Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?

Problem A. 530. (March 2011)

A. 530. There are given 2n+1 points A_1,A_3,\ldots,A_{2n+1},A_{2n},A_{2n-2},\ldots,
A_2 (n\ge2) lying on a circle in this order such that \angle A_1A_2A_3 = \angle
A_2A_3A_4 = \ldots = \angle A_{2n-1}A_{2n}A_{2n+1} = \frac{90^\circ}{n}. The lines A_2A_3,A_3A_4,,\ldots, A_{2n-1}A_{2n} split the line segment A1A2n+1 into 2n-1 part; denote their lengths by \ell_1,\ell_2,\ldots,\ell_{2n-1}, respectively. Prove that \ell_1^2-\ell_2^2+\ell_3^2-\ell_4^2+-\ldots-\ell_{2n-2}^2+\ell_{2n-1}^2 = 0.

(Proposed by Zoltán Gyenes, Budapest)

(5 pont)

Deadline expired on April 11, 2011.


12 students sent a solution.
5 points:Backhausz Tibor, Damásdi Gábor, Frankl Nóra, Gyarmati Máté, Janzer Olivér, Lenger Dániel, Mester Márton, Nagy 235 János, Nagy 648 Donát, Strenner Péter, Szabó 928 Attila.
2 points:1 student.

Problems in Mathematics of KöMaL, March 2011