Question

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 $\frac{p\sqrt{r}}{q},$ where $p,$ $q,$ and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p + q + r.$

Answer

Solution:

Since $5^2 + 12^2 = 13^2,$ angle $B$ is right. Place $A = (0, 0),$ $B = (5, 0),$ $C = (5, 12).$ The condition at $A$ says $\angle BAD = \angle CAD,$ so $D$ lies on the bisector of angle $BAC.$ Using the half-angle formula with $\tan \angle BAC = \frac{12}{5},$ $\tan \frac{\angle BAC}{2} = \frac{\sin \angle BAC}{1 + \cos \angle BAC} = \frac{12/13}{1 + 5/13} = \frac{2}{3},$ so $D$ lies on the line $y = \frac{2}{3}x.$

The condition at $C$ says $CB$ bisects angle $ACD,$ so ray $CD$ is the reflection of ray $CA$ over line $CB,$ which is the vertical line $x = 5.$ The reflection of $A$ is $(10, 0),$ so $D$ lies on the line through $C = (5, 12)$ and $(10, 0),$ namely $5y = 120 - 12x.$

Solving $y = \frac{2}{3}x$ and $5y = 120 - 12x$ gives $x = \frac{180}{23},$ $y = \frac{120}{23}.$ Then $AD = \frac{60}{23}\sqrt{3^2 + 2^2} = \frac{60\sqrt{13}}{23},$ so $p + q + r = 60 + 23 + 13 = 96.$

2009 AIME II Problem 10

Solution:

Problem 10 in Other Years