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/ 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 MBC.

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Solution: Set the triangle in a coordinate plane, with the origin at B, with side BC¯ along the x-axis, side AC¯ along the y-axis. Assign A = (0,a) and C = (c, 0). Let F = (c2,a2) be the midpoint of AC¯. Then M is located on the perpendicular bisector FM of the hypotenuse AC¯. In fact FM = FB. That is, the point M can be located as the tip of the vector FC rotated 90 counterclockwise about the point F = (c2,a2). The vector FC = (c2,a2). Rotating it 90 counterclockwise yields the vector (a2,c2). Adding this to the vector (c2,a2) gives M = ((a + c)2, (a + c)2). Then clearly MBC = 45.

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 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.