IMO 1978 SL 4

Origin: CZS

Problem

Let T1 be a triangle having a, b, c as lengths of its sides and let T2 be another triangle having u, v, w as lengths of its sides. If P, Q are the areas of the two triangles, prove that 16PQ \leqa2(−u2 + v2 + w2) + b2(u2 −v2 + w2) + c2(u2 + v2 −w2). When does equality hold?

Solution

Let \gamma, ϕ be the angles of T1 and T2 opposite to c and w respectively. By the cosine theorem, the inequality is transformed into a2(2v2 −2uv cos ϕ) + b2(2u2 −2uv cos ϕ) +2(a2 + b2 −2ab cos\gamma)uv cos ϕ \geq4abuv sin \gamma sin ϕ. This is equivalent to 2(a2v2 + b2u2) −4abuv(cos\gamma cos ϕ + sin \gamma sin ϕ) \geq0, i.e., to 2(av −bu)2 + 4abuv(1 −cos(\gamma −ϕ)) \geq0, which is clearly satisfied. Equality holds if and only if \gamma = ϕ and a/b = u/v, i.e., when the triangles are similar, a corresponding to u and b to v.