Problem:
Let △PQR be a triangle with ∠P=75∘ and ∠Q=60∘. A regular hexagon ABCDEF with side length 1 is drawn inside △PQR so that side AB lies on PQ​, side CD lies on QR​, and one of the remaining vertices lies on RP. There are positive integers a,b,c, and d such that the area of △PQR can be expressed in the form da+bc​​, where a and d are relatively prime, and c is not divisible by the square of any prime. Find a+b+c+d.
Solution:
Note that AB=BQ=QC=CD=1 and ∠RDE=60∘. If E lies on RP, then ∠REF=75∘+120∘=195∘ and F lies outside the triangle. Thus F must lie on RP, point E is in the interior of △PQR, and △RFD is a 45−45−90∘ triangle. Because FD=3​=DR, the length of RQ​ is 2+3​. Let the altitude from P to QR​ have length x. Then RQ can be expressed as x+3​x​=2+3​. Solving for x yields x=23+3​​, and the area of △PQR is 21​⋅23+3​​⋅(2+3​)=49+53​​. The requested sum is 9+5+3+4=21​.