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Difference between revisions of "2023 AIME I Problems/Problem 8"

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==Problem==
 
==Problem==
 
Rhombus <math>ABCD</math> has <math>\angle BAD < 90^\circ.</math> There is a point <math>P</math> on the incircle of the rhombus such that the distances from <math>P</math> to the lines <math>DA,</math> <math>AB,</math> and <math>BC</math> are <math>9,</math> <math>5,</math> and <math>16,</math> respectively. Find the perimeter of <math>ABCD.</math>
 
Rhombus <math>ABCD</math> has <math>\angle BAD < 90^\circ.</math> There is a point <math>P</math> on the incircle of the rhombus such that the distances from <math>P</math> to the lines <math>DA,</math> <math>AB,</math> and <math>BC</math> are <math>9,</math> <math>5,</math> and <math>16,</math> respectively. Find the perimeter of <math>ABCD.</math>
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==Diagram==
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Available soon. I am not gonna use the diagram from Math Jams. Instead, I will make one that's 100% to scale.
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 +
~MRENTHUSIASM
  
 
==Solution 1==
 
==Solution 1==

Revision as of 14:34, 9 February 2023

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

Available soon. I am not gonna use the diagram from Math Jams. Instead, I will make one that's 100% to scale.

~MRENTHUSIASM

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)

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=125/4 \\ 4WX=\boxed{125} \end{align*}

~Danielzh

See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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