2013 Mock AIME I Problems/Problem 5
Contents
[hide]Problem
In quadrilateral , . Also, , and . The perimeter of can be expressed in the form where and are relatively prime, and is not divisible by the square of any prime number. Find .
Solution
Let , as in the diagram. Thus, from the problem, . Because , by Power of a Point, we know that is cyclic. Thus, we know that , so, by the congruency of vertical angles and subsequently AA Similarity, we know that . Thus, we have the proportion , or, by substitution, . Solving this equation for yields . Similarly, we know that , so, like before, we can see that . Thus, we have the proportion , or, by substitution, . Solving for yields .
Now, we can use Ptolemy's Theorem on cyclic and solve for :
Solution 2
Consider the figure and notations from Solution 1.
Let . In triangle , by the cosine rule, In triangle , by the cosine rule, Since , we have: Solving for , we get . Now, using the cosine rule in triangles and with (substituting ), we can find and . After calculations, we get and . The perimeter of is given by: Substituting , we get: Therefore, , and $p + q + r = 13 + 22 + 2 = 37. Solution by Thunder Cloak