IMO 1988 · Problem 1

Consider two coplanar circles of radii \(R\) and \(r\) (\(R>r\)) with the same center. Let \(P\) be a fixed point on the smaller circle and \(B\) a variable point on the larger circle. The line \(BP\) meets the larger circle again at \(C\). The perpendicular \(l\) to \(BP\) at \(P\) meets the smaller circle again at \(A\). (If \(l\) is tangent to the circle at \(P\) then \(A=P\).)

(i) Find the set of values of \(BC^2+CA^2+AB^2\).

(ii) Find the locus of the midpoint of \(BC\).

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

Answer

(i) The singleton set \[ \boxed{\{6R^2+2r^2\}}. \]

(ii) The entire circle with diameter \(OP\), where \(O\) is the common center, including the points \(O\) and \(P\).

Solution

Let \(O\) be the common center. Since \(OP=r<R\), every line through \(P\) meets the larger circle at two distinct points, with \(P\) between them.

(i) For any choice of \(B\), choose orthonormal coordinates with \(P=(0,0)\) and \(BP\) as the horizontal axis. Write \(O=(u,v)\). Then \[ u^2+v^2=r^2. \]

The larger circle has equation \[ (x-u)^2+(y-v)^2=R^2. \] Consequently, its intersections with \(BP\) are, in some order, \[ B=(u+d,0),\qquad C=(u-d,0), \qquad d=\sqrt{R^2-v^2}>0. \]

The line \(l\) is \(x=0\). Its intersections with the smaller circle satisfy \[ u^2+(y-v)^2=r^2=u^2+v^2, \] so \(y(y-2v)=0\). Thus \[ A=(0,2v). \] When \(v=0\), this gives \(A=P\), exactly as prescribed for the tangent case.

It follows that \[ \begin{aligned} BC^2+CA^2+AB^2 &=4d^2+\bigl((u-d)^2+4v^2\bigr) +\bigl((u+d)^2+4v^2\bigr)\\ &=6d^2+2u^2+8v^2\\ &=6(R^2-v^2)+2u^2+8v^2\\ &=6R^2+2(u^2+v^2)\\ &=6R^2+2r^2. \end{aligned} \] Hence the requested set contains precisely this one value.

(ii) Let \(M\) be the midpoint of \(BC\). The midpoint of a chord is the orthogonal projection of the center onto the chord’s line. Therefore, if \(M\ne O,P\), then \[ OM\perp PM. \] By Thales’ theorem, \(M\) lies on the circle with diameter \(OP\). If \(M=O\) or \(M=P\), it also lies on that circle.

To prove that every point of this circle occurs, let \(X\) be any point on it.

  • If \(X\ne O,P\), Thales’ theorem gives \(OX\perp PX\). Choose \(B,C\) as the intersections of the line \(PX\) with the larger circle. Since \(OX\) is perpendicular to this chord, \(X\) is its midpoint.
  • If \(X=O\), choose \(B,C\) on the line \(OP\). The chord is a diameter, whose midpoint is \(O\).
  • If \(X=P\), choose \(B,C\) on the line through \(P\) perpendicular to \(OP\). The perpendicular from \(O\) to this chord meets it at \(P\), so its midpoint is \(P\).

All these lines meet the larger circle at two distinct points because they pass through its interior point \(P\). Thus the locus is exactly the entire circle with diameter \(OP\).

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