IMO 1968 SL 3

Origin: POL

Problem

Prove that in any tetrahedron there is a vertex such that the lengths of its sides through that vertex are sides of a triangle.

Solution

A triangle cannot be formed out of three lengths if and only if one of them is larger than the sum of the other two. Let us assume this is the case for all triplets of edges out of each vertex in a tetrahedron $ABCD$. Let w.l.o.g. $AB$ be the largest edge of the tetrahedron. Then $AB \geq AC + AD$ and $AB \geq BC + BD$, from which it follows that $2AB \geq AC + AD + BC + BD$. This implies that either $AB \geq AC + BC$ or $AB \geq AD + BD$, contradicting the triangle inequality. Hence the three edges coming out of at least one of the vertices $A$ and $B$ form a triangle.

Remark. The proof can be generalized to prove that in a polyhedron with only triangular surfaces there is a vertex such that the edges coming out of this vertex form a triangle.