## Geometry Problem 1 from the

1999 Bay Area Mathematical Olympiad

Steven R. Dunbar

Department of Mathematics

Univesity of Nebraska-Lincoln

Lincoln, Nebraska

www.math.unl.edu/$\sim $sdunbar1

sdunbar1 at unl dot edu

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
$0<a<b$. 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 $\angle BQP=\angle BOP$.

Solution: Consider the right $\u25b3PAB$.
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 $\u25b3PAB$ is
similar to $\u25b3OAQ$.
Then $AB\u2215AQ=AP\u2215AO$
and $\angle OAP=\angle QAB$.
Then $\u25b3OAP$ is
similar to $\u25b3QAB$.
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.