# Mock Geometry AIME 2011 Problems/Problem 13

## Problem 13

In acute triangle is the bisector of . is the midpoint of . a line through parallel to meets at respectively. Given that the sum of all possible values of can be expressed as where are positive integers. What is ?

## Solution

Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let be intersection of angle bisector with Let be , and is as well, since angle bisector. Since line through is parallel to , is also . Let then be , and by parallel lines, is also . Doing further angle chasing, we find that is isoceles with base . Using triangle ratio, we find

There are two possible configurations of the triangle, one such that is to the left of , and vice versa. In the first falls between and , with outside the triangle, and in the second between and , with outside the triangle. Using Law of Sines then:

Plugging in values, we find for acute and obtuse triangles denoted as and , respectively,

, and

Using Law of Sines again and substituting the expression for the and for ,

, and

Solving for the ratio of on both triangles, and then applying Angle Bisector theorem yields a with included angle for and with included angle for . Solving using Law of Cosine yields answer of and , or .