## A Geometry Problem from the

2002 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 31, 2009

Problem: Let $ABC$ be a right
triangle with right angle at $B$.
Let $ACDE$ be a square drawn
exterior to triangle $ABC$.
If $M$
is the center of this square, find the measure of
$\angle MBC$.

Solution: Set the triangle in a coordinate plane, with the origin at
$B$, with side
$\overline{BC}$ along
the $x$-axis,
side $\overline{AC}$ along
the $y$-axis.
Assign $A=\left(0,a\right)$
and $C=\left(c,0\right)$. Let
$F=\left(c\u22152,a\u22152\right)$ be the midpoint
of $\overline{AC}$. Then
$M$ is located on the
perpendicular bisector $\stackrel{\u20e1}{FM}$
of the hypotenuse $\overline{AC}$. In
fact $FM=FB$. That is, the point
$M$ can be located as the
tip of the vector $\stackrel{\u20d7}{FC}$ rotated
$9{0}^{\circ}$ counterclockwise
about the point $F=\left(c\u22152,a\u22152\right)$. The
vector $\stackrel{\u20d7}{FC}=\left(c\u22152,-a\u22152\right)$. Rotating it
$9{0}^{\circ}$ counterclockwise yields
the vector $\left(a\u22152,c\u22152\right)$. Adding
this to the vector $\left(c\u22152,a\u22152\right)$
gives $M=\left(\left(a+c\right)\u22152,\left(a+c\right)\u22152\right)$. Then
clearly $\angle MBC=4{5}^{\circ}$.

Commentary: This problem seems to be very easy, too easy in fact to require
a coordinate solution. At first glance, it seems that angle chasing should be enough to
find $\angle MBC$.
However, since the figure and construction depends more on slopes than on
angles, it seems reasonable to use a solution that uses slope information
more. A coordinate or vector proof seems to be the easiest way to do
that.