28 Nov 2012 ... Construct two smaller circles with diameters AC and. CB as in the figure. ....
Those of the form m × n, where mn is not divisible by 3. 4. Find the ...
Russian Math Circle Homework Problems Nov 28, 2012 Instructions: Work as many problems as you can. Even if you can’t solve a problem, try to learn as much as you can about it. 1. Let x, y, z, v, and w be nonnegative integers. If 2x+y+z + 2y+z+v + 2z+v+w + 2v+w+x = 1089 show that x + y + z + v + w is a perfect square. (Warning: This problem is flawed, so if you’re looking for problems to use in a contest of your own, I’d skip this one!) 2. A group of airplanes is based on a small island. The tank of each plane holds just enough fuel to take it halfway around the world. Any desired amount of fuel can be transferred from the tank of one plane to the tank of another while the planes are in flight. The only source of fuel is on the island, and we assume that there is no time lost in refueling either in the air or on the ground. What is the smallest number of planes that will ensure the flight of one plane around the world on a great circle, assuming that the planes have the same constant ground speed and rate of fuel consumption and that all planes return safely to the island base? 3. An ell is an L-shaped tile made from three 1 × 1 squares (see picture below). For what positive integers a, b is it possible to completely tile an a × b rectangle only using ells? For example, it is clear that you can tile a 2 × 3 rectangle with ells, but (draw a picture) you cannot tile a 3 × 3 rectangle with ells.
4. Find the maximum number of crossings of the diagonals of a non-regular n-sided convex polygon. Prove that your result is correct. 5. (a) In how many ways can the number 10 be represented as the sum of exactly 5 non-negative numbers, if the order is taken into account? In other words, 4 + 0 + 0 + 3 + 3 is considered different from 4 + 3 + 0 + 0 + 3. (b) In how many ways can the number n be represented as the sum of exactly m non-negative numbers with order taken into account? 6. Let C be any point on the diameter AB of a circle. Construct two smaller circles with diameters AC and CB as in the figure. Let D and E be the points of intersection of a common external tangent to the two smaller circles, and let F be the intersection with the larger circle of the line perpendicular to AB passing through C that is on the same side of AB as the points D and E. Show that CDF E is a rectangle. F
D E
C A
1
B
Russian Math Circle Homework Solutions 1. Let x, y, z, v, and w be nonnegative integers. If 2x+y+z + 2y+z+v + 2z+v+w + 2v+w+x = 1089 show that x + y + z + v + w is a perfect square. (Warning: This problem is flawed, so if you’re looking for problems to use in a contest of your own, I’d skip this one!) Solution: Every term is a perfect power of 2, and in binary we have: 1089 = 1024 + 64 + 1 = 210 + 26 + 20 . This means that one of the exponents has to be zero, and either two of them are 9 and one is 6 or two of them are 5 and one is 10. There are basically two different cases: • Case 1: x + y + z = 0 (or v + w + x = 0). The first yields a system where one of the exponents is just v and the other two are v + w. (The second generates one that looks the same, but with y taking the place of v and z taking the place of w.) Since all the variables are non-negative, this means that v + w = 9 and v = 6, so w = 3 and x + y + z + v + w = 9: a perfect square. • Case 2: y + z + v = 0 (or equivalently, z + v + w = 0). This yields the equations x = w = 5 and x + w = 10, but this makes x + y + z + v + w = 10 which is not a perfect square, so the problem is flawed. 2. A group of airplanes is based on a small island. The tank of each plane holds just enough fuel to take it halfway around the world. Any desired amount of fuel can be transferred from the tank of one plane to the tank of another while the planes are in flight. The only source of fuel is on the island, and we assume that there is no time lost in refueling either in the air or on the ground. What is the smallest number of planes that will ensure the flight of one plane around the world on a great circle, assuming that the planes have the same constant ground speed and rate of fuel consumption and that all planes return safely to the island base? Solution: Let’s let 1 unit of fuel be one full tank, which can fly 1 unit of distance which is half-way around the world. Here is a method. Planes 1, 2 and 3 take off together in the same direction, but after traveling 1/4 unit, plane 3 fills the tanks of planes 1 and 2 and returns. Now planes 1 and 2 have full tanks and are 1/4 unit out. They fly together for 1/4 more unit at which point plane 2 fills plane 1’s tank and returns. Now plane 1 is 1/2 unit out and can fly for a full additional unit so will be out of gas after traveling 3/2 units. At an appropriate time, planes 2 and 3 take off in the opposite direction. At 1/3 of the way out, plane 3 fills 2’s tank (so 2 is full) and 3 returns to the island. Plane 2 continues another 1/6 out to meet plane 1 just as plane 1 runs out of fuel, at which point plane 2 splits its gas in half, leaving both with 5/12 of a tank: not quite enough to get back. But when plane 3 returns to the island, it can refill and take off to meet planes 1 and 2 1/6 of the way out, at which point, with 5/6 of a tank remaining, there is plenty of fuel for all three to return. So the circumnavigation can be accomplished with 3 planes. It cannot be done with 2 planes, since the support plane can never get out farther than 1/2 unit, and at that point, it has no gas to spare. Thus there is no way for the circumnavigating plane to travel the distance between 1/2 and 3/2 since it needs a full tank, starting at 1/2 to do that. 3. An ell is an L-shaped tile made from three 1 × 1 squares (see picture below). For what positive integers a, b is it possible to completely tile an a × b rectangle only using ells? For example, it is clear that you can tile a 2 × 3 rectangle with ells, but (draw a picture) you cannot tile a 3 × 3 rectangle with ells.
2
Solution: Clearly rectangles of the form 1 × n are impossible, and if an a × b rectangle is possible, then ab must be a multiple of 3, so assume that at least a is a multiple of 3. If then b is a multiple of 2 then there’s an easy solution: break the rectangle into 2 × 3 smaller rectangles and each is easy to tile. If a is a multiple of 6 and b > 1 there is a solution. The cases 6 × 2, 6 × 3, 6 × 4 are easy and covered above. It is not hard to work out a 6 × 5 example, and by adding rows of size 6 × 2 we can extend the 6 × 5 example to any 6 × n, where n is odd. If n is even, it is trivial. So if a is a multiple of 3 and ab contains no multiples of 2, are there any solutions? Note that 3 × n is impossible for n odd, since it has to start with two interlocked ell units. It is possible to tile a 5 × 9 rectangle:
The only rectangles we don’t know about yet have an odd multiple of 3 as the length of one side and an odd size larger than 3 as the length of the other. We can take the rectangle above and increase the width by any odd multiple of 3 by adding 6 × 5 blocks to the right side. We can also take any of those and add 2 × n blocks, where n is a multiple of 3 to the top to make any odd size in that direction. Therefore, here is the final result: Every rectangle can be tiled by ell tiles except: • Those of the form 1 × n, for any n. • Those of the form 3 × n, for odd n. • Those of the form m × n, where mn is not divisible by 3. 4. Find the maximum number of crossings of the diagonals of a non-regular n-sided convex polygon. Prove that your result is correct. Solution: In an n-sided polygon, from each vertex there are n − 3 diagonals. It is not hard to count the maximum number for small polygons, and we obtain for n = 4, 5, 6, 7 the results 1, 5, 15, 35 which look like the binomial coefficients n4 = n(n − 1)(n − 2)(n − 3)/24. Using the above observation, the easiest solution is this: There are n vertices on the polygon, and there are n 4 ways to choose exactly 4 of them. For any 4 vertices, there is exactly one way to arrange diagonals connecting them so that they cross, so there are exactly n4 diagonal intersections, maximum. But we can also solve the problem by directly counting the crossings, although it is a bit messy: As motivation, consider a polygon with 9 sides. If we start drawing diagonals from one vertex and advance clockwise through the vertices, the first set of 9 − 3 = 6 diagonals will make no crossings. The 5 remaining diagonals from the second vertex will cross 1, 2, 3, 4, 5 others. From the third, there will be 2, 4, 6, 8. From the fourth, 3, 6, 9 crossings. Then 4, 8, and the final diagonal will make 5 crossings. Arrange these numbers in a table: 1 2 3 4 5
2 4 6 8
3
3 6 9
4 8
5
And the sum of the row sums is 15 + 20 + 18 + 12 + 5 = 70 =
8 4
.
If we read along the diagonals, we can express the numbers in the table as, since what we’re looking at is basically the top corner of an elementary school multiplication table: 1·1 2·1
1·2
3·1 4 ·1 5·1
2·2 3·2
1·3 2·3
4·2
3·3
1·4 2·4
1·5
The pattern continues for larger n. One nice way to see why this is true is to think of a large polygon and to consider successive diagonals from a given vertex. The first diagonal will have one vertex on one side and all the others (say k of them) on the other, so it will be crossed by all diagonals connecting the one to the many; 1 · k. Now advance one vertex. There will be two vertices on one side and one fewer (k − 1) on the other. But now there are 2 · (k − 1) diagonals connecting the 2 to the k − 1 vertices. As we move around, these numbers change to 3 and k − 2, then 4 and k − 3, et cetera, making the exact pattern we see above. It’s not hard to prove that the rows add to 30 , 41 , 52 , 63 , 74 , et cetera. These are just the numbers in a diagonal in Pascal’s triangle and by the ”hockeystick theorem,” the sums are 40 , 51 , 62 , 73 , et cetera, which is exactly what we were trying to show. 5. (a) In how many ways can the number 10 be represented as the sum of exactly 5 non-negative numbers, if the order is taken into account? In other words, 4 + 0 + 0 + 3 + 3 is considered different from 4 + 3 + 0 + 0 + 3. (b) In how many ways can the number n be represented as the sum of exactly m non-negative numbers with order taken into account? Solution: Imagine that there are 10 of the x symbols and 4 of the | symbols arranged in a line in some order like this: xxx|xx||xxxxx| Think of the vertical lines as being boundaries separating the line of x’s into 5 groups: those to the left of the first |, those between the first and second |, et cetera, and finally, those to the right of the final |. The grouping above is equivalent to the pattern: 3 + 2 + 0 + 5 + 0. Every possible arrangement of 10 of the x’s and 4 of the |’s yields a different way to represent 10 as such a sum. There are 10 + 4 = 14 slots and 4 of them must contain a |, so there are 14 4 = 1001 ways to do it. To represent n as a sum of m non-negative numbers, using the same argument, there are n+m−1 ways to m−1 do it. 6. Let C be any point on the diameter AB of a circle. Construct two smaller circles with diameters AC and CB as in the figure. Let D and E be the points of intersection of a common external tangent to the two smaller circles, and let F be the intersection with the larger circle of the line perpendicular to AB passing through C that is on the same side of AB as the points D and E. Show that CDF E is a rectangle.
4
F
D E
B
C A
Solution: W.L.O.G. assume AC is the larger circle. (If the two circles are equal, the problem is trivial.) First construct lines from D and E to the centers of the circles and then through E draw a line parallel to AB intersecting DX at W as in the figure below: F
D O E W
A
X
C
Y
B
If AX = x and CY = y by the Pythagorean theorem we have: DE 2 + (x − y)2 = (x + y)2 . This yields DE 2 = 4xy. We also know that F C is the geometric mean of AC and CB, so F C 2 = 4xy, so F C = DE. Since OD = OC and OC = OE (common external tangents) we know that DE and F C bisect each other, making CDF E a rectangle.
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