IMO 1993 SL 21

Origin: GBR

Problem

A circle S is said to cut a circle \Sigma diametrally if their common chord is a diameter of \Sigma. Let SA, SB, SC be three circles with distinct centers A, B, C respectively. Prove that A, B, C are collinear if and only if there is no unique circle S that cuts each of SA, SB, SC diametrally. Prove further that if there exists more than one circle S that cuts each of SA, SB, SC diametrally, then all such circles pass through two fixed points. Locate these points in relation to the circles SA, SB, SC.

Solution

Assume that S is a circle with center O that cuts Si diametrically in points Pi, Qi, i \in{A, B, C}, and denote by ri, r the radii of Si and S respectively. Since OA is perpendicular to PAQA, it follows by Pythagoras’s theorem that OA2 +AP 2 A = OP 2 A, i.e., r2 A +OA2 = r2. Analogously r2 B +OB2 = r2 and r2 C + OC2 = r2. Thus if OA, OB, OC are the feet of perpendiculars from O to BC, CA, AB respectively, then OCA2 −OCB2 = r2 B −r2 A. Since the left-hand side is a monotonic function of OC \inAB, the point OC is uniquely determined by the imposed conditions. The same holds for OA and OB. If A, B, C are not collinear, then the positions of OA, OB, OC uniquely determine the point O, and there- fore the circle S also. On the other hand, if A, B, C are collinear, all one can deduce is that O lies on the lines lA, lB, lC through OA, OB, OC, per- pendicular to BC, CA, AB respec- tively. By this, lA, lB, lC are paral- lel, so O can be either anywhere on the line if these lines coincide, or A B C O OA OB OC SA SB SC lA lB lC S nowhere if they don’t coincide. So if there exists more than one circle S, A, B, C lie on a line and the foot O′ of the perpendicular from O to the line ABC is fixed. If X, Y are the intersection points of S and the line ABC, then r2 = OX2 = OA2 + r2 A and consequently O′X2 = O′A2 + r2 A, which implies that X, Y are fixed.