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 $ABCD.$

Diagram

~MRENTHUSIASM

Solution 1

This solution refers to the Diagram section.

Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrightarrow{AB},$ and $\overleftrightarrow{BC},$ respectively, such that $\overline{RT}$ intersects $\odot O$ at $P$ and $Q.$

We obtain the following diagram:

Solution 2

Label the points of the rhombus to be $X$ , $Y$ , $Z$ , and $W$ and the center of the incircle to be $O$ so that $9$ , $5$ , and $16$ are the distances from point $P$ to side $ZW$ , side $WX$ , and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\frac{25}{2}$ .

Call the feet of the altitudes from P to side $ZW$ , side $WX$ , and side $XY$ to be $A$ , $B$ , and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$ , side $WX$ , and side $XY$ to be $D$ , $E$ , and $F$ respectively.

Draw a line segment from $P$ to $\overline{OD}$ so that it is perpendicular to $\overline{OD}$ . Notice that this segment length is equal to $AD$ and is $\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12$ by Pythagorean Theorem.

Similarly, perform the same operations with side $WX$ to get $BE=10$ .

By equal tangents, $WD=WE$ . Now, label the length of segment $WA=n$ and $WB=n+2$ .

Using Pythagorean Theorem again, we get

$\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\ n^2+9^2&=(n+2)^2+5^2 \\ n&=13. \end{align*}$

Which also gives us $\tan{\angle{OWX}}=\frac{1}{2}$ and $OW=\frac{25\sqrt{5}}{2}$ .

Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that

$\begin{align*} \frac{OX}{OW}&=\tan{\angle{OWX}} \\ OX&=\frac{25\sqrt{5}}{4} \\ WX^2&=OW^2+OX^2 \\ WX&=\frac{125}{4} \\ 4WX&=\boxed{125}. \end{align*}$

~Danielzh

Solution 3

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

Because $P$ is on the incircle of $ABCD$ , $OP = \frac{25}{2}$ . Plugging this into $(\bigstar)$ , 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 $4d = \boxed{125}$ .

~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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