IMO 1973 SL 15

Origin: CUB

Problem

Prove that for all n \inN the following is true: 2n n

k=1 sin k\pi 2n + 1 = \sqrt 2n + 1.

Solution

If z = cos \theta + i sin \theta, then z −z−1 = 2i sin \theta. Now put z = cos \pi 2n+1 + i sin \pi 2n+1. Using de Moivre’s formula we transform the required equality into A = n

k=1 (zk −z−k) = in\sqrt 2n + 1. (1) On the other hand, the complex numbers z2k (k = −n, −n + 1, . . ., n) are the roots of x2n+1 −1, and hence n

k=1 (x −z2k)(x −z−2k) = x2n+1 −1 x −1 = x2n + \cdot \cdot \cdot + x + 1. (2) Now we go back to proving (1). We have (−1)nzn(n+1)/2A = n

k=1 (1 −z2k) and z−n(n+1)/2A = n

k=1 (1 −z−2k). Multiplying these two equalities, we obtain (−1)nA2 = $n k=1(1−z2k)(1− z−2k) = 2n+1, by (2). Therefore A = \pmi−n\sqrt2n + 1. This actually implies that the required product is \pm\sqrt2n + 1, but it must be positive, since all the sines are, and the result follows.