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Difference between revisions of "2023 AIME I Problems/Problem 12"

(→‎Solution 2 (way quicker))
(→‎Solution 2 (way quicker))
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Drop the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{AC}</math>, <math>\overline{BC}</math>, and call them <math>Q,R,</math> and <math>S</math> respectively. This gives us three similar right triangles <math>FQP</math>, <math>ERP</math>, and <math>DSP.</math>  
 
Drop the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{AC}</math>, <math>\overline{BC}</math>, and call them <math>Q,R,</math> and <math>S</math> respectively. This gives us three similar right triangles <math>FQP</math>, <math>ERP</math>, and <math>DSP.</math>  
  
−
The sum of the perpendiculars to a point P within an equilateral triangle is always constant, so we have that <math>PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.</math>  
+
The sum of the perpendiculars to a point <math>P</math> within an equilateral triangle is always constant, so we have that <math>PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.</math>  
  
−
The sum of the lengths of the alternating segments split by the perpendiculars from a point P is always equal to half the perimeter, so <math>QA+RC+SB = \dfrac{165}{2},</math> which means that <math>FQ+ER+DS = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.</math>
+
The sum of the lengths of the alternating segments split by the perpendiculars from a point <math>P</math> within an equilateral triangle is always equal to half the perimeter, so <math>QA+RC+SB = \dfrac{165}{2},</math> which means that <math>FQ+ER+DS = QA+RC+SB - CE - AF - BD = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.</math>
  
 
Finally, <math>\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.</math>  
 
Finally, <math>\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.</math>  

Revision as of 20:55, 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)

Solution 2 (way quicker)

Drop the perpendiculars from $P$ to $\overline{AB}$, $\overline{AC}$, $\overline{BC}$, and call them $Q,R,$ and $S$ respectively. This gives us three similar right triangles $FQP$, $ERP$, and $DSP.$

The sum of the perpendiculars to a point $P$ within an equilateral triangle is always constant, so we have that $PQ+PR+PS=\dfrac{55 \sqrt{3}}{2}.$

The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is always equal to half the perimeter, so $QA+RC+SB = \dfrac{165}{2},$ which means that $FQ+ER+DS = QA+RC+SB - CE - AF - BD = \dfrac{165}{2} - 30 - 40 - 7 = \dfrac{11}{2}.$

Finally, $\tan AEP = \dfrac{PQ}{FQ} = \dfrac{PR}{ER} = \dfrac{PS}{DS} = \dfrac{PQ+PR+PS}{FQ+ER+DS} = 5 \sqrt{3}.$

Thus, $\tan^2 AEP = \boxed{075.}$

~anon

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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