IMO 1992 SL 5

Origin: COL

Problem

Let ABCD be a convex quadrilateral such that AC = BD. Equilateral triangles are constructed on the sides of the quadrilat- eral. Let O1, O2, O3, O4 be the centers of the triangles constructed on AB, BC, CD, DA respectively. Show that O1O3 is perpendicular to O2O4.

Solution

Denote by K, L, M, and N the midpoints of the sides AB, BC, CD, and DA, respectively. The quadrilateral KLMN is a rhombus. We shall prove that O1O3 \parallelKM. Similarly, O2O4 \parallelLN, and the desired result follows immediately. We have −−−\to O1O3 = −−\to KM+ −−−\to O1K + −−−\to MO3

. Assume that ABCD is positively oriented. A rotational homothety R with angle −90◦and coefficient 1/ \sqrt takes the vectors −−\to BK and −−\to CM into −−−\to O1K and −−−\to MO3 respectively. Therefore −−−\to O1O3 = −−\to KM + (−−−\to O1K + −−−\to MO3) = −−\to KM + R(−−\to BK + −−\to CM) = −−\to KM + 1 2R(−−\to BA + −−\to CD) = −−\to KM + R(−−\to LN). Since LN \perpKM, it follows that R(LN) is parallel to KM and so is O1O3.