IMO 1978 SL 2

Origin: BUL

Problem

Two identically oriented equilateral triangles, ABC with center S and A′B′C, are given in the plane. We also have A′ ̸= S and B′ ̸= S. If M is the midpoint of A′B and N the midpoint of AB′, prove that the triangles SB′M and SA′N are similar.

Solution

Consider the transformation \varphi of the plane defined as the homothety H with center B and coefficient 2 followed by the rotation R about the center O through an angle of 60◦. Being direct, this mapping must be a rotational homothety. We also see that H maps S into the point symmetric to S with respect to OA, and R takes it back to S. Hence S is a fixed point, and is consequently also the center of \varphi. Therefore \varphi is the rotational homo- thety about S with the angle 60◦ O A B S A′ B′ M N and coefficient 2. (In fact, this could also be seen from the fact that \varphi preserves angles of triangles and maps the segment SR onto SB, where R is the midpoint of AB.) Since \varphi(M) = B′, we conclude that \angleMSB′ = 60◦and SB′/SM = 2. Similarly, \angleNSA′ = 60◦and SA′/SN = 2, so triangles MSB′ and NSA′ are indeed similar. Second solution. Probably the simplest way here is using complex num- bers. Put the origin at O and complex numbers a, a′ at points A, A′, and denote the primitive sixth root of 1 by \omega. Then the numbers at B, B′, S and N are \omegaa, \omegaa′, (a + \omegaa)/3, and (a + \omegaa′)/2 respectively. Now it is easy to verify that (n −s) = \omega(a′ −s)/2, i.e., that \angleNSA′ = 60◦and SA′/SN = 2.