2023 AIME I Problems/Problem 8: Difference between revisions

Revision as of 17:09, 8 February 2023

There is a rhombus ABCD with an incircle. A point P is chosen somewhere on the incircle, and the distances from P to sides AB, CD, and BC, are 9, 16, and 5, respectively. Figure out the perimeter.

Solution

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)

@@ Line 1: / Line 1: @@
 There is a rhombus ABCD with an incircle. A point P is chosen somewhere on the incircle,
 and the distances from P to sides AB, CD, and BC, are 9, 16, and 5, respectively. Figure out the perimeter.
+==Solution==
+Denote by <math>O</math> the center of <math>ABCD</math>.
+We drop an altitude from <math>O</math> to <math>AB</math> that meets <math>AB</math> at point <math>H</math>.
+We drop altitudes from <math>P</math> to <math>AB</math> and <math>AD</math> that meet <math>AB</math> and <math>AD</math> at <math>E</math> and <math>F</math>, respectively.
+We denote <math>\theta = \angle BAC</math>.
+We denote the side length of <math>ABCD</math> as <math>d</math>.
+Because the distances from <math>P</math> to <math>BC</math> and <math>AD</math> are 16 and 9, respectively, and <math>BC \parallel AD</math>, the distance between each pair of two parallel sides of <math>ABCD</math> is <math>16 + 9 = 25</math>.
+Thus, <math>OH = \frac{25}{2}</math> and <math>d \sin \theta = 25</math>.
+We have
+<cmath>
+\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*}
+</cmath>
+Thus, <math>BH = OH \tan \angle BOH = \frac{25}{2} \tan \frac{\theta}{2}</math>.
+In <math>FAEP</math>, we have <math>\overrightarrow{FA} + \overrightarrow{AE} + \overrightarrow{EP} + \overrightarrow{PF} = 0</math>.
+Thus,
+<cmath>
+\[
+AF + AE e^{i \left( \pi - \theta \right)} + EP e^{i \left( \frac{3 \pi}{2} - \theta \right)}
+- PF i .
+\]
+</cmath>
+Taking the imaginary part of this equation and plugging <math>EP = 5</math> and <math>PF = 9</math> into this equation, we get
+<cmath>
+\[
+AE = \frac{9 + 5 \cos \theta}{\sin \theta} .
+\]
+</cmath>
+We have
+<cmath>
+\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*}
+</cmath>
+Because <math>P</math> is on the incircle of <math>ABCD</math>, <math>OP = \frac{25}{2}</math>. Plugging this into (1), we get the following equation
+<cmath>
+\[
+\sin \theta - 15 \cos \theta = 7 .
+\]
+</cmath>
+By solving this equation, we get <math>\sin \theta = \frac{4}{5}</math> and <math>\cos \theta = \frac{3}{5}</math>.
+Therefore, <math>d = \frac{25}{\sin \theta} = \frac{125}{4}</math>.
+Therefore, the perimeter of <math>ABCD</math> is <math>4d = \boxed{\textbf{(125) }}</math>.
+~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

Page

Toolbox

Search

2023 AIME I Problems/Problem 8: Difference between revisions

Revision as of 17:09, 8 February 2023

Solution

See also