2023 AIME I Problems/Problem 8

Problem

Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,$ $AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$

Solution 1

Denote by $O$ the center of $ABCD$ . We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$ . We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$ , respectively. We denote $\theta = \angle BAC$ . We denote the side length of $ABCD$ as $d$ .

Because the distances from $P$ to $BC$ and $AD$ are 16 and 9, respectively, and $BC \parallel AD$ , the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$ . Thus, $OH = \frac{25}{2}$ and $d \sin \theta = 25$ .

We have $\begin{align*} \angle BOH & = 90^\circ - \angle HBO \\ & = 90^\circ - \angle HBD \\ & = 90^\circ - \frac{180^\circ - \angle C}{2} \\ & = 90^\circ - \frac{180^\circ - \theta}{2} \\ & = \frac{\theta}{2} . \end{align*}$

Thus, $BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}$ .

In $FAEP$ , we have $\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0$ . Thus, $AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)} - PF i .$

Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get $AE = \frac{9 + 5 \cos \theta}{\sin \theta} .$

We have $\begin{align*} OP^2 & = \left( OH - EP \right)^2 + \left( AH - AE \right)^2 \\ & = \left( \frac{25}{2} - 5 \right)^2 + \left( d - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) \\ & = \left( \frac{15}{2} \right)^2 + \left( \frac{25}{\sin \theta} - \frac{25}{2} \tan \frac{\theta}{2} - \frac{9 + 5 \cos \theta}{\sin \theta} \right) . \hspace{1cm} (1) \end{align*}$

Because $P$ is on the incircle of $ABCD$ , $OP = \frac{25}{2}$ . Plugging this into (1), we get the following equation $20 \sin \theta - 15 \cos \theta = 7 .$

By solving this equation, we get $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$ . Therefore, $d = \frac{25}{\sin \theta} = \frac{125}{4}$ .

Therefore, the perimeter of $ABCD$ is $4d = \boxed{\textbf{(125) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2023 AIME I Problems/Problem 8

Problem

Solution 1

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