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Difference between revisions of "2023 AIME I Problems/Problem 12"
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==Problem 12==
 
==Problem 12==
 
Let <math>ABC</math> be an equilateral triangle with side length <math>55</math>. Points <math>D</math>, <math>E</math>, and <math>F</math> lie on sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively, such that <math>BD=7</math>, <math>CE=30</math>, and <math>AF=40</math>. A unique point <math>P</math> inside <math>\triangle ABC</math> has the property that <cmath>\measuredangle AEP=\measuredangle BFP=\measuredangle CDP.</cmath> Find <math>\tan^{2}\measuredangle AEP</math>.
 
Let <math>ABC</math> be an equilateral triangle with side length <math>55</math>. Points <math>D</math>, <math>E</math>, and <math>F</math> lie on sides <math>BC</math>, <math>CA</math>, and <math>AB</math>, respectively, such that <math>BD=7</math>, <math>CE=30</math>, and <math>AF=40</math>. A unique point <math>P</math> inside <math>\triangle ABC</math> has the property that <cmath>\measuredangle AEP=\measuredangle BFP=\measuredangle CDP.</cmath> Find <math>\tan^{2}\measuredangle AEP</math>.
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==Solution==
 
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Miquel point and law of cosines, answer should be 75
 
  
 
==Solution==
 
==Solution==

Revision as of 19:21, 8 February 2023

Problem 12

Let $ABC$ be an equilateral triangle with side length $55$. Points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that $BD=7$, $CE=30$, and $AF=40$. A unique point $P$ inside $\triangle ABC$ has the property that \[\measuredangle AEP=\measuredangle BFP=\measuredangle CDP.\] Find $\tan^{2}\measuredangle AEP$.

Solution

Denote $\theta = \angle AEP$.

In $AFPE$, we have $\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0$. Thus, \[ AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)} + EA e^{- i 120^\circ} = 0. \]

Taking the real and imaginary parts, we get \begin{align*} AF + FP \cos \theta + PE \cos \left( \theta + 60^\circ \right) + EA \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (1) \\ FP \sin \theta + PE \sin \left( \theta + 60^\circ \right) + EA \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (2) \end{align*}

In $BDPF$, analogous to the analysis of $AFPE$ above, we get \begin{align*} BD + DP \cos \theta + PF \cos \left( \theta + 60^\circ \right) + FB \cos \left( - 120^\circ \right) & = 0 \hspace{1cm} (3) \\ DP \sin \theta + PF \sin \left( \theta + 60^\circ \right) + FB \sin \left( - 120^\circ \right) & = 0 \hspace{1cm} (4) \end{align*}

Taking $(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)$, we get \[ AF \sin \left( \theta + 60^\circ \right) + \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5) \]

Taking $(3) \cdot \sin \theta - (4) \cdot \cos \theta$, we get \[ BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6) \]

Taking $(5) + (6)$, we get \[ AF \sin \left( \theta + 60^\circ \right) - EA \sin \theta + BD \sin \theta + FB \sin \left( \theta + 120^\circ \right) . \]

Therefore, \begin{align*} \tan \theta & = \frac{\frac{\sqrt{3}}{2} \left( AF + FB \right)} {\frac{FB}{2} + EA - \frac{AF}{2} - BD} \\ & = 5 \sqrt{3} . \end{align*}

Therefore, $\tan^2 \theta = \boxed{\textbf{(075) }}$.

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

See also

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

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