a problem solving approach to some applications of

INTERNATIONAL JOURNAL OF GEOMETRY Vol. 1 (2012), No. 2, 46 - 60

A PROBLEM SOLVING APPROACH TO SOME APPLICATIONS OF THE ROOTS OF UNITY TO REGULAR POLYGONS MARCELINA MOCANU

Abstract. The aim of this note is to emphasize the unity between geometry, algebra and analysis, by means of two applications of complex numbers to the study of regular polygons. We use an heuristic problem-solving approach to …nd a complete description of lattice regular polygons and a formula for the minimum of the sum of squared reciprocals of the distances from a point on a circle to the vertices of a regular polygon inscribed in that circle. We also give an example of the use of mathematical software in arriving at a conjecture for calculating the above-mentioned sum. 1. Introduction The connection between regular polygons and the roots of unity plays a fundamental role in algebra and geometry. The images of the n-th roots of unity in the complex plane are the vertices of a regular n-gon inscribed in the unit circle. Moreover, a positively oriented n gon in the complex plane is regular if and only if the a¢ xes of its vertices can be obtained as the images of the n-th roots of unity under a polynomial function of …rst degree. The multiplicative group of nonzero complex numbers contains only one subgroup with n elements, which is the group Un of the n-th roots of unity. There are some corresponding isomorphic groups of geometric transformations, the group of rotations about the origin in plane, respectively its subgroup preserving each regular n gon centered at the origin. This connections between regular polygons and roots of unity explain, for example, why the problem of constructibility of regular polygons by ruler and compass, as well as the derivation of several trigonometric identities involving integer multiples of 2n are in essence algebraic problems. ————————————– Keywords and phrases: regular polygon, lattice polygon, roots of unity (2010)Mathematics Subject Classi…cation: 97G40, 51M04. Received: 21.06.2012. In revised form: 3.07.2012. Accepted: 10.07.2012.

A problem solving approach to some applications of the roots of unity to regular polygons 47

On the other hand, in the process of solving algebraic problems involving the roots of unity, the geometric representation is meaningful and enlightening. For example, the property: "The sum of the n distinct n roots of unity is zero", that usually is proved using the Viète relations for the polynomial z n 1 or the sum of a geometric progression, shows that the sum of the vectors from the center of a regular polygon to its vertices is the null vector. Note that, for every prime p; the sum of k distinct vectors joining the center of a regular p gon to its vertices cannot be null if 1 k p 1, a fact that is a consequence of the irreducibility in Z [X] of any cyclotomic polynomial in X. In this note we outline a problem solving approach to two applications of complex numbers to the study of regular polygons. The …rst application leads to a complete description of lattice regular polygons. The second application provides a formula for the minimum of the sum of squared reciprocals of the distances from a point on a circle to the vertices of a regular polygon, which is inscribed in that circle. We prefer to use an heuristic approach: rather than to write down a short and elegant solution, we try to explore several paths to a solution. We also give an example of the use of mathematical software in arriving at a conjecture for calculating the abovementioned sum. Our aim is not only to solve two speci…c problems, but also to give perspectives on other related problems. 2. The existence of lattice regular polygons Given a Cartesian coordinates system in a plane, it is said that a point of the plane is latticial if it has integer coordinates. A plane polygon is called a lattice polygon, with respect to a Cartesian coordinates system, if its vertices are latticial points. The following problem, although a classical one, was on the shortlist for the International Mathematics Olympiad 1985. Problem 1. For which integers n 3 does there exist a regular n gon in the plane such that all its vertices have integer coordinates in a Cartesian coordinates system? We give three solutions to this problem, some of them opening the way to other related more general issues. Necessary condition. Assuming that there exists a lattice regular n gon, we try to …nd a subset of N, as small as possible, where n belongs. First solution. We provide a formula for the a¢ xes of the vertices of a regular n gon, assuming that the a¢ xes of two consecutive vertices are known. Let A1 A2 ::An be a regular n gon in the plane and let zk be the a¢ x of Ak , k = 1; n. We may assume that A1 A2 ::An is positively oriented, i.e. A1 ! A2 ! ::: ! An ! A1 gives the trigonometric orientation. At a …rst step, we determine the center C of the given regular n gon, by calculating its a¢ x z0 in terms of z1 and z2 . At a second step, we take into account that Ak is the image of A1 under a rotation by the angle 2kn around C.

48

Marcelina Mocanu

First step. Denote by l the length of the polygon’s sides and let 2 2 ! := cos + i sin : n n 1

Then CAk = l 2 sin n . Performing a rotation by the angle 2 1 A1 , then a homotethy of center A1 and ratio 2 sin n , we get z0

z1 = 2 sin

Denoting a =

1 2

1

cos + i sin n 2 n 2 1 + i cot n , this yields z0 = (1

(z2

n

n

around

z1 ) :

a) z1 + az2 :

Second step. We have zk z0 = ! k 1 (z1 z0 ) for k = 1; n, hence zk = 1 ! k 1 z0 + ! k 1 z1 . We obtain the following Lemma 2.1. Let A1 A2 ::An be a positively oriented regular n gon in the plane and let zk be the a¢ x of Ak , k = 1; n. Then zk = 1

a + a! k

1

z1 + a

a! k

1

for k = 1; 2; ::; n, where ! := cos 2n + i sin 2n and a =

1 2

z2 , 1 + i cot n .

We will not use the full power of our assumption that Re zk , Im zk 2 Z for all k 2 f1; 2; :::; ng. Instead, we will analyze the coordinates of the center C of the regular polygon. The following well-known result is proved for the sake of completness. Lemma 2.2. The a¢ x of the center of a regular polygon is the arithmetic mean of the a¢ xes of polygon’s vertices. Proof. With the above notations, zk z0 = ! k Then we have n n X X (zk z0 ) = (z1 z0 ) ! k 1 = (z1 k=1

hence 0 =

n X k=1

k=1

(zk

z0 ) =

n X

zk

1 (z

1

z0 )

z0 ) for k = 1; n. 1 !n = 0; 1 !

nz0 , q.e.d.

k=1

Assuming that A1 A2 ::An is a lattice polygon, it follows that its center has rational coordinates. Denote xk := Re zk , yk := Im zk 2 Z for k 2 f0; 1; 2; :::; ng. We have x0 = 12 (x1 + x2 )+ 12 (y1 y2 ) cot n , y0 = 21 (y1 + y2 ) 12 (x1 x2 ) cot n and xk , yk 2 Q for k 2 f0; 1; 2g. Taking into account that x1 6= x2 or y1 6= y2 , we deduce that cot n 2 Q, which is equivalent to tan n 2 Q. We arrive to the following auxiliary problem. Problem 2. Determine the values n 2 N, n 3 for which tan n 2 Q. Solution. Let us note that tan 4 = 1 2 Q, while tan 3 , tan 6 2 R n Q. It su¢ ces to assume n 5. We will make use of the fact that t = n implies tan nt = 0, where n 2 N, n 3.

A problem solving approach to some applications of the roots of unity to regular polygons 49

Using de Moivre’s formula and Newton’s binomial Theorem we get (1)

cos nt =

[ n2 ] X

( 1)k Cn2k cosn

2k

t sin2k t

k=0

and (2)

sin nt =

For t 2 Rn

n

n 1 [X 2 ]

( 1)k Cn2k+1 cosn

2k 1

t sin2k+1 t:

k=0

(2k+1) 2n

o : k 2 Z , we have sin t = cos t tan t, hence

cos nt = cosn t

[ n2 ] X

( 1)k Cn2k tan2k t

k=0

and n

sin nt = cos t

n 1 [X 2 ]

( 1)k Cn2k+1 tan2k+1 t:

k=0

Both functions tan t and tan nt are well-de…ned for t 2 Dn , where (2k + 1) (2k + 1) Dn := Rn :k2Z [ :k2Z : 2 2n Using the above formulas we get the following Lemma 2.3. For each n 2 N there exist two polynomials Pn ; Qn with Pn (tan t) for every t 2 Dn . Moreover, integer coe¢ cients such that tan nt = Q n (tan t) n 1 [ n2 ] [X 2 ] X k 2k+1 2k+1 ( 1) Cn X and Qn (X) = ( 1)k Cn2k X 2k . Pn (X) = k=0

k=0

Let n 2 N, n 5. Since tan n n = 0, it follows by Lemma 2.3 that Pn tan n = 0, Now assume that tan n 2 Q. Write tan n = pq , where p; q 2 N are relatively prime. Since tan n < tan 4 = 1 for all n 5, we have p q 1. The constant term of Pn is Cn1 and the leading coe¢ cient of Pn is ( 1)m Cn2m+1 , where we denoted m = n 2 1 . We have pj Cn1 and qj ( 1)m Cn2m+1 . Case 1. If n 5 is odd, then n = 2m + 1 and the leading coe¢ cient of Pn is ( 1)m . Since q divides the leading coe¢ cient of Pn , it follows that q = 1, a contradiction with 1 p q 1. Case 2. If n 8 is even, then we can write n = 2r (2s + 1), where r 2 N and s 2 N. Subcase 2.1. Assume s = 0. Then n = 2r , where r 3. According to Lemma 2.3 or simply due to formula tan 2 = 1 2 tan 2 , tan 2r 2 Q implies p p p tan p 1 r 3 tan 2 2 2 and cos 8 = 12 2 + 2, 2r = tan 8 2 Q. But sin 8 = 2 hence s p 2 2 p p = 2 1 2 R n Q: tan = 8 2+ 2

50

Marcelina Mocanu

We obtain a contradiction. Subcase 2.2. Assume s

1. According to Lemma 2.3, tan 2r (2s+1) 2 Q

implies tan 2r 2r (2s+1) = tan 2s+1 2 Q. For s = 1 it follows that tan 3 2 Q, which is false. For s 2 the conclusion tan 2s+1 2 Q is false, according to Case 1. In all cases, the assumption that tan n 2 Q, where n 2 N, n a contradiction. This proves the following Lemma 2.4. Let n 2 N , n

5 leads to

3. Then tan n 2 Q if and only if n = 4.

We conclude that a necessary condition for the existence of a lattice regular n gon is that n = 4. Remark 1. The rational values of the circular trigonometric functions at the rational multiples of have been well-studied by Underwood (1921), Lehmer (1933), Olmsted (1945), Niven (1956), Carlitz and Thomas (1962), see [2]. A thorough study of the rationality of the values of the tangent function at the rational multiples of has been done in [2], using more advanced algebraic concepts and methods than in this note. Second solution. The idea is to use the area of the desired polygonal region. By Pick’s Theorem, the area of a lattice polygonal region is given by B 1; S=I+ 2 where I and B denote the the number of latticial points belonging to the interior of the polygon, respectively to the polygon. An alternative way to express the area S of the region bounded by the polygon A1 A2 :::An is to use the general formula n

X 1 S = Im zk zk+1 , 2 k=1

where zk denotes the a¢ x of Ak , k = 1; 2; :::; n and An+1 = A1 . Note that each of the above formulas shows that the area of a lattice polygonal region is a rational number, moreover its double is a positive integer. But the area of the region bounded by a regular n gon with side length L is Sn = 41 nL2 cot n . For the region bounded by a lattice regular n gon the area Sn is rational, but the squared side length L2 is a positive integer, hence tan n 2 Q. We go back to Lemma 2.4 and conclude that it is necessary to have n = 4. Third solution. Here we explain in detail the "o¢ cial solution" from IMO 1985, see [4]. This solution involves more ingenuity than the previous ones, since some auxiliary construction and the extremal principle are used. In addition, this solution has a more geometric character than the above solutions.

A problem solving approach to some applications of the roots of unity to regular polygons 51

First suppose that for some n there exists at least one lattice regular n gon A1 A2 ::An . Step 1. There exist a family of lattice regular n gons, some of these being obtained as images of A1 A2 ::An under homotethies of integer ratios, with center at the origin O. Since the square of the side length is a positive integer, there exists a lattice regular n gon S1 S2 ::Sn with a smallest side length a. Let Bj be ! ! the point such that OBj = Aj 1 Aj , for j = 1; 2; :::; n, where we denote A0 = An . Then each Bj is a latticial point. The n gon B1 B2 :::Bn is ! ! regular, since the angle between OBj and OBj+1 is supplementary to the angle ]Aj 1 Aj Aj+1 (where j = 1; :::; n, An+1 = A1 ) and the measure of ]Aj 1 Aj Aj+1 is n1 (n 2). Then the side length of B1 B2 :::Bn is 2a sin n . By the minimality of the side length of S1 S2 ::Sn , we have 2a sin

n

a,

hence n 6. We conclude that there is no lattice regular n gon with n 7. Step 2. Assume that n 6. The center of every lattice regular n gon has rational coordinates. Using a homotethy we may …nd another lattice regular n gon whose center C has integer coordinates. Now rotate the vertices of around C with angles 2 , and 32 and take the union between the set of the vertices of and its images under this rotations. We obtain the vertices of a regular N gon, where N = 12 for n = 3 and n = 6, while N = 20 for n = 5. If n = 6 it su¢ ces to rotate the vertices of around C with 2 , since the set of the vertices of is preserved under a rotation with . See …gures 1, 2 and 3 corresponding to n = 3, n = 6 and n = 5, respectively. Obviously, for n = 4 the set of the vertices of is invariant to the above rotations.

Figure 1

Figure 2

52

Marcelina Mocanu

Figure 3 Step 3. Now we show that rotating a latticial point with 2 around another latticial point we get a third latticial point, a property that is easy to notice by graphical representation. Lemma 2.5. The image of a latticial point under a rotation around another latticial point, with angle 2 , is also a latticial point. Proof. Let M (x; y) and M0 (x0 ; y0 ) be latticial points and let M (x ; y ) be the point obtained as we rotate M around M0 with 2 . Since x

x0 + i (y

and cos 2 + i sin respectively:

2

y0 ) = cos

+ i sin [(x x0 ) + i (y y0 )] 2 2 = i, we get by identifying the real and imaginary parts,

x = x0 + y0

y and y = y0

x0 + x.

Now rotate M (x; y) around M0 (x0 ; y0 ) with 2 and denote the point we obtain by M (x ; y ). Then M0 is the midpoint of the line segment [M M ], hence x = 2x0 x = x0 y0 + y and y

= 2y0

y = x0 + y0

x:

By the above formulas, x0 ; y0 ; x; y 2 Z implies x ; y ; x ,y

2 Z.

Remark 2. We can …nd (x ; y ) without using complex numbers, but using graphical representations. In this case, we need to discuss the signs of x x0 and y y0 . If x = x0 , then (x ; y ) = (x0 + y0 y; y0 ) and if y = y0 , then (x ; y ) = (x0 ; y0 x0 + x). Assume that x 6= x0 and y 6= y0 . The parallels to Ox through M and M intersect the parallel to Oy through M0 at N and N , respectively. For all four choices of signs (+; +), (+; ), ( ; ) and ( ; +) for (x x0 , y y0 ), we can see that the right triangles N M0 M and N M M0 are congruent and positively oriented, hence x x0 = (y y0 ) and y y0 = x x0 , q.e.d.

A problem solving approach to some applications of the roots of unity to regular polygons 53

Assume that our problem has a solution for n = 3 or n = 5 or n = 6. It follows that the N gon obtained at Step 2, with N = 12, N = 20; respectively N = 12 is a lattice regular polygon, a contradiction with the conclusion of Step 1. We conclude that a lattice regular n gon could exist only if n = 4. Su¢ cient condition. Now we check if the condition n = 4 is also su¢ cient for the existence of a lattice regular n gon. Assuming that n = 4, Lemma 2.5 shows that for each pair of latticial points A and B we obtain a latticial point C by rotating A around B with angle 2 , then another latticial point D by rotating B around C with . Obviously, ABCD is a positively oriented lattice square . 2 Remark 3. A lattice square region whose sides are not parallel to the coordinate axes can be decomposed, using parallels through its vertices to the axes, into a smaller square region and four congruent right triangular regions, as in a well-known proof of Pythagorean Theorem that makes use of areas. See Figure 4.

Figure 4 We proved in three di¤erent ways the following known result Theorem 2.6. There exists a lattice regular n gon if and only if n = 4. Remark 4. Several authors have considered the problem of determining all the lattice regular polygons or generalizations of this problem to higher dimensions. We mention here only I. J. Schoenberg (1937) [10], M. S. Klamkin and H. E. Chrestenson (1963) [6], D. J. O’Loughlin (2002) [8]. Unfortunately, the above cited articles are not available online, except on

54

Marcelina Mocanu

JSTOR. For more references see [3]. This problem has attracted much interest because it admits various elementary solutions, but is connected to several non-elementary problems. Note that lattice polygons play an important role in discrete and computational geometry, a …eld which is at the cutting-edge frontier of mathematics and computer science.

3. The sum of squared reciprocals of the distances from a point on a circle to the vertices of a regular polygon A well-known formula shows that the sum of the squared distances from a point on a circle of radius R to the vertices of a regular n gon, which is inscribed in that circle, is nR2 . We will study the sum of squared reciprocals of the distances from a point on a circle to the vertices of a regular polygon inscribed in that circle. Problem 3. Let A1 A2 :::An be a regular polygon inscribed in a circle C of radius R. n X 1 attains a minimum value when P moves on Find out if the sum P A2 k

k=1

the circle C, avoiding the vertices of the given polygon, and calculate the minimum value, if this exists. Solution Step 1. We may assume, without loss of generality, that the circle C is centered at the origin and that Ak has the a¢ x zk = cos

2 (k

1) n

+ i sin

2 (k

1) n

; for k = 1; n:

Denote by z = cos t + i sin t the a¢ x of P , where t 2 [0; 2 ). Then P A2k = jz zk j2 = (z zk )(z zk ), hence P A2k = 2R2

(3)

(zk z + zk z)

Using the trigonometric form of complex numbers and denoting (3) implies P A2k = 2R2 (1

(4)

cos (t

(k

:=

2 n

1) )) .

for k = 1; n. Remark 5. (3) implies n X

P A2k =

k=1

But

n X

n X

2R2

(zk z + zk z) = 2nR2

z

k=1

n X k=1

zk = 0, as sum of the roots of the equation z n

k=1

n X k=1

P A2k = 2nR2 :

zk

z

n X

zk :

k=1

1 = 0, therefore

,

A problem solving approach to some applications of the roots of unity to regular polygons 55

Moreover, if OP =

, then

n X

P A2k = n

2

+ R2 (see [9] or [1] for a

k=1

problem solving approach to this formula) By (4) we have n X k=1

n 1 1 X = 2R2 1 P A2k k=1

1 cos (t

(k

1) )

, t 2 [0; 2 ]nf(k

Step 2. We study the existence of a minimum for the sum

n X k=1

Let us de…ne a function fn : by fn (t) =

n X k=1

n

! R, where

n

cos (t

(k

1) )

;t 2

: k = 1; :::; ng .

1 . P A2k

:= [0; 2 ]nf(k

1 1

1)

1)

: k = 1; :::; ng,

n:

The function fn is continuous on n . The set n is not closed, therefore it is not compact. It is easy to see that fn has no maximum, since lim fn (t) = +1 for k = 1; n. We have to prove that fn has a global

t!(k 1)

minimum, but we cannot take advantage by the well-known boundedness property of real continuous functions on compact topological spaces. The geometrical symmetry of the problem shows that it su¢ ces to assume t 2 (0; ). In fact, extending fn to a periodic function ff n : R n fk : k 2 Zg 2 with period 2 , we see that ff has also the period = n n .

In Figure 5 and Figure 6 we represent a part of the graph of fn in the cases n = 3 and n = 5, respectively: 150

100

50

1

2

3

Figure 5 Com p ut ed by Wolfram Alp ha

4

5

6

7

56

Marcelina Mocanu

400

300

200

100

1

2

3

4

5

6

7

Figure 6 Com p ut ed by Wolfram Alp ha Since fn is continuous on (0; ) and (5)

limfn (t) = lim fn (t) = +1,

t&0

t%

the function fn attains its global minimum at a point in (0; ). In order to prove this, consider m = inf fn ((0; )). It is clear that m 0, since fn (t) > 0 for every t 2 (0; ). Let s > m. By (5), we can …nd > 0 with < 2 , such that fn (t) > s whenever t 2 (0; ) [ ( ; ). Since fn is continuous on the compact interval [ ; ], there exists tmin 2 [ ; ] such that fn (tmin ) fn (t) for every t 2 [ ; ]. We conclude that fn (tmin ) fn (t) for every t 2 (0; ), hence fn attains its minimum on (0; ) at tmin . The …rst derivative or fn is fn0 (t) =

n X k=1

[1

sin (t (k cos (t (k

1) ) : 1) )]2

It is di¢ cult to solve the equation fn0 (t), t 2 (0; ) in the general case. Instead of trying to solve this equation, we will study the intervals of convexity/ concavity of fn . The second derivative of fn is fn00 (t) =

n X 2 + cos (t [1 cos (t k=1

(k (k

1) ) : 1) )]2

Since fn00 (t) > 0 for every t 2 (0; ), the function fn is strictly convex on (0; ), i.e. fn ((1 ) t1 + t2 ) < (1 ) fn (t1 ) + fn (t2 ) whenever 2 (0; 1) and t1 6= t2 are in (0; ). The strict convexity of fn implies the uniqueness of the minimum point. Indeed, assuming that t1 6= t2 are two minimum points of fn on (0; ), that is fn (t1 ) = fn (t2 ) = m, we get fn ((1 ) t1 + t2 ) < m whenever 2 (0; 1), which is a contradiction.

A problem solving approach to some applications of the roots of unity to regular polygons 57

Again, the geometric symmetry of the problem inspires us to …nd the n X 1 unique minimum point. Note that the expression S (P ) := is inP A2 k

k=1

variant to a symmetry with respect to the perpendicular bisector b of the line segment [A1 A2 ]. Assuming that S (P ) attains its minimum at a point P1 2 C\IntA\ 1 OA2 and that P1 and P2 are symmetric to each other with respect to b, it follows that S (P ) attains its minimum also at P2 . If P1 and P2 are distinct, this contradicts the uniqueness in (0; )of the minimum point of fn , otherwise P1 = P2 is the midpoint of the arc C\IntA\ 1 OA2 . In analytical terms, the graph of fn is symmetric with respect to the line t) for all t 2 (0; ) and from the existence and t = 2 , i.e. fn (t) = fn ( uniqueness of the minimum point we get min ffn (t) : t 2 (0; )g = fn 2 . We should have fn0 2 = 0, according to Fermat’s Theorem. Indeed, fn0

=

2

n X

sin 1

k=1

cos

But k + h = n + 1 implies cos 2 = sin 2 (h sin 2 (k 1) fn0

2 2

(k 1) 2

(k

1)

(k

1)

2

1) = cos , therefore

(h

2

1)

and

= 0.

Since fn0 2 = 0, fn00 2 > 0, the function fn has a local minimum at 2 , by the second derivative test. Since fn has a unique point of global minimum, by the above discussion, it follows that this minimum point is 2 . The global minimum of fn is

n

where

=

2 n

:= fn

2

=

n X k=1

1

1 cos 2k2

3

;

, hence n

=

n X1 h=0

The minimum of

n X k=1

1

1 P A2k

1 cos 2hn 1

=

n X k=1

1

1 : cos 2kn 1

for P 2 C n Ak : k = 0; n is

1 2R2

n.

Step 3. In order to …nd a closed form for n we can use an heuristical approach, with the aid of mathematical software. In our case, we used Maple. We calculate n (by using "Evaluate", then "Simplify") for n = 49 3; 4; 5; 6; 7; 8; 9; 10::We …nd 3 = 92 , 4 = 8, 5 = 25 2 , 6 = 18, 7 = 2 , 81 121 8 = 32, 9 = 2 , 10 = 50, 11 = 2 . The rough values provided by the computer could look rather complicated, sometimes it is necessary to use a combinationation of commands as in the following case, where we used "Evaluate"+"Simplify"+"Evaluate numerically", or better, "Evaluate"+Evaluate numerically", as in the following

58

Marcelina Mocanu

example: 9 X

1 1

k=1

cos

=

2 cos

2 9

+ 24 cos

4 9

(2k 1) 9

2 +1

cos

1 24 cos

2 9

24 cos

1 9

1 9

1

188 cos

+2

+

2 9

2 cos

4 9

+1

188 cos

+

9 = 2

1 4 + 188 cos + 81 9 9

=

81 : 2

Similarly, 10 X

1

1 4

p p 2 5

2

p

1 4

5+1

and

p p 2 5

2

p

5

11 X

2 cos

2 + 1 cos

1

1 cos

2 p pp 2 5+5

2

+

3 11

2 + p pp +2 = 50:0 1 1 4 2 5+5+1

=

cos (2k111)

2

1 11

1 4

1

k=1 1

2 11

=

cos (2k101)

1

k=1

1 cos

2

4 11

+ 1 cos

1 + = 60: 5 1 2

5 11

The "empirical data" lead us to conjecture that n X

(6)

k=1

1

=

cos (2k n 1)

1

n2 : 2

It is not of great value to continue to test this conjecture for the next values of n, although this may help. Instead of getting "concrete proofs" we need a rigorous deductive proof. We have 1 1

cos

(2k 1) n

for k 2 f0; 1; :::; n (7)

=

1 2 sin2 (2k n 1)

=

1 2

1 + cot2

(2k

1) n

1g. The relation (6) is equivalent to n X

cot2

(2k

k=1

1) n

= n (n

1) :

A similar identity, namely (8)

n X k=1

csc2

k 2n (n + 1) = 2n + 1 3

is proved in the book "Non-elementary problems in an elementary exposition" by A.M. Iaglom and I.M. Iaglom. The idea for the proof of (8) is to k start with sin (2n + 1) 2n+1 = 0 and to apply the formula of sin (2n + 1) as a polynomial of sin and cos (see (2)).

A problem solving approach to some applications of the roots of unity to regular polygons 59

Here we use the identity cos n (2k2n1) = 0, k = 1; 2; :::; n. For each n o 2 (2k2n1) : k = 1; n it follows from (1) that 0 = cos n =

[ n2 ] X

k

( 1)

Cn2k

cos

n 2k

n

2k

sin

= sin

k=0

Since sin

(2k 1) 2n

2k

6= 0 for k = 1; 2; :::; n, the equation ( 1)k Cn2k xn

2k

k=0

admits the distinct roots xk = cot X

( 1)k Cn2k cotn

k=0

[ n2 ] X

and

[ n2 ] X

(2k 1) 2n

= 0, x 2 R , k = 1; 2; :::; n. Since

n X

xk = 0

k=1

Cn2 , it follows that

xk xh =

1 k