IMO 1987 SL 12

Origin: POL

Problem

Given a nonequilateral triangle ABC, the vertices listed coun- terclockwise, find the locus of the centroids of the equilateral triangles A′B′C′ (the vertices listed counterclockwise) for which the triples of points A, B′, C′; A′, B, C′; and A′, B′, C are collinear.

Solution

Here all angles will be oriented and measured counterclockwise. Note that ∡CA′B = ∡AB′C = ∡BC′A = \pi/3. Let a′, b′, c′ denote respectively the inner bisectors of angles A′, B′, C′ in triangle A′B′C′. The lines a′, b′, c′ meet at the cen- troid X of A′B′C′, and ∡(a′, b′) = ∡(b′, c′) = ∡(c′, a′) = 2\pi/3. Now let K, L, M be the points such that KB = KC, LC = LA, MA = MB, and ∡BKC = ∡CLA = ∡AMB = 2\pi/3, and let C1, C2, C3 be the cir- cles circumscribed about triangles A B C L M K C′ A′ B′ P X BKC, CLA, and AMB respectively. These circles are characterized by C1 = {Z | ∡BZC = 2\pi/3}, etc.; hence we deduce that they meet at a point P such that ∡BPC = ∡CPA = ∡APB = 2\pi/3 (Torricelli’s point). Points A′, B′, C′ run over C1 ∖{P}, C2 ∖{P}, C3 ∖{P} respectively. As for a′, b′, c′, we see that K \ina′, L \inb′, M \inc′, and also that they can take all possible directions except KP, LP, MP respectively (if K = P, KP is assumed to be the corresponding tangent at K). Then, since ∡KXL = 2\pi/3, X runs over the circle defined by {Z | ∡KZL = 2\pi/3}, without P. But analogously, X runs over the circle {Z | ∡LZM = 2\pi/3}, from which we can conclude that these two circles are the same, both equal to the circumcircle of KLM, and consequently also that triangle KLM is

equilateral (which is, anyway, a well-known fact). Therefore, the locus of the points X is the circumcircle of KLM minus point P.