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+3x=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.