IMO 1979 · Problem 3

Two circles in a plane intersect. Let \(A\) be one of the points of intersection. Starting simultaneously from \(A\) two points move with constant speeds, each point travelling along its own circle in the same sense. The two points return to \(A\) simultaneously after one revolution. Prove that there is a fixed point \(P\) in the plane such that, at any time, the distances from \(P\) to the moving points are equal.

Official text of the problem. Official problem paper (PDF)

Solution

Let the circles have centres \(O_1,O_2\), and let \(T\) be the common time for one revolution. Because the speeds are constant and the senses of motion agree, at any time \(t\) both moving points have turned through the same signed angle \(\theta\) about their respective centres.

If \(O_1=O_2\), the circles coincide because both contain \(A\). Their moving points therefore coincide at every time, and any fixed point \(P\) works.

Suppose now that \(O_1\ne O_2\). Let \(s\) be reflection in the perpendicular bisector of \(O_1O_2\), and define \[ P=s(A). \] We prove that this fixed point has the required property.

At an arbitrary time, denote the moving points by \(X\) and \(Y\), and put \(X'=s(X)\). Reflection sends \(O_1\) to \(O_2\), sends \(A\) to \(P\), and reverses directed angles. Thus, since \(X\) is obtained from \(A\) by rotation through \(\theta\) about \(O_1\), the point \(X'\) is obtained from \(P\) by rotation through \(-\theta\) about \(O_2\).

Consequently, rotation through \(\theta\) about \(O_2\) sends \(X'\) to \(P\). The same rotation sends \(A\) to \(Y\). Since rotations preserve distances, \[ AX'=YP. \] Also, reflection sends \(P\) to \(A\) and \(X\) to \(X'\), so \[ PX=AX'. \] Combining these equalities gives \[ \boxed{PX=PY}. \]

The time was arbitrary, so the equality holds throughout the motion. The construction and proof also apply when the circles are tangent or when \(P=A\).

The solution was checked against John Scholes' IMO solution archive (Kalva).