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

Revision as of 23:56, 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 (no trig)

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

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

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

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 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

@@ Line 68: / Line 68: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Solution 2 (way quicker)==
+==Solution 2 (no trig)==
 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>

2023 AIME I (Problems • Answer Key • Resources)
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