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

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<math>x^2+y^2+xy=42^2</math>
 
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<math>y^2+z^2+yz=35^2</math>  
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<math>z^2+x^2+zx=13^2</math>  
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<math>y^2+z^2+yz=35^2</math>
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<math>z^2+x^2+zx=13^2</math>
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By using Law of Cosines for <math>\triangle{DPE}</math>, <math>\triangle{EPF}</math>, and <math>\triangle{FPD}</math> respectively
 
By using Law of Cosines for <math>\triangle{DPE}</math>, <math>\triangle{EPF}</math>, and <math>\triangle{FPD}</math> respectively
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<math>(x+y+z)^2=\frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2}</math>
 
<math>(x+y+z)^2=\frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2}</math>
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<math>(x+y+z)^2=\frac{42^2+35^2+13^2+3*2^5*3*5}{2}</math>
 
<math>(x+y+z)^2=\frac{42^2+35^2+13^2+3*2^5*3*5}{2}</math>
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<math>(x+y+z)^2=2299</math>
 
<math>(x+y+z)^2=2299</math>
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<math>x+y+z=11\sqrt{19}</math>
 
<math>x+y+z=11\sqrt{19}</math>
  
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<math>\tan{\alpha}=-5\sqrt{3}</math>
 
<math>\tan{\alpha}=-5\sqrt{3}</math>
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<math>\tan^2{\alpha}=\boxed{075}</math>
 
<math>\tan^2{\alpha}=\boxed{075}</math>
  

Revision as of 22:56, 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

Solution 3

Draw line segments from P to points A, B, and C. And label the angle measure of $\angle{BFP}$, $\angle{CDP}$, and $\angle{AEP}$ to be $\alpha$

Using Law of Cosines (note that $cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180°-\alpha}=-cos{\alpha}$)

(1) $BP^2=FP^2+15^2-2*FP*15*cos(\alpha)$

(2) $BP^2=DP^2+7^2+2*DP*7*cos(\alpha)$

(3) $CP^2=DP^2+48^2-2*DP*48*cos(\alpha)$

(4) $CP^2=EP^2+30^2+2*EP*30*cos(\alpha)$

(5) $AP^2=EP^2+25^2-2*EP*25*cos(\alpha)$

(6) $AP^2=FP^2+40^2+2*FP*40*cos(\alpha)$

We can perform this operation: (1) - (2) + (3) - (4) + (5) - (6)

Leaving us with (after combining and simplifying)

$\cos{\alpha}=\frac{-11}{2*(DP+EP+FP)}$

Therefore, we want to solve for $DP+EP+FP$

Notice that $\angle{APC}=\angle{APC}=\angle{APC}=120°$

We can use Law of Cosines again to solve for the sides of DEF, which have side lengths of 13, 42, and 35 and area $120\sqrt{3}$.

Label the lengths of $PD$, $PE$, and $PF$ to be x, y, and z.

Therefore, using area formula,

$\area{\triangle{DEF}} = \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg)

$xy+yz+zx=2^5*3*5$

In addition, we know that

$x^2+y^2+xy=42^2$

$y^2+z^2+yz=35^2$

$z^2+x^2+zx=13^2$


By using Law of Cosines for $\triangle{DPE}$, $\triangle{EPF}$, and $\triangle{FPD}$ respectively

Because we want $DP+EP+FP$, which is $x+y+z$, we see that

$(x+y+z)^2=\frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2}$

$(x+y+z)^2=\frac{42^2+35^2+13^2+3*2^5*3*5}{2}$

$(x+y+z)^2=2299$

$x+y+z=11\sqrt{19}$

So plugging the results back into the equation before, we get

$\cos{\alpha} = \frac{-1}{2\sqrt{19}}$ leading to $\sin{\alpha} = \frac{5\sqrt{3}}{2\sqrt{19}}$

Giving us

$\tan{\alpha}=-5\sqrt{3}$

$\tan^2{\alpha}=\boxed{075}$

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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