IMO 2001 SL G4

Origin: FRA | Category: Geometry

Problem

Let M be a point in the interior of triangle ABC. Let A′ lie on BC with MA′ perpendicular to BC. Define B′ on CA and C′ on AB similarly. Define p(M) = MA′ \cdot MB′ \cdot MC′ MA \cdot MB \cdot MC . Determine, with proof, the location of M such that p(M) is maximal. Let µ(ABC) denote the maximum value. For which triangles ABC is the value of µ(ABC) maximal?

Solution

Let \alpha1, \beta1, \gamma1, \alpha2, \beta2, \gamma2 denote the angles \angleMAB, \angleMBC, \angleMCA, \angleMAC, \angleMBA, \angleMCB respectively. Then MB′\cdotMC′ MA2 = sin \alpha1 sin \alpha2, MC′\cdotMA′ MB2 = sin \beta1 sin \beta2, MA′\cdotMB′ MC2 = sin \gamma1 sin \gamma2; hence p(M)2 = sin \alpha1 sin \alpha2 sin \beta1 sin \beta2 sin \gamma1 sin \gamma2. Since sin \alpha1 sin \alpha2 = 1 2(cos(\alpha1 −\alpha2) −cos(\alpha1 + \alpha2) \leq1 2(1 −cos \alpha) = sin2 \alpha 2 , we conclude that p(M) \leqsin \alpha 2 sin \beta 2 sin \gamma 2 . Equality occurs when \alpha1 = \alpha2, \beta1 = \beta2, and \gamma1 = \gamma2, that is, when M is the incenter of \triangleABC. It is well known that µ(ABC) = sin \alpha 2 sin \beta 2 sin \gamma 2 is maximal when \triangleABC is equilateral (it follows, for example, from Jensen’s inequality applied to ln sin x). Hence max µ(ABC) = 1 8.