2023 AIME I Problems/Problem 12

Problem

Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that $\angle AEP = \angle BFP = \angle CDP.$ Find $\tan^2(\angle AEP).$

Solution (Coord bash)

By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$ , $EF=35$ , and $FD=13$ . Toss the points on the coordinate plane; let $B=(-7, 0)$ , $D=(0, 0)$ , and $C=(48, 0)$ , where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$ .

By the extended law of sines, the radius of circle $(BFD)$ is $\frac{13}{2\sin 60^{\circ}}=\frac{13}{3}\sqrt{3}$ . Its center lies on the line $x=-\frac{7}{2}$ , and the origin is a point on it, so $y=\frac{23}{6}\sqrt{3}$ .

The radius of circle $(CDE)$ is $\frac{42}{2\sin 60^{\circ}}=14\sqrt{3}$ . The origin is also a point on it, and its center is on the line $x=24$ , so $y=2\sqrt{3}$ .

The equations of the two circles are $\begin{align*}(BFD)&:\left(x+\tfrac{7}{2}\right)^{2}+\left(y-\tfrac{23}{6}\sqrt{3}\right)^{2}=\tfrac{169}{3} \\ (CDE)&:\left(x-24\right)^{2}+\left(y-2\sqrt{3}\right)^{2}=588\end{align*}$ These equations simplify to $\begin{align*}(BFD)&:x^{2}+7x+y^{2}-\tfrac{23}{3}\sqrt{3}y=0 \\ (CDE)&: x^{2}-48x+y^{2}-4\sqrt{3}y=0\end{align*}$ Subtracting these two equations gives that both their points of intersection, $D$ and $P$ , lie on the line $55x-\tfrac{11}{3}\sqrt{3}y=0$ . Hence, $\tan^{2}\left(\measuredangle AEP\right)=\tan^{2}\left(\measuredangle CDP\right)=\left(\frac{55}{\tfrac{11}{3}\sqrt{3}}\right)^{2}=3\left(\tfrac{55}{11}\right)^{2}=\boxed{075}$ . To scale, the configuration looks like the figure below.

Solution 2 (Vectors/Complex)

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 3 (Synthetic)

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 4 (LOC)

This solution is heavily inspired by AIME 1999 Problem 14 Solution 2 (a similar question)

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 $\cos{\angle{AFP}}=\cos{\angle{BDP}}=\cos{\angle{CEP}}=\cos{180^\circ-\alpha}=-\cos{\alpha}$ )

$\begin{align*} (1) BP^2 &= FP^2+15^2-2*FP*15*\cos(\alpha)\\ (2) BP^2 &= DP^2+7^2+2*DP*7*\cos(\alpha)\\ (3) CP^2 &= DP^2+48^2-2*DP*48*\cos(\alpha)\\ (4) CP^2 &= EP^2+30^2+2*EP*30*\cos(\alpha)\\ (5) AP^2 &= EP^2+25^2-2*EP*25*\cos(\alpha)\\ (6) AP^2 &= FP^2+40^2+2*FP*40*\cos(\alpha)\\ \end{align*}$

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

Leaving us with (after combining and simplifying)

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

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

Notice that $\angle{APC}=\angle{APC}=\angle{APC}=120^\circ$

We can use Law of Cosines again to solve for the sides of $\triangle{DEF}$ , 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 the $\sin$ area formula,

$\begin{align*} [\triangle{DEF}] &= \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3} \\ xy+yz+zx &= 2^5*3*5 \end{align*}$

In addition, we know that

$\begin{align*} x^2+y^2+xy=42^2\\ y^2+z^2+yz=35^2\\ z^2+x^2+zx=13^2\\ \end{align*}$

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

$\begin{align*} (x+y+z)^2 &= \frac{x^2+y^2+x+y^2+z^2+yz+z^2+x^2+3(zx+xy+yz+zx)}{2} \\ (x+y+z)^2 &= \frac{42^2+35^2+13^2+3*2^5*3*5}{2} \\ (x+y+z)^2 &= 2299 \\ x+y+z &= 11\sqrt{19} \end{align*}$

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

$\begin{align*} \cos{\alpha} = \frac{-1}{2\sqrt{19}}\\ \sin{\alpha} = \frac{5\sqrt{3}}{2\sqrt{19}} \end{align*}$

Giving us

$\begin{align*} \tan^2{\alpha}=\boxed{075} \end{align*}$

~Danielzh

Solution 5

This diagram isn't to scale

By the law of cosines, $FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.$ Similarly we get $FD=13$ and $DE=42$ . $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$ , $BDPF$ , and $CDPE$ are three cyclic quadrilaterals. Using the law of sines in each, $\frac{AP}{35}=\frac{AP}{FE}=\frac{CP}{42}=\frac{CP}{ED}=\frac{BP}{13}=\frac{BP}{DF}=\frac{\sin\theta}{\sin\frac\pi3}.$ So we can set $AP=35k$ , $BP=13k$ , and $CP=42k$ . Let $PD=d$ , $PE=e$ , and $PF=f$ . Applying Ptolemy theorem in the cyclic quadrilaterals, $\begin{cases}AP\cdot FE=AF\cdot PE+AE\cdot PF,\\CP\cdot ED=CE\cdot PD+CD\cdot PE,\\BP\cdot DF=BD\cdot PF+BF\cdot PD.\end{cases} \implies \begin{cases} 1225k=40e+25f,\\1764k=30d+48e,\\169k=15d+7f, \end{cases}$ We can solve out $d=\frac{54k}5$ , $e=30k$ , $f=k$ . By the law of cosines in $\triangle PEF$ , $FE=\sqrt{900k^2+k^2-60k\cdot\left(\frac{-1}2\right)}=\sqrt{931}k$ . The law of sines yield $\frac{\sin\angle AEP}{\sin\angle FAE}=\frac{AP}{FE}=\frac{35k}{\sqrt{931}k}=\frac{35}{\sqrt{931}}$ . Lastly, $\sin\angle AEP=\frac{5\sqrt{57}}{38}$ , then $\tan\angle AEP=5\sqrt3$ . The answer is $\left(5\sqrt3\right)^2=\boxed{75}.$

Animated Video Solution

https://youtu.be/5d98iXeyu4E

~Star League (https://starleague.us)

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
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