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 bythe 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 qprββ, 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:
Note that β³ABC has a right angle at B. Place β³ABC on the coordinate plane with A at the origin, B at (5,0),C at (5,12), and D at a point (x,y). Let point E be at (10,0); then β ACB equals β ECB. This means that point D lies on the line segment CE. Then tanβ CEB=512β=10βxyβ, so 5y=120β12x. Thus
It follows that 12x=18y. Combining this with 5y=120β12x yields y=23120β and x=23180β. The distance from A to D is then (23180β)2+(23120β)2β=2360β32+22β=236013ββ. The requested answer is then 60+23+13=96β.