2009 AIME II Problems/Problem 10

Revision as of 19:53, 17 April 2009 by Aimesolver (talk | contribs) (Solution)

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 $60$ + $13$ + $23$ = $\boxed{096}$.

See Also

2009 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions