IMO 2012 Shortlist G5

Category: Geometry

Problem

Let ABC be a triangle with ∠BCA = 90◦ , and let C0 be the foot of the altitude from C. Choose a point X in the interior of the segment CC0, and let K,L be the points on the segments AX,BX for which BK = BC and AL = AC respectively. Denote by M the intersection of AL and BK. Show that MK = ML. Solution. Let C′ be the reflection of C in the line AB, and let ω1 and ω2 be the circles with centers A and B, passing through L and K respectively. Since AC′ = AC = AL and BC′ = BC = BK, both ω1 and ω2 pass through C and C′ . By ∠BCA = 90◦ , AC is tangent to ω2 at C, and BC is tangent to ω1 at C. Let K1 6= K be the second intersection of AX and ω2, and let L1 6= L be the second intersection of BX and ω1. A K M L1 K1 ω3 C L B C0 C′ X ω2 ω1 By the powers of X with respect to ω2 and ω1, XK · XK1 = XC · XC′ = XL · XL1, so the points K1, L, K, L1 lie on a circle ω3. The power of A with respect to ω2 gives AL2 = AC2 = AK · AK1, indicating that AL is tangent to ω3 at L. Analogously, BK is tangent to ω3 at K. Hence MK and ML are the two tangents from M to ω3 and therefore MK = ML.34