . . 2} 1| . . . 1{z2 . . . 2} − |2 .{z 2n

n+1

. . 0} − 2| .{z . . 2} 0| .{z . . 0} = 10k (1 . . 1} − |2 .{z . . 2}). (1) = |1 .{z . . 1} 0| .{z | .{z 2n−k

k

n+1−k

k

2n−k

n+1−k

This number ends in k zeros. As we explained above, this number can be a perfect square only if k is even. Set l = k/2. Clearly, the number in (1) is a perfect square if and only if A = |1 .{z . . 1} − |2 .{z . . 2} 2n−2l

n+1−2l

is a perfect square. Notice that
2 1 1 . . 9} − · 9| .{z . . 9} = 102n−2l − 1 − 2(10n+1−2l − 1) ; A = · |9 .{z 9 9 9 2n−2l

n+1−2l

see Fact 11. Let B = 9A. Then A is a perfect square if and only if B is. Write B as B = 102n−2l − 2 · 10n+1−2l + 1 = (10n−l )2 − 2 · 10n−l · 101−l + 1. For l = 1, the right-hand side of this equality coincides with the square of a difference: B = (10n−1 )2 − 2 · 10n−1 + 1 = (10n−1 − 1)2 .

Now suppose that l > 1. We remark that if X = Y 2 is a perfect square (with Y > 0), the perfect square closest to X is (Y − 1)2 = Y 2 − 2Y + 1. That is, if a number Z satisfies Y 2 − 2Y + 1 < Z < Y 2 ,

page 237

YEAR 1999 OLYMPIAD

185

then Z is not a perfect square. We apply this to Y = 10n−l and Z = B. Clearly, Z < Y 2 . In addition, Z = (10n−l )2 − 2 · 10n−l · 101−l + 1 > (10n−l )2 − 2 · 10n−l + 1.

By the remark, this number cannot be a perfect square; therefore, only l = 1 satisfies our requirement, implying that k = 2. A

Problem 5. Denote the lengths of the sides AB, BC, and AC by c, a, and b, respectively. Assume for definiteness that R lies on AC and S on BC. Then (see the remark) P R b a+b−c c−a Q RQ = |RC − QC| = − . = T 2 2 2 Since △TRQ is similar to the isosceles triangle PAQ, we have RQ = R T . Therefore, B S C c c−a a ST = RS − RT = RS − RQ = − = = BS. 2 2 2 It follows that T SB is an isosceles triangle, and \SBT = \ST B = \T BA, which means that BT is the bisector of the angle ABC.

page 238

Remark. In any triangle, the distance from a vertex to the tangency point of the incircle with either side incident on that vertex is equal to p − a, where p is the semiperimeter and a is the length of the opposite side. Proof. Let the incircle touch the sides AB, BC, and AC at C ′ , A′ , and B ′ , respectively. The lengths of the two tangents to a circle drawn from the same point are equal, so the triangle’s perimeter can be written as 2AB ′ + 2BA′ + 2A′ C = 2AB ′ + 2BC. Our statement follows readily.

˜

Problem 6. Each part of this problem is solved by induction (see Fact 24), but the complexity of the reasoning increases sharply. To streamline the exposition, we define an (k, 2n )-ranking to be a set of rankings of 2n players in k events: in other words, an array of k columns, each corresponding to one event and ranking the 2n players according to how well they’re expected to perform in it. The typical situation, then, is this: given an (n, 2n )-ranking and some player, can the events be scheduled in such an order that the player wins the competition — assuming, of course, that the (n, 2n )-ranking is accurate? Part (a) uses induction on n. The base of induction, with n = 1, is obvious: the winner of the single event is one-half of the two players. Suppose we have solved the problem for an n-event competition, so we have an (n, 2n )-ranking Cn admitting 2n−1 potential winners. After adding a new event, we must describe an (n+1, 2n+1 )-ranking Cn+1 admitting 2n potential winners. We do this as follows. First divide the 2n+1 players into two equal groups A and A′ . The defining conditions for Cn+1 are:

∗ ∗

page 239

186

SOLUTIONS

(1) In both groups, the ranking with respect to the old events is given by the (n, 2n )-ranking Cn . (2) Any member of A′ ranks higher in the new event than any member of A, whereas in the old events any member of A ranks higher than any member of A′ . (3) The ranking of members of A with respect to the new event is the same as the ranking for some fixed old event, which we denote by α. (The ranking of members of A′ in the new event is arbitrary.) Now, if we start the competition with the new event, only members of A′ will remain. By the induction hypothesis, half this group are potential winners. If we start the competition with event α, only members of A will remain. We claim that half of them are potential winners as well. By the induction hypothesis, for half the members of A we can find a schedule of events allowing them to become winners. But this includes the event α, which has already taken place! To fix this, we replace event α in the schedule by the new event and use the fact that the ranking within A is the same for the new event as for α. Thus, half of the players in A, as well as half of those in A′ , are potential winners. This completes the induction step. (b) In this case we fix n and an (n, 2n )-ranking. To prove there are n players who are not potential winners, we will find inductively for each event a different player who gets beaten either before or in this event, irrespective of the schedule of events. For the induction we must imagine the n events ordered in a certain way, once and for all. We will call this their alphabetical order. This has nothing to do with the scheduling order. Base of the induction. For the alphabetically first event α, we simply take the player rated lowest in α. This person cannot stay past this event, regardless of when it takes place. Induction step. Suppose that we have selected a set Ak = {a1 , . . . , ak } of players such that ai leaves after the (alphabetically) i-th event or earlier. Of the remaining players, let ak+1 be the lowest-rated in the alphabetically (k +1)-st event. We prove that ak+1 leaves right after this event or earlier, irrespective of the schedule of events. Suppose the alphabetically (k +1)-st event is the r-th in the schedule, and that w players from the set Ak have left after the first r−1 events. In the r-th event, 2n−r players must leave. Therefore, ak+1 can only go on to the next event if 2n−r ≤ k −w (since only k −w of the remaining players can be weaker than ak+1 in (k +1)-th event). But after the alphabetically (k+1)-st event, at least k−w of the k-th alphabetically first events are still to be held. Therefore, k−w ≤ n−r < 2n−r , where we have used that 2l > l for all l. This contradiction concludes the induction step.

∗

∗ multiple times

∗∗ ∗ ∗

page 240

∗ ∗

YEAR 1999 OLYMPIAD

187

(c) To perform the induction in this case we need an auxiliary construction, also inductive. It involves a ranking of 2n players in n+1 events, that is, a (n+1, 2n )-ranking. We borrow the notion of alphabetical order from (b). Lemma 1. There exists an (n+1, 2n )-ranking En such that every player but one is a potential winner in some n-event subcompetition that includes the alphabetically last event. The exception is the player ranked lowest in this event. Before proving this statement, we explain how it is used to solve part (c) of the problem. We use induction on n. The base of induction, for n = 1, is just as in part (a). For the induction step, suppose that we have an (n, 2n )-ranking Dn such that 2n −n players are potential winners. We add a new (alphabetically last) event and describe an (n+1, 2n+1 )-ranking that continues the induction. The construction is similar to the one used in item (a). We again divide the 2n+1 players into two groups of 2n players, A and A′ . The defining conditions for Dn+1 are: (1) The members of A′ are ranked with respect to the old events as in the (n, 2n )-ranking Dn . (2) Any member of A′ ranks higher in the new event than any member of A, whereas in the old events any member of A ranks higher than any member of A′ . (3) The members of A are ranked according to En , the (n+1, 2n )-ranking provided by the lemma. If we start our (n+1)-athlon with the new event, only members of A′ will remain. By the induction hypothesis, 2n −n players from this group are potential winners. Next we show that a member of A that is potential winners in the ranking En is also a potential winner in the ranking Dn+1 . Given such a player, Lemma 1 says that there is a schedule of n events, including the new one, which makes this player win. Let α be the event omitted in this schedule. If we hold α first, it eliminates members of A′ , leaving those of A; now we play out the schedule of n events that ensures the win of this player in the ranking En . Thus, there are 2n −n potential winners in group A′ and 2n −1 potential winners in A. This gives 2n −n+2n −1 = 2n+1 −(n+1) potential winners in the ranking En+1 , as needed. The proof of the lemma also involves induction. In fact we need a slightly stronger statement (the change is highlighted): Lemma 2. There exists an (n+1, 2n )-ranking En such that every player but one is a potential winner in some n-event subcompetition that includes the alphabetically last event. The exception is the player ranked lowest in this event; this player, called the sure loser, nonetheless can progress to the final.

∗

∗

page 241

∗ ∗

188

SOLUTIONS

Base of the induction. We take for E1 the (2, 2)-ranking that says that the winner in the alphabetically first event, denoted by α, is the loser in the second. Induction step. We construct an (n+2, 2n+1 )-ranking En+1 , assuming the existence of En , where we have removed the alphabetically last event. Divide the 2n+1 players into two equal groups, B and B ′ . The defining conditions for En+1 are: (1) For 2 ≤ j ≤ n+2, the members of B are ranked with respect to the alphabetically j-th event in the same order as they are with respect to the (j−1)-st event in En . The same is true for members of B ′ . (2) Any member of B ′ ranks higher in event α than any member of B, whereas in the other events any member of B ranks higher than any member of B ′ . (3) In B, the sure loser according to En is ranked first in α. If we start with α, we have 2n −1 potential winners from B ′ , and we can organize the competition so that the sure loser from B ′ reaches the final, by the induction hypothesis. If we do not hold event α at all, then all players from B ′ leave after the first event in the schedule, and then n events are held. According to the induction hypothesis, by choosing the first event and the sequence of subsequent events, we can guarantee that any of 2n −1 members of B wins. We still have to explain how to arrange a win for the sure loser from B. To this end, we start with the event that is the last in the schedule of events that lets the sure loser reach the final. After this round only members of group B remain. Then we proceed in the sequence that leads the sure loser to the final, and we hold event α last. The sure loser then wins the competition.

∗ ∗ ∗ ∗ ∗ ∗ several times

page 242

∗

∗ ∗

Level C Problem 1. First solution. If a = 0, then b 6= 0 (otherwise c 2 < 0). Then b2 > 0 = 4ac and we’re done. If a 6= 0, we consider the quadratic trinomial f (x) = ax2 + bx + c. We have f (1) = a + b + c and f (0) = c. Hence, by assumption, f (1)f (0) = (a + b + c)c < 0. It follows that one of the numbers f (1) and f (0) is negative and the other is positive. Therefore, the parabola y = f (x) intersects the x-axis, which means that the discriminant of this quadratic polynomial is positive: b2 − 4ac > 0. Second solution. We can avoid the consideration of two cases. Consider the quadratic polynomial g(x) = x2 + bx + ac. It follows from the assumption that g(c) = c2 + bc + ac = (a + b + c)c < 0. Since the branches of the parabola y = g(x) are directed upward and g(c) < 0, this parabola meets the

∗ removed “The...

c”

YEAR 1999 OLYMPIAD

189

x-axis at two points, i.e., it has two distinct roots. Hence the discriminant of this quadratic polynomial is positive: b2 − 4ac > 0. Problem 2. The triangles AP B and DP C are isosceles, since AP , BP , CP , and DP are radii of the third circle. Denote the angles at their bases by \ABP = \BAP = α and \DCP = \CDP = β. The quadrilaterals B ABQP and DCQP are cyclic; hence the angles C Q AQP and ABP both equal α, while \DQP and α ββ β ββ \DCP equal β (see figure). D We have \AQD = \AQP + \DQP = α + β.

A

∗

page 243

P

Further, \BQP = π − \BAP = π − α, similarly, \CQP = π − β. Thus, \BQC = 2π − \BQP − \CQP = α + β.

Problem 3. We first prove that x and y are coprime. Suppose otherwise; that is, let x and y be divisible by some prime number p. Let a ≥ 1 and b ≥ 1 be the exponents of p in the prime factorizations of x and y, respectively. Without loss of generality we can assume that a ≥ b. Then the exponent of the highest power of p that divides x3 + y is equal to b (since x3 is divisible by p3a and hence by pb+1 , whereas y is divisible by pb and not by pb+1 ). On the other hand, x2 + y 2 is divisible by p2b . It follows that x3 + y is not divisible by x2 + y 2 . This contradiction shows that x and y are coprime. Further, x(x2 + y 2 ) − (x3 + y) = y(xy − 1) must be divisible by x2 + y 2 . Notice that y and x2 + y 2 cannot have a common factor greater than 1 (because x and y are coprime); therefore, xy − 1 is divisible by x2 + y 2 (see Fact 9). But this is impossible whenever xy − 1 > 0, because x2 + y 2 ≥ 2xy > xy − 1. Problem 4. The numbers will be called red and blue according to the colors of sectors in which they are written. All the numbers are paired up with their equals; we take a pair of equal numbers that are closest, meaning that the number of sectors in the smaller arc ω lying strictly between the two numbers is minimal. Only the cyclic order of the numbers from 1 to n matters in the problem; a cyclic shift of the numbering through the same amount of sectors for both red and blue numbers replaces the problem by an equivalent one. Therefore, we can assume that the chosen pair consists of 1s. ω We can also assume that the arc ω between them nn 11111 nnnnn n n−1 n n runs counterclockwise from the red sector to the blue 11 1 11 one; see the figure. Further, all the sectors in ω, if there are any, have the same color. Indeed, if both colors are represented, the red and the blue n sectors are both in ω; but then this pair is closer together than the pair of 1s, contrary to our assumption.

∗

page 244

∗

190

SOLUTIONS

Suppose that all the numbers on this arc (if there are any) are blue; the case of red numbers is similar. Draw the diameter separating the blue 1 from its clockwise neighbor (this is either blue n or red 1); we will show that this is the diameter we need. Indeed, consider the half-disk containing the blue 1. The blue numbers in this half-disk read up, counterclockwise, from 1 through some positive integer l. Now imagine reading the red numbers, also counterclockwise. Since there are no red numbers in the arc ω, the first number in our half-disk is n. It follows that the red numbers we read are the numbers n, n − 1, . . . , n − m, where m is also a positive integer. Thus, there are l blue and m + 1 red numbers in this half-disk. Since all in all there are n numbers in the half-disk, we have l + (m + 1) = n; that is, n − m = l + 1. Therefore, the numbers in the half-disk are the blue numbers from 1 to l and the red numbers from n to l+1. Together these amount to all the numbers from 1 to n, each taken once.

∗

Problem 5. Let

√ f : [0, 1] → [0, 1], f (x) = x/ 3 and √ √ g : [0, 1] → [0, 1], g(x) = x/ 3 + (1 − 1/ 3)

be the functions describing the jumps of the grasshopper. The range of f √ √ is the interval [0, 1/ 3],√while the range of g is [1 − 1/ 3, 1]. Each of these intervals is of length 1/ 3 and together they cover all of [0, 1]. Let n be a positive integer. Consider all possible compositions of functions of the form

page 245

h1 (h2 (. . . (hn (x)) . . . )) : [0, 1] → [0, 1], where each function hi is either f or g. It can be readily √ seen that the range of any of these functions is an interval of length (1/ 3)n . We’ll prove by induction on n (see Fact 24) that these intervals cover all of [0, 1]. For n = 1 this statement has already been verified. Suppose that the ranges of all possible functions h1 (h2 (. . . (hk−1 (x)) . . . )) cover [0, 1]. But the range of a particular h1 (h2 (. . . (hk−1 (x)) . . . )) is covered by the ranges of the functions h1 (h2 (. . . (hk−1 (f (x))) . . . )) and h1 (h2 (. . . (hk−1 (g(x))) . . . )). This proves the statement. Now suppose that a point a is chosen on the interval [0, 1]. Consider the interval (a − 0, 01, a + 0, 01). We show that the grasshopper can hit √ it. Choose n large enough to satisfy the inequality (1/ 3)n < 0, 01 (for instance, n = 10). As we have proved, it is possible to find a function h1 (h2 (. . . (hn (x)) . . . )) whose range contains a. Then the entire range of this √ function — an interval of length (1/ 3)n — lies inside (a − 0, 01, a + 0, 01). This means that starting anywhere in [0, 1] the grasshopper will hit the interval (a − 0, 01, a + 0, 01) after making the jumps corresponding to the functions hn , hn−1 , . . . , h1 . Remarks. (a) Compare with Problems 99114 and 99117.

REF 99114 REF 99117

YEAR 1999 OLYMPIAD

191

(b) This problem, as well as Problem 99114, are based on properties of contracting mappings. A differentiable mapping is said to be contracting if its derivative REF 99114 has absolute value everywhere less than some fixed number δ < 1. Let f : [0, 1] → [0, 1] and g : [0, 1] → [0, 1] be two contracting mappings, and let x be a point on the interval [0, 1]. It is natural to ask: What can we say about the position of this point after it undergoes a large number of the consecutive mappings f and g? More exactly, consider an infinite sequence of mappings f and g, and let xn be the image of x after the first n mappings. page 246 Since the mappings are contracting, limn→∞ xn does not depend on x. However, it depends on the order in which the mappings f and g are applied. If the ranges of the mappings f and g cover the entire interval [0, 1], then xn can approach any point in the interval [0, 1] (in fact, this is just the statement of our problem). Otherwise, the set of possible limits is like the Cantor set. For instance, the functions f (x) = 31 x and g(x) = 13 (x + 2) generate exactly the usual Cantor set.

Problem 6. Lemma. Suppose that 1999 distinct positive numbers a1 , a2 , . . . , a1999 , are arranged around a circle, and that a1 > a1998 . For all i = 2, 3, . . . , 999, we perform the following operation: the numbers ai and a1999−i are swapped if ai < a1999−i , and stay in place otherwise. If at least one pair is swapped, then the sum of all products of ten consecutive numbers increases. Proof. Consider two groups of 10 numbers in a row, arranged symmetrically: ai , . . . , ai+9 and a1999−i , . . . , a1990−i . Take the sums of the products of the numbers in either group. We first show that this sum can never decrease. Consider the product z of the numbers appearing in both groups. (If there are no such numbers, we set z = 1.) Also let x and x′ be the products of the numbers that belong only to the first or only to the second group, respectively, and that don’t get swapped; while y and y ′ are the products of the numbers that belong only to the first or only to the second group and do get swapped. (Again, if no numbers satisfy a specified condition, the corresponding product is defined as 1.) The sum of products of numbers in the two groups before the operation is s1 = zxy + zx′ y ′ ; after the operation, it is s2 = zxy ′ + zx′ y. We have s1 − s2 = z(x − x′ )(y − y ′ ). It is easily seen that x′ ≤ x and y ′ ≥ y. Hence s1 − s2 ≤ 0. Now we consider all the numbers again. We show that if some of the numbers are swapped when the operation of the lemma is performed, the difference s1 − s2 is strictly negative for at least one of the symmetric pairs of groups of 10 numbers. This will imply the lemma. Clearly, if at least one pair of numbers (in a given group) was swapped, then y ′ > y. And if at least one pair of numbers stayed in place, then x′ < x, since all the numbers are different. So it suffices to show that we can find two symmetric groups of ten numbers each for which at least one pair was swapped and at least one pair was not. But this is obvious, because by assumption some pair was swapped and the pair (a1 , a1998 ) was not. ˜

∗

page 247

∗

192

SOLUTIONS

To apply the lemma to the problem, we assume that the numbers from 1 through 1999 are placed at the vertices of an 1999-gon, and that the desired condition is satisfied: the sum of all products of numbers taken 10 in a row is maximal. Draw the diameter through some number k. We claim that, for all pairs symmetric about this diameter, the smaller number lies in one semicircle and the larger one in the other. Indeed, denote by a1 , . . . , a1999 the numbers around the circle, starting from the greater of the neighbors of k and ending with k. Then a1 > a1998 , and we can apply the lemma: since the arrangement is optimal, we cannot increase our sum of products by swapping symmetric numbers, hence all numbers on one side of the diameter are greater than the corresponding numbers on the other side. Ignoring rotations and reflections, there is only one arrangement of numbers satisfying this property for all diameters. We show this by induction (see Fact 24). First, 2 must be next to 1; otherwise we can find a diameter separating 2 from 1 and such that the two numbers are not symmetric about this diameter. Denote by A and B the numbers symmetric to 1 and 2 about this diameter. Then A > 1 and 2 < B, contradicting the claim in the previous paragraph. 2 1 Suppose that we have proved that 1, 2, . . . , 2k, where 1 ≤ k ≤ 998, must be arranged as in the answer, that 2k−1 is, in the order 2k, 2k − 2, . . . , 2, 1, 3, . . . , 2k − 1 2k B (say clockwise, for definiteness). Denote by A and B A the numbers following 2k counterclockwise and 2k −1 2k+1 clockwise; see figure. Suppose that the number 2k+1 is C distinct from A and B. Then let C be the clockwise neighbor of 2k +1. The number C is distinct from 1, 2, . . . , 2k. The numbers C and 2k−1, as well as 2k+1 and B are symmetric about the diameter, but C > 2k−1 and 2k +1 < B. This is a contradiction, which means that either A = 2k +1 or B = 2k + 1. But the assumption A = 2k + 1 immediately leads to a contradiction: it suffices to consider the diameter which is the symmetry axis of the numbers 2k and 2k−1. Hence B = 2k+1. In a similar way we show that A = 2k+2, completing the induction step.

∗ ∗ page 248

Level D Problem 1. By the triangle inequality, a > |b − c|. Squaring both sides we get a2 > (b − c)2 . Hence a2 + 2bc > b2 + c2 . The right-hand side here is positive, so we can divide both sides by it. After the division, we see that the first term in the inequality we are proving is greater than 1. The same is true for the other two terms. Therefore, their sum is greater than 3.

∗

YEAR 1999 OLYMPIAD

193

Problem 2. (a) It is not difficult to construct the midpoint of the broken line BAC, that is, the point that divides the broken line into two broken lines of equal length. Clearly, the line joining this point to the midpoint of the arc is the desired one. D l (b) Let D be the midpoint of the given arc. The B shaded caps shown in the figure are clearly congruent. Therefore, it suffices to draw the line through D F A E bisecting the area of the quadrilateral ABDC. Let F be the midpoint of the diagonal BC, and let C l be the line drawn through F parallel to the diagonal AD. For definiteness, assume that the line l intersects the segment AC (the case in which l intersects the segment AB is considered similarly). If E is the intersection point, then DE is the line to be constructed. Indeed, the sum of the areas of triangles ABF and BDF is half the area of ABDC. But this sum is also equal to the area of quadrilateral ABDE, because the triangles AED and AF D have a common side AD and equal altitudes dropped on this side, and so have equal area.

Problem 3. First solution. The planes containing four faces of the same color bound congruent regular tetrahedra. To grasp why, imagine a cube ABCDEF GH and the two regular tetrahedra ACFH B and BDEG. The intersection of these tetrahedra C is a regular octahedron. Indeed, the vertices of A D D D this intersection are the centers of the cube’s faces, and the centers of a cube’s faces are vertices of a regular octahedron. (See the figure.) The black faces of the given octahedron lie on F G one tetrahedron, and the white faces on the other. E The result in the problem follows from the fact H that the sum of distances from a point in a regular tetrahedron to its faces is constant and equals three times the volume of the tetrahedron divided by its face area. To prove this last statement, let A, B, C, and D be the vertices of the tetrahedron. Denote by hA , hB , hC , and hD the distances from a point O inside the tetrahedron to the planes BCD, ACD, ABD, and ABC, respectively, and by S the area of any face of the tetrahedron. Then the volumes of the tetrahedra BCDO, ACDO, ABDO, and ABCO are equal to 31 ShA , 31 ShB , 13 ShC , and 13 ShD , respectively. Therefore, the volume of the tetrahedron ABCD is equal to

∗ page 249 ∗ ∗ ∗∗

∗ removed “altitude”

S (h + hB + hC + hD ), 3 A which is equivalent to our statement. Second solution. (Above grade level.) Consider the directed distances from a point to the face planes. More exactly, if the point and the octahedron are on the same side of the plane, then the distance is taken with the plus sign

page 250

194

SOLUTIONS

(so the directed distance for these points is just the ordinary one); otherwise, the sign is reversed. For brevity, the distance to the plane containing a face will be called the distance to that face. We prove a claim stronger than that of the problem: The sum of the directed distances from any point to the black faces is equal to the sum of the directed distances to the white faces, even for points lying outside the octahedron. The directed distance to a plane is a linear functional in space (see Fact 25 on page 208). Hence the sum of distances to the faces of a certain color is also a linear functional. Denote the sum of distances from a point X to white faces by lw (X) and a similar sum for the black faces by lb (X). We want to prove that lb − lw = 0. If this were not so, the set of zeros of the linear functional lb − lw is a plane. But it is readily seen that it vanishes at all the vertices of the octahedron. Hence lb − lw is identically zero. Problem 4. Suppose the square has side length 1. Divide each side into 2n equal segments and draw the lines parallel to the sides through all partition points. They cut the square into small squares with side length 2−n . If n is sufficiently large, then one of these small squares will be completely inside the hole (for instance, for n such that 2−n is less than half the radius of the hole, we can take the small square containing the center of the hole). Therefore, it will suffice to prove that for any n the grasshopper can hit any of the 22n small squares. We will prove this by induction (see Fact 24). For n = 0 the claim is trivial. Let us describe the induction step from n to n + 1. Consider a 2−n−1 × 2−n−1 square Q. Cut the initial square into four squares with side length 12 . Without loss of generality we can assume that Q lies in the bottom left square, whose outer corner we denote by A (see figure). The dilation with center A and ratio 2 maps the chosen small square Q onto a square Q′ with side length 2−n . Clearly, this is one of the squares obtained by cutting the initial A square into 22n squares with side length 2−n . By the induction hypothesis, the grasshopper can get into Q′ . Now, if it jumps half the distance to the vertex A, it will hit the target square Q. Remark. Compare with Problems 99105 and 99117.

Problem 5. We number the colors of the nodes from 1 to k. Then we take the nodes of color 2 not adjacent with nodes of color 1 and reassign them color 1. The new coloring is regular as well; therefore, it involves k colors. This means that some of the nodes of color 2 were not recolored; hence they are adjacent with nodes of color 1. Next we take the nodes of color 3 not adjacent with nodes of color 2 (that were not recolored at the first step), and we reassign them color 2. We

∗

page 251

∗ ∗ ∗ ∗ REF 99105 REF 99117

∗

YEAR 1999 OLYMPIAD

195

continue in this fashion with all the colors; at the last step we reassign color k − 1 to certain nodes of color k. After that, consider any node of color k. It has not been recolored; therefore, it is adjacent with a node of color k − 1. This node has not been recolored either: otherwise, its initial color would be k and it would be adjacent to a node of the same color, which is impossible by the assumption of regularity. Since the last node preserved its initial color after the recoloring, it is adjacent with a node of color k − 2, and so on. This process eventually yields a path of length k visiting the nodes of all the k colors that were not changed.

removed “See also”

Remark. We have, in fact, proved a stronger statement: there exists a path that visits nodes of all k colors, each color once, in a given order.

Problem 6. First solution. Let p be a prime factor of l. Since nm = (1 + nk )l − 1, the number nm is divisible by (1 + nk )p − 1 (see Fact 8). But, by the binomial theorem (see Remark below), we have (1 + nk )p − 1 = nk p + n2k

p(p − 1) + n3k r, 2

where r is a nonnegative integer. Dividing both sides by nk , we see that nm is divisible by p(p − 1) p + nk + n2k r. 2 If n is divisible by p, then this expression is coprime with n, and nm cannot be divisible by it. Hence p is a divisor of n (see Fact 9). Then 1 + nk

page 252

n2k

p−1 + r 2 p

is a positive integer greater than 1. If k > 1 or p is odd, then the second term is divisible by n (the third term is always divisible by n); therefore, the sum is coprime with n, and so is not a divisor of nm . This contradiction shows that k = 1 and p = 2. Therefore, 2 is the only prime factor of the number l and l can be written as 2s . Using the binomial theorem again, we have nm = (1 + nk )l − 1 = (1 + n)l − 1 = ln +

∗

∗

l(l − 1) 2 n + · · · + nl . 2

After the first, all terms on the right-hand side are divisible by n2 . Since m > 1, it follows that l is divisible by n. Therefore, n, as well as l, is a power of two. Since l is even, (1 + n)l − 1 is divisible by (1 + n)2 − 1 = n(n + 2). Since m n is a power of two, the number n + 2 is also a power of two. Since n and n + 2 are powers of 2, we see that n = 2. If l ≥ 4, then (1 + n)l − 1 is divisible by (1 + n)4 − 1 = 80, and cannot be a power of two. Therefore, l = 2, whence m = 3.

∗ ∗

196

SOLUTIONS

Outline of second solution. Since 1 + nm is divisible by 1 + nk , m is divisible by k for n 6= 1 (see Fact 8). Therefore, replacing nk by n and m/k by m, we reduce the problem to the case k = 1. Suppose that n is divisible by pt , but not by pt+1 , where p is a prime number (t > 0). Let ps be the greatest power of p that divides l. Now use the binomial theorem (see Remark below) to write

∗ page 253

ln + Cl2 n2 + · · · + nl = nm .

Assume that p 6= 2 or t > 1. One can show that the right-hand side and and removed earlier all summands except the first are divisible by pt+s+1 . The contradiction ∗ paragraph obtained shows that n = 2. Remark. The formulas for (a + b)2 and (a + b)3 are well known. A similar formula for (a + b)n , for any n, is the subject of the binomial theorem. To write this formula, we’ll need the notion of combinations. The number of combinations of n elements taken k at a time is the number of ways of picking k things out of a set of n different things, without regard to order. This number is denoted
by nCk or nk , and is sometimes read “n choose k”. We stress that the order in which the k elements are picked does not matter; in other words, nCk is the number of k-element subsets in an n-element set. This number is given by n Ck

=

n(n − 1)(n − 2) . . . (n − k + 1) n! = . k!(n − k)! k (k − 1) · · · 2 · 1

The binomial theorem (or binomial formula) now says that (a + b)n =

n X

nCk a

k n−k

b

k=0 n

= a + nC1 an−1 b + nC2 an−2 b2 + · · · + nCn−2 a2 bn−2 + nCn−1 abn−1 + bn .

For this reason the nCk are also called binomial coefficients.

Problem 7. If we think of a circle of length 1 as the interval [0, 1] with endpoints identified (compare solution to Problem 56.10.4), then the fractional part fm of the number log10 (2m ) = m log10 2 can be viewed as a point on this circle. Consider the points

REF 56.10.4

0, log10 2, . . . , log10 9 on the circle and the nine half-open intervals into which they divide the circle. Denote these intervals by I1 = [0, log10 2), . . . , I9 = [log10 9, 0). The first digit of the number 2m is equal to s if and only if fm belongs to the interval Is . For instance, if 2m begins with 7, then 7 · 10l ≤ 2m < 8 · 10l

for a certain positive integer l. The fractional part of m log10 2 is equal to m log10 2 − l and lies between log10 7 and log10 8. n Suppose that the first digits of 22 repeat with period k, after some preperiod of length n0 . Then the fractional parts of 2n log10 2 and 2n+k log10 2 hit the same interval Is for any n > n0 .

page 254

YEAR 1999 OLYMPIAD

197

It can be readily seen that the longest of the intervals is the first one, and its length is log10 2 < 13 . Step 1. Mark the fractional parts of two positive numbers A and B on the circle. Suppose these fractional parts are distinct and are not antipodal points of the circle. Let x be the shorter of the two arcs into which these points divide the circle. Then the length of one of the arcs joining the fractional parts of the numbers 2A and 2B is equal to 2x. (Explain why.) Step 2. Now suppose that the fractional parts of the numbers A and B lie in the same interval Is , and consider the pairs 2A and 2B, 4A and 4B, etc. It follows from Step 1 that the length of the shorter arc joining the fractional parts of a pair is doubled until it becomes greater than or equal to 12 . Therefore, at a certain step, one of the arcs joining the fractional parts of a pair will become greater than 31 , but less than 23 . Then these fractional parts belong to different interval on the circle. Step 3. Consider the numbers A = 2n0 log10 2 and B = 2n0 +k log10 2. These numbers considered as points on the circle are different and are not antipodal, since log10 2 is irrational (see below). Therefore, we can apply Step 1 to these numbers, which yields a contradiction with the periodicity assumption. It remains to prove that log10 2 is an irrational number. If log10 2 = p/q, then 2p = 10q ; this is obviously impossible, by prime factorization.

∗

Remarks. (a) We have in fact proved that if α is not a rational power of ten, the n sequence of the first digits of α2 is nonperiodic. n On the other hand, if we look instead at the sequence α10 , it is possible page 255 to find α that is not a rational power of ten and the sequence of first digits is periodic. For instance, let log10 α = 0.101001000100001 . . . (the number of zeros between consecutive ones increases). This decimal is not periodic; hence α is not a rational power of ten (see Fact 13). But for any n we ∗ have {10n log10 α} < 0.11 < log10 2. Therefore, for all n, the first digit of the n number α10 is 1! ∗ (b) Problems 99105 and 99114 are about contracting mappings of an interval into REF 99105 itself (see the remarks to those probelms). The problem we have just considered REF 99114 is a problem about expanding mappings.