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Difference between revisions of "2023 AIME I Problems/Problem 8"
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Unofficial problem: There is a rhombus ABCD with an incircle. A point P is chosen somewhere on the incircle,
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==Problem==
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and the distances from P to sides AB, CD, and BC, are 9, 16, and 5, respectively. Figure out the perimeter.
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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 P to the lines <math>DA, AB</math> and <math>BC</math> are <math>9, 5,</math> and <math>16</math>, respectively. Find the perimeter of <math>ABCD</math>.
  
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==Solution==
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==Solution 1==
  
 
Denote by <math>O</math> the center of <math>ABCD</math>.
 
Denote by <math>O</math> the center of <math>ABCD</math>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==See also==
 
==See also==

Revision as of 21:55, 8 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 $ABCD$.

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)


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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