IMO 1979 SL 22

Origin: USS

Problem

There are two circles in the plane. Let a point A be one of the points of intersection of these circles. Two points begin moving simultaneously with constant speeds from the point A, each point along its own circle. The two points return to the point A at the same time. Prove that there is a point P in the plane such that at every moment of time the distances from the point P to the moving points are equal.

Solution

Let the centers of the two circles be denoted by O and O1 and their respective radii by r and r1, and let the positions of the points on the circles at time t be denoted by M(t) and N(t). Let Q be the point such that OAO1Q is a parallelogram. We will show that Q is the point P we are looking for, i.e., that QM(t) = QN(t) for all t. We note that OQ = O1A = r1, O1Q = OA = r and A O O1 P = Q M(t) N(t) \omegat \omegat \varphi \varphi \angleQOA = \angleQO1A = \varphi. Since the two points return to A at the same time, it follows that \angleM(t)OA = \angleN(t)O1A = \omegat. Therefore \angleQOM(t) = \angleQO1N(t) = \varphi+\omegat, from which it follows that \triangleQOM(t) ∼= \triangleQO1N(t). Hence QM(t) = QN(t), as we claimed.