AMC Practice #1 - Problem 11

Let ABCDE be a regular pentagon. Let there be a point P in the triangle ACD such that P, D, and B are collinear. We extend AP to intersect side CD at point F, where \frac{DF}{FC} = \frac{1}{2}. We extend CP to intersect side ED at point G and segment AD at point H. Then, the ratio \frac{HD}{AH} can be written as \frac{\sqrt{p} - q}{r}, where p is square free. Find p+q+r.