# 2009 AIME II Problems/Problem 10

## Problem

Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$, the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac {p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p$ + $q$ + $r$.

## Solution 1

Let $O$ be the intersection of $BC$ and $AD$. By the Angle Bisector Theorem, $\frac {5}{BO}$ = $\frac {13}{CO}$, so $BO$ = $5x$ and $CO$ = $13x$, and $BO$ + $OC$ = $BC$ = $12$, so $x$ = $\frac {2}{3}$, and $OC$ = $\frac {26}{3}$. Let $P$ be the foot of the altitude from $D$ to $OC$. It can be seen that triangle $DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\frac {26}{3}$, $y$ = $\frac {13}{69}$, and $AD$ = $\frac {60\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $\boxed{096}$.

## Solution 2

Extend $AB$ and $CD$ to intersect at $P$. Note that since $\angle ACB=\angle PCB$ and $\angle ABC=\angle PBC=90^{\circ}$ by ASA congruency we have $\triangle ABC\cong \triangle PBC$. Therefore $AC=PC=13$.

By the angle bisector theorem, $PD=\dfrac{130}{23}$ and $CD=\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$:

\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\ AD^2&=130-\dfrac{130\cdot 169}{23^2}\\ AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\ AD^2&=\dfrac{130(23^2-169)}{23^2}\\ AD^2&=\dfrac{130(360)}{23^2}\\ AD&=\dfrac{60\sqrt{13}}{23}\\ \end{align*}

and our final answer is $60+13+23=\boxed{096}$.

## Solution 3

Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\dfrac{BF}{AB}=\dfrac{FC}{AC}$ we have $FC=\dfrac{26}{3}$. Then the equation of the line $AF$ through the points $(12,5)$ and $\left(\frac{26}{3},0\right)$ is $y=\frac32 x-13$. Hence the intersection point of $AF$ and $CD$ is the point $D$ at the coordinates $\left(\dfrac{156}{23},-\dfrac{65}{23}\right)$. Using the distance formula, $AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}$ for an answer of $60+13+23=\fbox{096}$.

## Solution 4

After drawing a good diagram, we reflect $ABC$ over the line $BC$, forming a new point that we'll call $A'$. Also, let the intersection of $AD$ and $BC$ be point $E$. Point $D$ lies on line $A'C$. Since line $AD$ bisects $\angle{CAB}$, we can use the Angle Bisector Theorem. $AA'=10$ and $AC=13$, so $\frac{CD}{A'D}=\frac{13}{10}$. Letting the segments be $13x$ and $10x$ respectively, we now have $13x+10x=13$. Therefore, $x=\frac{13}{23}$. By the Pythagorean Theorem, $AE=\frac{5\sqrt{13}}{3}$. Using the Angle Bisector Theorem on $\angle{ACD}$, we have that $ED=\frac{5x\sqrt{13}}{3}$. Substituting in $x=\frac{13}{23}$, we have that $AD=(\frac{5\sqrt{13}}{3})(1+x)=\frac{60\sqrt{13}}{23}$, so the answer is $60+13+23=\boxed{096}$.