IMO 1977 LL USS48

Origin: USS

Problem

The intersection of a plane with a regular tetrahedron with edge a is a quadrilateral with perimeter P. Prove that 2a \leqP \leq3a.

Solution

Let a plane cut the edges AB, BC, CD, DA at points K, L, M, N respec- tively. Let D′, A′, B′ be distinct points in the plane ABC such that the triangles BCD′, CD′A′, D′A′B′ are equilateral, and M ′ \in[CD′], N ′ \in[D′A′], and K′ \in[A′B′] such that CM ′ = CM, A′N ′ = AN, and A′K′ = AK. The perimeter P of the quadrilat- eral KLMN is equal to the length of the polygonal line KLM ′N ′K′, which is not less than KK′. It fol- lows that P \geq2a. A C A′ B D′ B′ K L M ′ N ′ K′ Let us consider all quadrilaterals KLMN that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane \alpha. The lengths of the segments KL, LM, MN, NK are linear functions in AK, and so is P. Thus P takes its maximum at an endpoint of the interval, i.e., when the plane KLMN passes through one of the vertices A, B, C, D, and it is easy to see that in this case P \leq3a.