## A Geometry Problem from the 2002 Bay Area Mathematical Olympiad

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 $\angle MBC$.

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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 $F=\left(c∕2,a∕2\right)$ be the midpoint of $\overline{AC}$. Then $M$ is located on the perpendicular bisector $\stackrel{⃡}{FM}$ of the hypotenuse $\overline{AC}$. In fact $FM=FB$. That is, the point $M$ can be located as the tip of the vector $\stackrel{⃗}{FC}$ rotated $9{0}^{\circ }$ counterclockwise about the point $F=\left(c∕2,a∕2\right)$. The vector $\stackrel{⃗}{FC}=\left(c∕2,-a∕2\right)$. Rotating it $9{0}^{\circ }$ counterclockwise yields the vector $\left(a∕2,c∕2\right)$. Adding this to the vector $\left(c∕2,a∕2\right)$ gives $M=\left(\left(a+c\right)∕2,\left(a+c\right)∕2\right)$. 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.