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:
https://web2.0calc.com/questions/in-the-diagram-square-abcd-has-sides-of-length-4
+591
triangle PQC is a 45-45-90, so QC=x
triangle PQB is a 30-60-90 triangle, so BQ = x$\sqrt3$
QC+BQ=4=x+x$\sqrt3$
x($\sqrt3$+1)=4
x=$\dfrac4{\sqrt3+1}$
we can multiply top and bottom of the fraction by $\sqrt3-1$:
x=$\dfrac{4\sqrt3-4}{3-1}=2\sqrt3-2$
PC=x$\sqrt2=2\sqrt6-2\sqrt2$
AP=AC-PC=$4\sqrt2-2\sqrt6+2\sqrt2=\boxed{6\sqrt2-2\sqrt6}$
