IMO 1989 SL 21

Origin: NET

Problem

Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than 120◦.

Solution

In order to obtain a triangle as the intersection we must have three points P, Q, R on three sides of the tetrahedron passing through one vertex, say T . It is clear that we may suppose w.l.o.g. that P is a vertex, and Q and R lie on the edges TP1 and TP2 (P1, P2 are vertices) or on their extensions respectively. Suppose that −\to TQ = \lambda−−\to TP1 and −\to TR = µ−−\to TP2, where \lambda, µ > 0. Then cos \angleQPR = −−\to PQ \cdot −\to PR PQ \cdot PR = (\lambda −1)(µ −1) + 1 \sqrt \lambda2 −\lambda + 1

µ2 −µ + 1 . In order to obtain an obtuse angle (with cos < 0) we must choose µ < 1 and \lambda > 2−µ 1−µ > 1. Since \sqrt \lambda2 −\lambda + 1 > \lambda −1 and

µ2 −µ + 1 > 1 −µ, we get that for (\lambda −1)(µ −1) + 1 < 0,

cos \angleQPR > 1 −(1 −µ)(\lambda −1) 2(1 −µ)(\lambda −1)

−1 2; hence \angleQPR < 120◦. Remark. After obtaining the formula for cos\angleQPR, the official solution was as follows: For fixed µ0 < 1 and \lambda > 1, cos \angleQPR is a decreasing function of \lambda: indeed, \partialcos \angleQPR \partial\lambda = µ −(3 −µ)\lambda 4(\lambda2 −\lambda + 1)3/2(µ2 −µ + 1)1/2 < 0. Similarly, for a fixed, sufficiently large \lambda0, cos \angleQPR is decreasing for µ decreasing to 0. Since lim\lambda\to0,µ\to0+ cos \angleQPR = −1/2, we conclude that \angleQPR < 120◦.