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2009 AIME II Problems/Problem 10

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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 (ProblemsResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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