Mock Geometry AIME 2011 Problems/Problem 4
In triangle . Let and be points on such that is equilateral. The perimeter of can be expressed in the form where are relatively prime positive integers. Find
Let be the midpoint of . It follows that is perpendicular to and to . The area of can then be calculated two different ways: , and .
By the Law of Cosines, and so . Therefore, . Solving for yields .
Let be the side length of . The height of an equilateral triangle is given by the formula . Then . Solving for yields . Then the perimeter of the triangle is and .
Let and . By the Law of Cosines, . It is easy to see that . Since , by AA similarity. From this, we have: Solving, we find that , so the perimeter is , and our answer is