IMO 1987 SL 5

Origin: GBR

Problem

Find, with proof, the point P in the interior of an acute-angled triangle ABC for which BL2 +CM2 +AN2 is a minimum, where L, M, N are the feet of the perpendiculars from P to BC, CA, AB respectively.

Solution

Assuming the notation a = BC, b = AC, c = AB; x = BL, y = CM, z = AN, from the Pythagorean theorem we obtain (a −x)2 + (b −y)2 + (c −z)2 = x2 + y2 + z2 = x2 + (a −x)2 + y2 + (b −y)2 + z2 + (c −z)2 . Since x2+(a−x)2 = a2/2+(a−2x)2/2 \geqa2/2 and similarly y2+(b−y)2 \geq b2/2 and z2 + (c −z)2 \geqc2/2, we get x2 + y2 + z2 \geqa2 + b2 + c2 . Equality holds if and only if P is the circumcenter of the triangle ABC, i.e., when x = a/2, y = b/2, z = c/2.