IMO 1967 LL ITA29

Origin: ITA

Problem

The triangles A0B0C0 and A′B′C′ have all their angles acute. Describe how to construct one of the triangles ABC, similar to A′B′C′ and circumscribing A0B0C0 (so that A, B, C correspond to A′, B′, C′, and AB passes through C0, BC through A0, and CA through B0). Among these triangles ABC, describe, and prove, how to construct the triangle with the maximum area.

Solution

Let arc la be the locus of points A lying on the opposite side from A0 with respect to the line B0C0 such that \angleB0AC0 = \angleA′. Let ka be the circle containing la, and let Sa be the center of ka. We similarly define lb, lc, kb, kc, Sb, Sc. It is easy to show that circles ka, kb, kc have a common point S inside \triangleABC. Let A1, B1, C1 be the points on the arcs la, lb, lc diametrically opposite to S with respect to Sa, Sb, Sc respectively. Then A0 \inB1C1 because \angleB1A0S = \angleC1A0S = 90◦; similarly, B0 \inA1C1 and C0 \inA1B1. Hence the triangle A1B1C1 is circumscribed about \triangleA0B0C0 and similar to \triangleA′B′C′. Moreover, we claim that \triangleA1B1C1 is the triangle ABC with the desired properties having the maximum side BC and hence the maximum area.

Indeed, if ABC is any other such triangle and S′ b, S′ c are the projections of Sb and Sc onto the line BC, it holds that BC = 2S′ bS′ c \leq2SbSc = B1C1, which proves the maximality of B1C1.