Maths Olympiad Prep

Track / Stage 5 / 292 of 400 #892 of 1964

Problem 892

AIME late
Geometry Difficulty 5.8 Prove it

Let ω1\omega_{1} and ω2\omega_{2} be two circles, tt and tt^{\prime} two tangents to the circles. We denote T1T_{1} and T2T_{2} the points of tangency of tt with ω1\omega_{1} and ω2\omega_{2} respectively, and T1T_{1}^{\prime} and T2T_{2}^{\prime} the points of tangency with tt^{\prime}. Let MM be the midpoint of [T1T2]\left[T_{1} T_{2}\right], P1P_{1} and P2P_{2} the intersections of (MT1)\left(M T_{1}^{\prime}\right) and (MT2)\left(M T_{2}^{\prime}\right) with ω1\omega_{1} and ω2\omega_{2} respectively. Show that the quadrilateral P1P2T1T2P_{1} P_{2} T_{1}^{\prime} T_{2}^{\prime} is cyclic.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

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MM is on the radical axis of ω1\omega_{1} and ω2\omega_{2} because Pω1(M)=MT12=MT22=Pω2(M)P_{\omega_{1}}(M)=M T_{1}^{2}=M T_{2}^{2}=P_{\omega_{2}}(M). Therefore, MP1MT1=Pω1(M)=Pω2(M)=MP2MT2M P_{1} \cdot M T_{1}^{\prime}=P_{\omega_{1}}(M)=P_{\omega_{2}}(M)=M P_{2} \cdot M T_{2}^{\prime}, which concludes by the power of point MM.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.