## Geometry Problem 1 from the 1999 Bay Area Mathematical Olympiad

January 29, 2009

Problem: Let $k$ be a circle in the $xy$-plane with center on the $y$-axis and passing through $A=\left(0,a\right)$ and $B=\left(0,b\right)$ with $0. Let $P$ be any other point on the circle, let $Q$ be the intersection of the line through $P$ and $A$ with the $x$-axis and let $O=\left(0,0\right)$. Prove that $\angle BQP=\angle BOP$.

Sorry, the GeoGebra Applet could not be started. Please make sure that Java 1.4.2 (or later) is installed and active in your browser (Click here to install Java now)

Solution: Consider the right $△PAB$. It is a right triangle because $P$ is on the circle and $\overline{AB}$ is a diameter. By opposite angles $\angle PAB=\angle OAQ$. Hence $△PAB$ is similar to $△OAQ$. Then $AB∕AQ=AP∕AO$ and $\angle OAP=\angle QAB$. Then $△OAP$ is similar to $△QAB$. Then $\angle BQP=\angle BOP$.

Commentary: The problem requests us to show that 2 angles are equal. The main practical tool for showing that two angles are equal is to show that two triangles are similar. To show two triangles are similar without knowing something about the angles, we need to show proportionality of sides. That means we should find some other triangles with some proportional sides. The fact that the problem is embedded in the coordinate plane suggests that we should use the right angle at the intersection of the $x$ and $y$ axes. In turn that suggests that another right angle is needed. The logical place to look for a right angle is in a semi-circle bounded by the $y$-axis.