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

(→‎Solution 3)
(→‎Solution 3)
Line 86: Line 86:
 
Using Law of Cosines (note that <math>cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180°-\alpha}=-cos{\alpha}</math>)
 
Using Law of Cosines (note that <math>cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180°-\alpha}=-cos{\alpha}</math>)
  
−
<math></math>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
  
−
(1) <math>BP^2=FP^2+15^2-2*FP*15*cos(\alpha)</math>
+
(1) BP^2=FP^2+15^2-2*FP*15*cos(\alpha)
  
−
(2) <math>BP^2=DP^2+7^2+2*DP*7*cos(\alpha)</math>
+
(2) BP^2=DP^2+7^2+2*DP*7*cos(\alpha)
  
−
(3) <math>CP^2=DP^2+48^2-2*DP*48*cos(\alpha)</math>
+
(3) CP^2=DP^2+48^2-2*DP*48*cos(\alpha)
  
−
(4) <math>CP^2=EP^2+30^2+2*EP*30*cos(\alpha)</math>
+
(4) CP^2=EP^2+30^2+2*EP*30*cos(\alpha)
  
−
(5) <math>AP^2=EP^2+25^2-2*EP*25*cos(\alpha)</math>
+
(5) AP^2=EP^2+25^2-2*EP*25*cos(\alpha)
  
−
(6) <math>AP^2=FP^2+40^2+2*FP*40*cos(\alpha)</math>
+
(6) AP^2=FP^2+40^2+2*FP*40*cos(\alpha)
  
 
\end{align*}
 
\end{align*}
−
<cmath>
+
</cmath>
  
 
We can perform this operation: (1) - (2) + (3) - (4) + (5) - (6)
 
We can perform this operation: (1) - (2) + (3) - (4) + (5) - (6)
Line 108: Line 108:
 
Leaving us with (after combining and simplifying)
 
Leaving us with (after combining and simplifying)
  
−
</cmath>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
  
−
<math>\cos{\alpha}=\frac{-11}{2*(DP+EP+FP)}</math>
+
\cos{\alpha}=\frac{-11}{2*(DP+EP+FP)}
  
 
\end{align*}
 
\end{align*}
−
<math></math>
+
</cmath>
  
 
Therefore, we want to solve for <math>DP+EP+FP</math>
 
Therefore, we want to solve for <math>DP+EP+FP</math>
Line 126: Line 126:
 
Therefore, using area formula,
 
Therefore, using area formula,
  
−
<math></math>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
  
−
<math>[\triangle{DEF}] = \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3}</math>
+
[\triangle{DEF}] = \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3}
  
−
<math>xy+yz+zx=2^5*3*5</math>
+
xy+yz+zx=2^5*3*5
  
 
\end{align*}
 
\end{align*}
−
<cmath>
+
</cmath>
  
 
In addition, we know that
 
In addition, we know that
  
−
</cmath>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
  
−
<math>x^2+y^2+xy=42^2</math>
+
x^2+y^2+xy=42^2
  
−
<math>y^2+z^2+yz=35^2</math>
+
y^2+z^2+yz=35^2
  
−
<math>z^2+x^2+zx=13^2</math>
+
z^2+x^2+zx=13^2
  
 
\end{align*}
 
\end{align*}
−
<math></math>
+
</cmath>
  
 
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
Line 154: Line 154:
 
Because we want <math>DP+EP+FP</math>, which is <math>x+y+z</math>, we see that
 
Because we want <math>DP+EP+FP</math>, which is <math>x+y+z</math>, we see that
  
−
<math></math>
+
<cmath>
 
\begin{align*}
 
\begin{align*}
  
−
<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>
+
(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>(x+y+z)^2=\frac{42^2+35^2+13^2+3*2^5*3*5}{2}</math>
+
(x+y+z)^2=\frac{42^2+35^2+13^2+3*2^5*3*5}{2}
  
−
<math>(x+y+z)^2=2299</math>
+
(x+y+z)^2=2299
  
−
<math>x+y+z=11\sqrt{19}</math>
+
x+y+z=11\sqrt{19}
  
 
\end{align*}
 
\end{align*}
−
<cmath>
+
</cmath>
  
 
So plugging the results back into the equation before, we get
 
So plugging the results back into the equation before, we get
  
−
</cmath>
+
<math></math>
 
\begin{align*}
 
\begin{align*}
  
−
<math>\cos{\alpha} = \frac{-1}{2\sqrt{19}}</math> leading to <math>\sin{\alpha} = \frac{5\sqrt{3}}{2\sqrt{19}}</math>
+
\cos{\alpha} = \frac{-1}{2\sqrt{19}}<math> leading to </math>\sin{\alpha} = \frac{5\sqrt{3}}{2\sqrt{19}}
  
 
\end{align*}
 
\end{align*}
Line 183: Line 183:
 
\begin{align*}
 
\begin{align*}
  
−
<math>\tan^2{\alpha}=\boxed{075}</math>
+
\tan^2{\alpha}=\boxed{075}
  
 
\end{align*}
 
\end{align*}

Revision as of 00:01, 9 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 $cos{\angle{AFP}}=cos{\angle{BDP}}=cos{\angle{CEP}}=cos{180°-\alpha}=-cos{\alpha}$) 

\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*} (Error compiling LaTeX. Unknown error_msg)

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*} (Error compiling LaTeX. Unknown error_msg)

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, 

\begin{align*}

[\triangle{DEF}] = \frac{1}{2}*\sin{120°}*(xy+yz+zx) = 120\sqrt{3}

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

\end{align*} (Error compiling LaTeX. Unknown error_msg)

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*} (Error compiling LaTeX. Unknown error_msg)

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*} (Error compiling LaTeX. Unknown error_msg)

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

$$ (Error compiling LaTeX. Unknown error_msg) \begin{align*}

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

\end{align*} \[Giving us\] \begin{align*}

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

\end{align*} $$ (Error compiling LaTeX. Unknown error_msg)

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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