IMO 1968 SL 2

Origin: ROM

Problem

Prove that there exists a unique triangle whose side lengths are consecutive natural numbers and one of whose angles is twice the measure of one of the others.

Solution

The sides a, b, c of a triangle ABC with \angleABC = 2\angleBAC satisfy b2 = a(a + c) (this statement is the lemma in (SL98-7)). Taking into account the remaining condition that a, b, c are consecutive integers with a < b, we obtain three cases: (i) a = n, b = n + 1, c = n + 2. We get the equation (n + 1)2 = n(2n + 2), giving us (a, b, c) = (1, 2, 3), which is not a valid triangle. (ii) a = n, b = n + 2, c = n + 1. We get (n + 2)2 = n(2n + 1) ⇒ (n −4)(n + 1) = 0, giving us the triangle (a, b, c) = (4, 6, 5). (iii) a = n + 1, b = n + 2, c = n. We get (n + 2)2 = (n + 1)(2n + 1) ⇒ n2 −n −3 = 0, which has no positive integer solutions for n. Hence, the only solution is the triangle with sides of lengths 4, 5, and 6.