IMO 1995 SL G2

Origin: GER | Category: Geometry

Problem

Let A, B, and C be noncollinear points. Prove that there is a unique point X in the plane of ABC such that XA2 + XB2 + AB2 = XB2 + XC2 + BC2 = XC2 + XA2 + CA2.

Solution

Let A′, B′, C′ be the points symmetric to A, B, C with respect to the midpoints of BC, CA, AB respectively. From the condition on X we have XB2 −XC2 = AC2 −AB2 = A′B2 −A′C2, and hence X must lie on the line through A′ perpendicular to BC. Similarly, X lies on the line through B′ perpendicular to CA. It follows that there is a unique position for X, namely the orthocenter of \triangleA′B′C′. It easily follows that this point X satisfies the original equations.