Talk:2001 AIME I Problems/Problem 7

A more elegant solution exists using the fact that an angle bisector divides the opposite leg in direct proportion to the ratio of the adjacent legs, and mass point geometry. Essentially, it goes something like:

Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this intersection F. Using the angle bisector theorem, we know the ractio BF:CF is 21:22, thus we shall assign a weight of 22 to point B and a weight of 21 to point C. Thus giving F a weight of 43. In the same manner, using another bisector, we find that A has a weight of 20. So, now we know P has a weight of 63, and the ratio of FP:PA is 20:43. Therefore, the smaller similar triangle ADE as it's base is 43/63 the height of the original triangle ABC. Therefore, DE is 43/63 the size of BC. Thus, DE is 860/63 = m/n. Therefore, m+n=923.