In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x. Compute AP.
Wow! This problem really is hard! I don't really know how to figure it out, but it might not be too hard once you've used a bunch of theorems about right triangles/cosines/sines and whatnot.
This question has already been answered though, so I'll leave you a link to the answer:
triangle PQC is a 45-45-90, so QC=x
triangle PQB is a 30-60-90 triangle, so BQ = x$\sqrt3$
we can multiply top and bottom of the fraction by $\sqrt3-1$: