# APMO 1992 Problem 2

Problem 2 of Asian Pacific Mathematical Olympiad 1992

In a circle C with center O and radius r, let C_1, C_2 be two circles with centers O_1, O_2 and radii r_1, r_2 respectively, so that each circle C_i is internally tangent to C at A_i and so that C_1, C_2 are externally tangent to each other at A. Prove that the three lines OA, O_1A_2, and O_2A_1 are concurrent.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (tặng bạn Kha Nguyen) Extend OA to meet A_1A_2 at B. From O_1 and O_2 draw altitudes O_1G and O_2H to OB, respectively. From O draw tangential lines to C_1 and C_2 and meet them at E and F, respectively. Use the law of the sine function, we have

\frac {A_1B}{sin∠A_1OB} = \frac {OB}{sin∠OA_1B} = \frac {OB}{sin∠OA_2B} = \frac {A_2B}{sin∠A_2OB} or

\frac {A_1B}{A_2B} = \frac {sin∠A_1OB}{sin∠A_2OB} = \frac {O_1G}{O_1O} / \frac {O_2H}{O_2O} = \frac {O_1G}{O_1O} / \frac {O_2O}{O_2H} (1)

Because the two triangles GO_1A and HO_2A are similar, we have

\frac {O_1G}{O_1A} = \frac {O_2H}{O_2A} or \frac {O_1G}{O_2H} = \frac {r_1}{r_2} and (1) becomes

\frac {A_1B}{A_2B} = \frac {r_1}{r_2} × \frac {O_2O}{O_1O} = \frac {r_1}{r_2} × \frac {r - r_2}{r - r_1} or

\frac {A_1B}{A_2B} × \frac {r_2}{r - r_2} × \frac {r - r_1}{r_1} = 1

Therefore, per Cevas theorem, the three lines OA, O_1A_2, and O_2A_1 are concurrent.