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Source: Indian Postal Coaching 2005; 18-th Korean Mathematical Olympiad 2005, final round, problem 4

October 27, 2005
ratiogeometrycircumcirclegeometry unsolved

Problem Statement

In the following, the point of intersection of two lines g g and h h will be abbreviated as gh g\cap h. Suppose ABC ABC is a triangle in which \angle A \equal{} 90^{\circ} and B>C \angle B > \angle C. Let O O be the circumcircle of the triangle ABC ABC. Let lA l_{A} and lB l_{B} be the tangents to the circle O O at A A and B B, respectively. Let BC \cap l_{A} \equal{} S and AC \cap l_{B} \equal{} D. Furthermore, let AB \cap DS \equal{} E, and let CE \cap l_{A} \equal{} T. Denote by P P the foot of the perpendicular from E E on lA l_{A}. Denote by Q Q the point of intersection of the line CP CP with the circle O O (different from C C). Denote by R R be the point of intersection of the line QT QT with the circle O O (different from Q Q). Finally, define U \equal{} BR \cap l_{A}. Prove that \frac {SU \cdot SP}{TU \cdot TP} \equal{} \frac {SA^{2}}{TA^{2}}.