IMO 1971 SL 12

Origin: POL

Problem

Two congruent equilateral triangles ABC and A′B′C′ in the plane are given. Show that the midpoints of the segments AA′, BB′, CC′ either are collinear or form an equilateral triangle.

Solution

Let us start with the case A = A′. If the triangles ABC and A′B′C′ are oppositely oriented, then they are symmetric with respect to some axis, and the statement is true. Suppose that they are equally oriented. There is a rotation around A by 60◦that maps ABB′ onto ACC′. This rotation also maps the midpoint B0 of BB′ onto the midpoint C0 of CC′, hence the triangle AB0C0 is equilateral. In the general case, when A ̸= A′, let us denote by T the translation that maps A onto A′. Let X′ be the image of a point X under the (unique) isometry mapping ABC onto A′B′C′, and X′′ the image of X under T . Furthermore, let X0, X′ 0 be the midpoints of segments XX′, X′X′′. Then X0 is the image of X′ 0 under the translation −(1/2)T . However, since it has already been proven that the triangle A′ 0B′ 0C′ 0 is equilateral, its image A0B0C0 under (1/2)T is also equilateral. The statement of the problem is thus proven.