## 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/$\sim $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 $\u25b3ABC$,
$\angle ACB=9{0}^{\circ}$,
$D$ is the foot of the
altitude from $C$
to $\overline{AB}$, and
$E$ is the point
on the side $\overline{BC}$
such that $CE=BD\u22152$.
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 $\u25b3ACD\sim \u25b3BCD\sim \u25b3ABC$.
Therefore, $AD={b}^{2}\u2215c$
and $CD=ab\u2215c$. Let
$F$ be the
midpoint of $\overline{DB}$.
Also $BD={a}^{2}\u2215c$.
Let $F$ be the
midpoint of $\overline{BD}$.
Then $DF=FB=CE={a}^{2}\u2215\left(2c\right)$. Then
$AD+CE={b}^{2}\u2215c+{a}^{2}\u2215\left(2c\right)$. By the Pythagorean
Theorem applied to $\u25b3AEC$,
$A{E}^{2}={\left({a}^{2}\u2215\left(2c\right)\right)}^{2}+{b}^{2}={a}^{4}\u2215\left(4{c}^{2}\right)+{b}^{2}$. Now
compare ${\left(AD+CE\right)}^{2}={\left({b}^{2}\u2215c+{a}^{2}\u2215\left(2c\right)\right)}^{2}={b}^{4}\u2215{c}^{2}+{a}^{2}{b}^{2}\u2215{c}^{2}+{a}^{4}\u2215\left(4{c}^{2}\right)={b}^{2}\left({b}^{2}+{a}^{2}\right)\u2215{c}^{2}+{a}^{4}\u2215\left(4{c}^{2}\right)={b}^{2}+{a}^{4}\u2215\left(4{c}^{2}\right)$.
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 $i=CE$.

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.