IMO 2004 SL G5

Origin: SMN | Category: Geometry

Problem

Let A1A2 . . . An be a regular n-gon. The points B1, . . . , Bn−1 are defined as follows: (i) If i = 1 or i = n −1, then Bi is the midpoint of the side AiAi+1. (ii) If i ̸= 1, i ̸= n −1, and S is the intersection point of A1Ai+1 and AnAi, then Bi is the intersection point of the bisector of the angle AiSAi+1 with AiAi+1. Prove the equality \angleA1B1An + \angleA1B2An + \cdot \cdot \cdot + \angleA1Bn−1An = 180◦.

Solution

We first establish the following lemma. Lemma. Let ABCD be an isosceles trapezoid with bases AB and CD. The diagonals AC and BD intersect at S. Let M be the midpoint of BC, and let the bisector of the angle BSC intersect BC at N. Then \angleAMD = \angleAND. Proof. It suffices to show that the points A, D, M, N are concyclic. The statement is trivial for AD \parallelBC. Let us now assume that AD and BC meet at X, and let XA = XB = a, XC = XD = b. Since SN is the bisector of \angleCSB, we have a −XN XN −b = BN CN = BS CS = AB CD = a b , and an easy computation yields XN = 2ab a+b. We also have XM = a+b 2 ; hence XM \cdot XN = XA \cdot XD. Therefore A, D, M, N are concyclic, as needed.

Denote by Ci the midpoint of the side AiAi+1, i = 1, . . . , n −1. By def- inition C1 = B1 and Cn−1 = Bn−1. Since A1AiAi+1An is an isosceles trapezoid with A1Ai \parallelAi+1An for i = 2, . . . , n −2, it follows from the lemma that \angleA1BiAn = \angleA1CiAn for all i. The sum in consideration thus equals \angleA1C1An+\angleA1C2An+\cdot \cdot \cdot+ \angleA1Cn−1An. Moreover, the trian- gles A1CiAn and An+2−iC1An+1−i are congruent (a rotation about the center of the n-gon carries the first one to the second), and conse- quently \angleA1CiAn = \angleAn+2−iC1An+1−i for i = 2, . . . , n −1. A1 A2 A3 A4 An B1 C2 B3 Bn−1 B2 . . . Hence \Sigma = \angleA1C1An+\angleAnC1An−1+\cdot \cdot \cdot+\angleA3C1A2 = \angleA1C1A2 = 180◦.