IMO 1986 SL 17

Origin: CHN

Problem

Let A, B, C be fixed points in the plane. A man starts from a certain point P0 and walks directly to A. At A he turns his di- rection by 60◦to the left and walks to P1 such that P0A = AP1. Af- ter he does the same action 1986 times successively around the points A, B, C, A, B, C, . . . , he returns to the starting point. Prove that \triangleABC is equilateral and that the vertices A, B, C are arranged counterclockwise.

Solution

We use complex numbers to represent the position of a point in the plane. For convenience, let A1, A2, A3, A4, A5, . . . be A, B, C, A, B, . . . respec- tively, and let P0 be the origin. After the kth step, the position of Pk will be Pk = Ak + (Pk−1 −Ak)u, k = 1, 2, 3, . . ., where u = e4\piı/3. We easily obtain Pk = (1 −u)(Ak + uAk−1 + u2Ak−2 + \cdot \cdot \cdot + uk−1A1). The condition P0 \equivP1986 is equivalent to A1986 +uA1985 +\cdot \cdot \cdot+u1984A2 + u1985A1 = 0, which, having in mind that A1 = A4 = A7 = \cdot \cdot \cdot , A2 = A5 = A8 = \cdot \cdot \cdot , A3 = A6 = A9 = \cdot \cdot \cdot , reduces to 662(A3 + uA2 + u2A1) = (1 + u3 + \cdot \cdot \cdot + u1983)(A3 + uA2 + u2A1) = 0. It follows that A3 −A1 = u(A1 −A2), and the assertion follows. Second solution. Let fP denote the rotation with center P through 120◦ clockwise. Let f1 = fA. Then f1(P0) = P1. Let B′ = f1(B), C′ = f1(C), and f2 = fB′. Then f2(P1) = P2 and f2(AB′C′) = A′B′C′′. Finally, let f3 = fC′′ and f3(A′B′C′′) = A′′B′′C′′. Then g = f3f2f1 is a translation sending P0 to P3 and C to C′′. Now P1986 = P0 implies that g662 is the identity, and thus C = C′′. Let K be such that ABK is equilateral and positively oriented. We observe that f2f1(K) = K; therefore the rotation f2f1 satisfies f2f1(P) ̸= P for P ̸= K. Hence f2f1(C) = C′′ = C implies K = C.