Maths Olympiad Prep

Track / Stage 8 / 119 of 180 #1819 of 1964

Problem 1819

IMO Shortlist mid-range; USAMO P2/P5
Geometry Difficulty 8.4 Prove it Team Selection Test for IMO 2011 · Turkey · 2011

Let KK be a point in the interior of an acute triangle ABCABC and ARBPCQARBPCQ be a convex hexagon whose vertices lie on the circumcircle Γ\Gamma of the triangle ABCABC. Let A1A_1 be the second point where the circle passing through KK and tangent to Γ\Gamma at AA intersects the line APAP. The points B1B_1 and C1C_1 are defined similarly. Prove that
min{PA1AA1,QB1BB1,RC1CC1}1. \min \left\{ \frac{PA_1}{AA_1}, \frac{QB_1}{BB_1}, \frac{RC_1}{CC_1} \right\} \le 1.

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

Let OO be the center of Γ\Gamma. Since ABCABC is an acute triangle OO lies inside ABCABC. Assume that KK lies on the same side of the lines AOAO and BOBO as CC, and on the same side of the bisector of the line segment ABAB as BB. Then KAOAKA \ge OA.

Let ω\omega be the circle passing through KK and tangent to Γ\Gamma at AA. Then Γ\Gamma and ω\omega are homothetic with center AA and ratio PA/A1APA/A_1A. Since KAOAKA \ge OA, OO lies inside ω\omega and the homothety ratio is at most 22. Hence PA1/AA11PA_1/AA_1 \le 1.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.