IMO 1971 SL 4

Origin: GBR

Problem

We are given two mutually tangent circles in the plane, with radii r1, r2. A line intersects these circles in four points, determining three segments of equal length. Find this length as a function of r1 and r2 and the condition for the solvability of the problem.

Solution

In the coordinate system in which the x-axis passes through the centers of the circles and the y-axis is their common tangent, the circles have equations x2 + y2 + 2r1x = 0, x2 + y2 −2r2x = 0. Let p be the desired line with equation y = ax+b. The abscissas of points of intersection of p with both circles satisfy one of (1 + a2)x2 + 2(ab + r1)x + b2 = 0, (1 + a2)x2 + 2(ab −r2)x + b2 = 0. Let us denote the lengths of the chords and their projections onto the x-axis by d and d1, respectively. From these equations it follows that d2 1 = 4(ab + r1)2 (1 + a2)2 − 4b2 1 + a2 = 4(ab −r2)2 (1 + a2)2 − 4b2 1 + a2 . (1) Consider the point of intersection of p with the y-axis. This point has equal powers with respect to both circles. Hence, if that point divides the segment determined on p by the two circles on two segments of lengths x and y, this power equals x(x + d) = y(y + d), which implies x = y = d/2. Thus each of the equations in (1) has two roots, one of which is thrice the other. This fact gives us (ab + r1)2 = 4(1 + a2)b2/3. From (1) and this we obtain ab = r2 −r1 , 4b2 + a2b2 = 3[(ab + r1)2 −a2b2] = 3r1r2; a2 = 4(r2 −r1)2 14r1r2 −r2 1 −r2 , b2 = 14r1r2 −r2 1 −r2 ; d2 1 = (14r1r2 −r2 1 −r2 2)2 36(r1 + r2)2 . Finally, since d2 = d2 1(1 + a2), we conclude that d2 = 1 12(14r1r2 −r2 1 −r2 2), and that the problem is solvable if and only if 7 −4 \sqrt 3 \leqr1 r2 \leq7 + 4 \sqrt 3.