Problem:
Let △ABC be a right triangle with right angle at C. Let D and E be points on AB with D between A and E such that CD and CE trisect ∠C. If BEDE=158, then tanB can be written as nmp, where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime. Find m+n+p.
Solution:
The Angle Bisector Theorem gives BCCD=BEDE=158, so without loss of generality assume BC=15 and CD=8. Applying the Law of Cosines to △BCD gives BD2=82+152−2⋅8⋅15⋅cos60∘=169, so BD=13. Again applying the Law of Cosines to △BCD gives 82=132+152−2⋅13⋅15⋅cos∠B, showing that cos∠B=1311. It then follows that sin∠B=1−(1311)2=1343. Then tan∠B=1143. The requested sum is 4+3+11=18.