IMO 1975 SL 13

Origin: ROM

Problem

Let A0, A1, . . . , An be points in a plane such that (i) A0A1 \leq1 2A1A2 \leq\cdot \cdot \cdot \leq 2n−1 An−1An and (ii) 0 < ∡A0A1A2 < ∡A1A2A3 < \cdot \cdot \cdot < ∡An−2An−1An < 180◦, where all these angles have the same orientation. Prove that the segments AkAk+1, AmAm+1 do not intersect for each k and n such that 0 \leqk \leq m −2 < n −2.

Solution

Suppose that AkAk+1 \capAmAm+1 ̸= \emptysetfor some k, m > k + 1. Without loss of generality we may suppose that k = 0, m = n −1 and that no two segments AkAk+1 and AmAm+1 intersect for 0 \leqk < m −1 < n −1 except for k = 0, m = n −1. Also, shortening A0A1, we may suppose that A0 \inAn−1An. Finally, we may reduce the problem to the case that A0 . . . An−1 is convex: Otherwise, the segment An−1An can be prolonged so that it intersects some AkAk+1, 0 < k < n −2. If n = 3, then A1A2 \geq2A0A1 implies A0A2 > A0A1, hence \angleA0A1A2 > \angleA1A2A3, a contradiction. Let n = 4. From A3A2 > A1A2 we conclude that \angleA3A1A2 > \angleA1A3A2. Using the inequality \angleA0A3A2 > \angleA0A1A2 we obtain that \angleA0A3A1 > \angleA0A1A3 implying A0A1 > A0A3. Now we have A2A3 < A3A0 +A0A1 + A1A2 < 2A0A1 + A1A2 \leq2A1A2 \leqA2A3, which is not possible. Now suppose n \geq5. If \alphai is the exterior angle at Ai, then \alpha1 > \cdot \cdot \cdot > \alphan−1; hence \alphan−1 < 360◦ n−1 \leq90◦. Consequently \angleAn−2An−1A0 \geq90◦and A0An−2 > An−1An−2. On the other hand, A0An−2 < A0A1+A1A2+\cdot \cdot \cdot+ An−3An−2 <
2n−2 + 2n−3 + \cdot \cdot \cdot + 1  An−1An−2 < An−1An−2, which contradicts the previous relation.