## A Geometry Problem from the 2008 Singapore Mathematical Olympiad

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 $△ABC$, $\angle ACB=9{0}^{\circ }$, $D$ is the foot of the altitude from $C$ to $\overline{AB}$, and $E$ is the point on the side $\overline{BC}$ such that $CE=BD∕2$. Prove that $AD+CE=AE$.

Solution: Let the lengths of the sides of the triangle be $a$ (opposite $\angle CAB$), $b$ (opposite $\angle CBA$) and $c$ (opposite $\angle ACB$). Then by the Pythagorean Theorem, ${a}^{2}+{b}^{2}={c}^{2}$. Note that $△ACD\sim △BCD\sim △ABC$. Therefore, $AD={b}^{2}∕c$ and $CD=ab∕c$. Let $F$ be the midpoint of $\overline{DB}$. Also $BD={a}^{2}∕c$. Let $F$ be the midpoint of $\overline{BD}$. Then $DF=FB=CE={a}^{2}∕\left(2c\right)$. Then $AD+CE={b}^{2}∕c+{a}^{2}∕\left(2c\right)$. By the Pythagorean Theorem applied to $△AEC$, $A{E}^{2}={\left({a}^{2}∕\left(2c\right)\right)}^{2}+{b}^{2}={a}^{4}∕\left(4{c}^{2}\right)+{b}^{2}$. Now compare ${\left(AD+CE\right)}^{2}={\left({b}^{2}∕c+{a}^{2}∕\left(2c\right)\right)}^{2}={b}^{4}∕{c}^{2}+{a}^{2}{b}^{2}∕{c}^{2}+{a}^{4}∕\left(4{c}^{2}\right)={b}^{2}\left({b}^{2}+{a}^{2}\right)∕{c}^{2}+{a}^{4}∕\left(4{c}^{2}\right)={b}^{2}+{a}^{4}∕\left(4{c}^{2}\right)$. So $A{E}^{2}={\left(AD+CE\right)}^{2}$ 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.