IMO 1988 SL 27

Origin: GBR

Problem

The triangle ABC is acute-angled. Let L be any line in the plane of the triangle and let u, v, w be the lengths of the perpendiculars from A, B, C respectively to L. Prove that u2 tan A + v2 tan B + w2 tan C \geq2∆, where ∆is the area of the triangle, and determine the lines L for which equality holds.

Solution

Consider a Cartesian system with the x-axis on the line BC and origin at the foot of the perpendicular from A to BC, so that A lies on the y-axis. Let A be (0, \alpha), B(−\beta, 0), C(\gamma, 0), where \alpha, \beta, \gamma > 0 (because ABC is acute-angled). Then tan B = \alpha \beta , tan C = \alpha \gamma and tan A = −tan(B + C) = \alpha(\beta + \gamma) \alpha2 −\beta\gamma ; here tan A > 0, so \alpha2 > \beta\gamma. Let L have equation x cos \theta + y sin \theta + p = 0. Then u2 tan A + v2 tan B + w2 tan C = \alpha(\beta + \gamma) \alpha2 −\beta\gamma (\alpha sin \theta + p)2 + \alpha \beta (−\beta cos \theta + p)2 + \alpha \gamma (\gamma cos \theta + p)2 = (\alpha2 sin2 \theta + 2\alphap sin \theta + p2)\alpha(\beta + \gamma) \alpha2 −\beta\gamma + \alpha(\beta + \gamma) cos2 \theta + \alpha(\beta + \gamma) \beta\gamma p2

\alpha(\beta + \gamma) \beta\gamma(\alpha2 −\beta\gamma)(\alpha2p2 + 2\alphap\beta\gamma sin \theta + \alpha2\beta\gamma sin2 \theta + \beta\gamma(\alpha2 −\beta\gamma) cos2 \theta)

\alpha(\beta + \gamma) \beta\gamma(\alpha2 −\beta\gamma) / (\alphap + \beta\gamma sin \theta)2 + \beta\gamma(\alpha2 −\beta\gamma) \geq\alpha(\beta + \gamma) = 2∆, with equality when \alphap + \beta\gamma sin \theta = 0, i.e., if and only if L passes through (0, \beta\gamma/\alpha), which is the orthocenter of the triangle.