IMO 1967 LL POL40

Origin: POL

Problem

Exactly one side of a tetrahedron is of length greater than

Solution

Suppose CD is the longest edge of the tetrahedron ABCD, AB = a, CK and DL are the altitudes of the triangles ABC and ABD respectively, and DM is the altitude of the tetrahedron ABCD. Then CK2 \leq1 −a2/4, since CK is a leg of the right triangle whose other leg has length not less than a/2 and whose hypotenuse has length not greater than 1 (AKC or BKC). In the similar way we can show that DL2 \leq1 −a2/4. Since DM \leqDL, then DM 2 \leq1 −a2/4. It follows that V = 1 a 2CK

DM \leq1 6a  1 −a2  = 1 24a(2 −a)(2 + a) = 1 24[1 −(a −1)2](2 + a) \leq1 24 \cdot 1 \cdot 3 = 1 8.