# 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 $\tan^2(\angle AEP).$

## Diagram

$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D); dot("A",A,1.5*dir(A),linewidth(4)); dot("B",B,1.5*dir(B),linewidth(4)); dot("C",C,1.5*dir(C),linewidth(4)); dot("D",D,1.5*S,linewidth(4)); dot("E",E,1.5*dir(30),linewidth(4)); dot("F",F,1.5*dir(150),linewidth(4)); dot("P",P,1.5*dir(-30),linewidth(4)); label("7",midpoint(B--D),1.5*S,red); label("30",midpoint(C--E),1.5*dir(30),red); label("40",midpoint(A--F),1.5*dir(150),red); [/asy]$ ~MRENTHUSIASM

## Solution 1 (Coordinates Bash)

By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles)$\text{*}$. 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: $[asy] /* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; filldraw(D--E--F--cycle,yellow); draw(A--B--C--cycle); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("A",A,1.5*dir(A),linewidth(4)); dot("B",B,1.5*dir(B),linewidth(4)); dot("C",C,1.5*dir(C),linewidth(4)); dot("D",D,1.5*S,linewidth(4)); dot("E",E,1.5*dir(30),linewidth(4)); dot("F",F,1.5*dir(150),linewidth(4)); dot("P",P,1.5*dir(-30),linewidth(4)); label("7",midpoint(B--D),1.5*S,red); label("30",midpoint(C--E),1.5*dir(30),red); label("40",midpoint(A--F),1.5*dir(150),red); [/asy]$

$\text{*}$ Basic angle chasing gives $\angle BDP=180-\angle CDP.$ Because $\angle BFP+\angle BDP=180+\angle BFP-\angle CDP=180,$ which means that $BFPD$ is cyclic, and that $P$ passes through the circumcircle of triangle $BFD.$ Similar reasoning leads us to the fact that $P$ also passes through the circumcircles of triangles $AEF$ and $CDE,$ which means that $P=(AEF)\cap(BFD)\cap(CDE).$ Continue as above.

## 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, $\tan^2 \theta = \boxed{075}$.

~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

Claim

a) Carnot's theorem. Given triangle $\triangle ABC$ and point $P.$ Let $PS \perp BC,$ $PR \perp AC, PQ \perp AB.$ $P$ doesn't have to be inside $\triangle ABC.$

Prove that $AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.$

b) Let $\triangle ABC$ be the equilateral triangle. Prove that $AQ + BS + CR = \frac {3}{2} AB.$ (The sum of the lengths of the alternating segments split by the perpendiculars from a point $P$ within an equilateral triangle is equal to half the perimeter.)

Proof

a) $AP^2 = AQ^2 + PQ^2 = AR^2 + PR^2,$ $$BP^2 = BQ^2 + PQ^2 = BS^2 + PS^2,$$ $$CP^2 = CR^2 + PR^2 = CS^2 + PS^2 \implies$$ $AQ^2 + PQ^2 + BS^2 + PS^2 + CR^2 + PR^2 = AR^2 + PR^2 + BQ^2 + PQ^2 + CS^2 + PS^2,$ $$AQ^2 + BS^2 + CR^2 = AR^2 + BQ^2 + CS^2.$$ b) $AQ + BQ = BS + CS = CR + AR = AB = AC = BC,$ $$AQ^2 + BQ^2 + 2 AQ \cdot BQ + BS^2 + CS^2 + 2 BS \cdot CS + AR^2 + CR^2+ 2 AR \cdot CR = 3 AB^2.$$

$$2AQ^2 + 2 AQ \cdot BQ + 2 BS^2 + 2 BS \cdot CS + 2 CR^2 + 2 AR \cdot CR = 3 AB^2.$$ $$AQ (AQ + BQ) + BS (BS + CS) + CR (AR + CR) = \frac {3}{2} AB^2.$$ $$AQ \cdot AB + BS \cdot BC + CR \cdot AC = \frac {3}{2} AB^2.$$ $$AQ + BS + CR = \frac {3}{2} AB.$$

## Solution 4 (Law of Cosines)

This solution is 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\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}

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

Leaving us with (after combining and simplifying) $$\cos{\alpha}=\frac{-11}{2\cdot(DP+EP+FP)}$$

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

Notice that $\angle{DPE}=\angle{EPF}=\angle{FPD}=120^\circ$

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}\cdot\sin{120°}\cdot(xy+yz+zx) = 120\sqrt{3} \\ xy+yz+zx &= 2^5\cdot3\cdot5 \end{align*}

\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\cdot2^5\cdot3\cdot5}{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 $$\tan^2{\alpha}=\boxed{075}$$

## Solution 5 (Combining Solutions 3 and 4)

We begin by using the fact stated in Solution 3 that, for any point in an equilateral triangle, the lengths of the three perpendicular lines dropped to the sides of the triangle add up to the altitude of that triangle. To make things simple, let's assign $\angle AEP = \angle BFP = \angle CDP = \alpha$. We can label these three perpendiculars as: $$PD\cdot\sin{\alpha} + PE\cdot\sin{\alpha} + PF\cdot\sin{\alpha} = \dfrac{55 \sqrt{3}}{2}$$ Simplifying, we get $$PD + PE + PF = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}$$ Now, as stated and quoting Solution 4, "Draw line segments from $P$ to points $A$, $B$, and $C$. [We know that] the angle measure of $\angle{AEP}$, $\angle{BFP}$, and $\angle{CDP}$ is $\alpha$

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\cdot FP\cdot15\cdot\cos(\alpha)\\ (2) \ BP^2 &= DP^2+7^2+2\cdot DP\cdot7\cdot\cos(\alpha)\\ (3) \ CP^2 &= DP^2+48^2-2\cdot DP\cdot48\cdot\cos(\alpha)\\ (4) \ CP^2 &= EP^2+30^2+2\cdot EP\cdot30\cdot\cos(\alpha)\\ (5) \ AP^2 &= EP^2+25^2-2\cdot EP\cdot25\cdot\cos(\alpha)\\ (6) \ AP^2 &= FP^2+40^2+2\cdot FP\cdot40\cdot\cos(\alpha)\\ \end{align*}

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

Leaving us with (after combining and simplifying) $$PD + PE + PF=\frac{11}{2\cdot\cos{\alpha}}$$".

Now, we can use our previous equation along with this one to get: $$\frac{11}{2\cdot\cos{\alpha}} = \dfrac{55 \sqrt{3}}{2\cdot\sin{\alpha}}$$.

This equation becomes: $$\tan{\alpha} = 5\sqrt{3}$$ As so, our answer is $$\left(5\sqrt3\right)^2=\boxed{075}.$$ ~Solution by armang32324 (Mathemagics Club)

## Solution 6

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, as shown below: $[asy] /* Made by MRENTHUSIASM */ size(400); pair A, B, C, D, E, F, P; A = 55*sqrt(3)/3 * dir(90); B = 55*sqrt(3)/3 * dir(210); C = 55*sqrt(3)/3 * dir(330); D = B + 7*dir(0); E = A + 25*dir(C-A); F = A + 40*dir(B-A); P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0]; draw(anglemark(A,E,P,20),red); draw(anglemark(B,F,P,20),red); draw(anglemark(C,D,P,20),red); add(pathticks(anglemark(A,E,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(B,F,P,20), n = 1, r = 0.2, s = 12, red)); add(pathticks(anglemark(C,D,P,20), n = 1, r = 0.2, s = 12, red)); draw(A--B--C--cycle^^P--E^^P--F^^P--D); draw(P--A^^P--B^^P--C,dashed); draw(circumcircle(A,E,F)^^circumcircle(B,D,F)^^circumcircle(C,D,E),blue); dot("A",A,1.5*dir(A),linewidth(4)); dot("B",B,1.5*dir(B),linewidth(4)); dot("C",C,1.5*dir(C),linewidth(4)); dot("D",D,1.5*S,linewidth(4)); dot("E",E,1.5*dir(30),linewidth(4)); dot("F",F,1.5*dir(150),linewidth(4)); dot("P",P,1.5*dir(-30),linewidth(4)); label("7",midpoint(B--D),1.5*S,red); label("30",midpoint(C--E),1.5*dir(30),red); label("40",midpoint(A--F),1.5*dir(150),red); [/asy]$ 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 $$\left(5\sqrt3\right)^2=\boxed{075}.$$

## Video Solution

~MathProblemSolvingSkills.com

## Animated Video Solution

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