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Difference between revisions of "2023 AIME I Problems/Problem 12"
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==Solution==
 
==Solution==
 
Miquel point and law of cosines, answer should be 75
 
Miquel point and law of cosines, answer should be 75
 +
 +
==Solution==
 +
 +
Denote <math>\theta = \angle AEP</math>.
 +
 +
In <math>AFPE</math>, we have <math>\overrightarrow{AF} + \overrightarrow{FP} + \overrightarrow{PE} + \overrightarrow{EA} = 0</math>.
 +
Thus,
 +
<cmath>
 +
\[
 +
AF + FP e^{i \theta} + PE e^{i \left( \theta + 60^\circ \right)}
 +
+ EA e^{- i 120^\circ} = 0.
 +
\]
 +
</cmath>
 +
 +
Taking the real and imaginary parts, we get
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
In <math>BDPF</math>, analogous to the analysis of <math>AFPE</math> above, we get
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Taking <math>(1) \cdot \sin \left( \theta + 60^\circ \right) - (2) \cdot \cos \left( \theta + 60^\circ \right)</math>, we get
 +
<cmath>
 +
\[
 +
AF \sin \left( \theta + 60^\circ \right)
 +
+ \frac{\sqrt{3}}{2} FP - EA \sin \theta = 0 . \hspace{1cm} (5)
 +
\]
 +
</cmath>
 +
 +
Taking <math>(3) \cdot \sin \theta - (4) \cdot \cos \theta</math>, we get
 +
<cmath>
 +
\[
 +
BD \sin \theta - \frac{\sqrt{3}}{2} FP + FB \sin \left( \theta + 120^\circ \right) . \hspace{1cm} (6)
 +
\]
 +
</cmath>
 +
 +
Taking <math>(5) + (6)</math>, we get
 +
<cmath>
 +
\[
 +
AF \sin \left( \theta + 60^\circ \right)
 +
- EA \sin \theta
 +
+ BD \sin \theta  + FB \sin \left( \theta + 120^\circ \right) .
 +
\]
 +
</cmath>
 +
 +
Therefore,
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore, <math>\tan^2 \theta = \boxed{\textbf{(075) }}</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==

Revision as of 14:49, 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

Miquel point and law of cosines, answer should be 75

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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