Equiangular Hexagon

Geometry Level 5

Equiangular hexagon ABCDEFABCDEF has AB=BC=DE=EF=4AB=BC=DE=EF=4 and AF=CD=1AF=CD=1. The hexagon is reflected across the segment EFEF to form hexagon ABCDEFA'B'C'D'EF. BC\overline{B'C} intersects EF\overline{EF} at point PP. EPFP\dfrac{EP}{FP} can be expressed as pq\dfrac{p}{q} for relatively prime positive integers p,qp,q. Find p×qp\times q.

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