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Problem 1441

AIME late

Geometry Difficulty 5.8 Prove it HMMT February · United States · 2024

Let $ABC$ be an acute triangle. Let $D$ , $E$ , and $F$ be the feet of altitudes from $A$ , $B$ , and $C$ to sides $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ , respectively, and let $Q$ be the foot of altitude from $A$ to line $EF$ . Given that $AQ = 20$ , $BC = 15$ , and $AD = 24$ , compute the perimeter of triangle $DEF$ .

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

Solution:

Note that $A$ is the excenter of $\triangle DEF$ and $AQ$ is the length of the exradius. Let $T$ be the tangency point of the $A$ -excircle to line $DF$ . We have $AQ = AT = 20$ . It is well known that the length of $DT$ is the semiperimeter of $DEF$ . Note that $\triangle ADT$ is a right triangle, so
$AT^2 + DT^2 = AD^2$
which implies
$DT = \sqrt{24^2 - 20^2} = 4\sqrt{11}$
Thus, the perimeter of $\triangle DEF$ is $2 \cdot 4\sqrt{11} = 8\sqrt{11}$ .

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