IMO 1981 SL 14

Origin: ROM

Problem

Prove that a convex pentagon (a five-sided polygon) ABCDE with equal sides and for which the interior angles satisfy the condition \angleA \geq\angleB \geq\angleC \geq\angleD \geq\angleE is a regular pentagon.

Solution

We need the following lemma. Lemma. If a convex quadrilateral PQRS satisfies PS = QR and \angleSPQ \geq \angleRQP, then \angleQRS \geq\anglePSR. Proof. If the lines PS and QR are parallel, then this quadrilateral is a parallelogram, and the statement is trivial. Otherwise, let X be the point of intersection of lines PS and QR. Assume that \angleSPQ+\angleRQP > 180◦. Then \angleXPQ \leq\angleXQP implies that XP \geqXQ, and consequently XS \geqXR. Hence, \angleQRS = \angleXRS \geq\angleXSR = \anglePSR. Similarly, if \angleSPQ + \angleRQP < 180◦, then \angleXPQ \geq\angleXQP, from which it follows that XP \leqXQ, and thus XS \leqXR; hence \angleQRS = 180◦−\angleXRS \geq180◦−\angleXSR = \anglePSR. Now we apply the lemma to the quadrilateral ABCD. Since \angleB \geq\angleC and AB = CD, it follows that \angleCDA \geq\angleBAD, which together with \angleEDA = \angleEAD gives \angleD \geq\angleA. Thus \angleA = \angleB = \angleC = \angleD. Analo- gously, by applying the lemma to BCDE we obtain \angleE \geq\angleB, and hence \angleB = \angleC = \angleD = \angleE.