A Geometry Problem from the
2008 Singapore Mathematical Olympiad

Steven R. Dunbar
Department of Mathematics
University of Nebraska-Lincoln
Lincoln, Nebraska
www.math.unl.edu/ sdunbar
sdunbar1 at unl dot edu

January 21, 2009

Problem: Problem 1 from the Singapore Mathematical Olympiad 2008 (Junior Section, Round 2) of June 28, 2008. Collected at the 2008 International Mathematical Olympiad, Madrid Spain, July 2008.

In ABC, ACB = 90, D is the foot of the altitude from C to AB¯, and E is the point on the side BC¯ such that CE = BD2. Prove that AD + CE = AE.

Solution: Let the lengths of the sides of the triangle be a (opposite CAB), b (opposite CBA) and c (opposite ACB). Then by the Pythagorean Theorem, a2 + b2 = c2. Note that ACD BCD ABC. Therefore, AD = b2c and CD = abc. Let F be the midpoint of DB¯. Also BD = a2c. Let F be the midpoint of BD¯. Then DF = FB = CE = a2(2c). Then AD + CE = b2c + a2(2c). By the Pythagorean Theorem applied to AEC, AE2 = (a2(2c))2 + b2 = a4(4c2) + b2. Now compare (AD+CE)2 = (b2c+a2(2c))2 = b4c2+a2b2c2+a4(4c2) = b2(b2+a2)c2+a4(4c2) = b2+a4(4c2). So AE2 = (AD + CE)2 and AE = AD + CE.

Observe in the GeoGebra applet that length g = AE is the sum of lengths h = AD and i = CE.

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)

Commentary: This problem depends on the fairly well-known theorem that the altitude from the right angle to the hypotenuse creates similar right triangles. Given that the problem statement involves a right triangle, using the Pythagorean Theorem to solve the problem is natural. After that, the rest of the of the problem follows quickly and easily.