Track / Stage 4 / 18 of 340 #278 of 1964

Problem 278

AMC 12 late, AIME early

Geometry Difficulty 4.3 Prove it Taiwan IMO Selection Camp · Taiwan

Let $\triangle ABC$ be an acute triangle, and let point $P$ be a point on side $BC$ . It is known that segment $PE$ is perpendicular to $AC$ , and segment $PD$ is perpendicular to side $AB$ . Find: the area of triangle $PDE$ is maximal if and only if $P$ is the midpoint of $BC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

First note that
$2\Delta PDE \text{ 的面積} = PD \times PE \times \sin \angle DPE = PD \times PE \times \sin(\angle B + \angle C),$
therefore $\triangle PDE$ 面積達到最大值若且唯若 $PD \times PE$ 達到最大值.

Next, let $x, y$ be the lengths of segments $PE, PD$ respectively, and let $b, c$ be the lengths of segments $AC, AB$ respectively. Let $d$ be the area of triangle $ABC$ , then clearly $xb + yc = 2d$ . Hence $y = (2d - xb)/c$ . Therefore,
$PD \times PE = xy = \frac{2dx - bx^2}{c} = \frac{1}{c} \left( \frac{d^2}{b} - b(x - \frac{d}{b})^2 \right),$
hence $PD \times PE$ 達到最大值若且唯若 $PE = x = d/b$ . Again let the altitude from $BH$ of triangle $ABC$ onto $AC$ , then $BH = 2d/b = 2PE$ , hence by similar triangles $P$ is the midpoint of $BC$ .

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty, ordering) added by this project.