Problem

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

Tags: ratio, geometry, circumcircle, geometry unsolved



Rushil

27.10.2005 10:00

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