2009 AIME II Problems/Problem 10

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

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 $be the altitude from$ 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} $. The answer is$ 60 $+$ 13 $+$ 23 $=$ \boxed{096}$.

2009 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2009 AIME II Problems/Problem 10

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