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Problem 1330

AIME late

Geometry Difficulty 5.5 Prove it Czech and Slovak Mathematical Olympiad · Czech Republic

Let $ABC$ be an acute triangle with altitudes $BD$ , $CE$ . Given that $AE \cdot AD = BE \cdot CD$ , what is the smallest possible measure of $\angle BAC$ ? (Patrik Bak)

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

We denote $\angle BAC$ by $\alpha$ and express lengths $AE$ , $AD$ , $BE$ , $CD$ in terms of $\cos \alpha$ and the side lengths $b = AC$ , $c = AB$ of triangle $ABC$ . The condition rewrites as

$b \cos \alpha \cdot c \cos \alpha = (c - b \cos \alpha)(b - c \cos \alpha),$
which simplifies to $bc = (b^2 + c^2) \cos \alpha$ . Hence
$\cos \alpha = \frac{bc}{b^2 + c^2} \le \frac{1}{2},$
where the last inequality is for any positive $b$ , $c$ equivalent with an obvious inequality $(b-c)^2 \ge 0$ (alternatively, one can use AM-GM inequality for $b^2$ and $c^2$ ). We proved that angle $BAC$ of any such triangle $ABC$ satisfies $\cos \angle BAC \le 1/2$ , therefore $\angle BAC \ge 60^\circ$ . Since for equilateral triangle, the condition is clearly satisfied (in that case $AE = AD = BE = CD$ ), the answer is $60^\circ$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.