Problem:
Let ABCDEF be a regular hexagon with side length 1 . Denote by X, Y, and Z the midpoints of sides AB,CD, and EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of â–³ACE and â–³XYZ?
Answer Choices:
A. 83​3​
B. 167​3​
C. 3215​3​
D. 21​3​
E. 169​3​
Solution:
Let O be the center of the regular hexagon. Points B,O,E are collinear and BE=BO+OE=2. Trapezoid FABE is isosceles, and XZ is its midline. Hence XZ=23​ and analogously XY=ZY=23​.
Denote by U1​ the intersection of AC and XZ and by U2​ the intersection of AC and XY. It is easy to see that △AXU1​ and △U2​XU1​ are congruent 30−60−90∘ right triangles.
By symmetry the area of the convex hexagon enclosed by the intersection of △ACE and △XYZ, shaded in the figure, is equal to the area of △XYZ minus 3 times the area of △U2​XU1​. The hypotenuse
of △U2​XU1​ is XU2​=AX=21​, so the area of △U2​XU1​ is
21​⋅43​​⋅(21​)2=321​3​
The area of the equilateral triangle XYZ with side length 23​ is equal to 41​3​⋅(23​)2=169​3​. Hence the area of the shaded hexagon is
Let U1​ and U2​ be as above, and continue labeling the vertices of the shaded hexagon counterclockwise with U3​,U4​,U5​, and U6​ as shown. The area of △ACE is half the area of hexagon ABCDEF. Triangle U2​U4​U6​ is the midpoint triangle of △ACE, so its area is 41​ of the area of △ACE, and thus 81​ of the area of ABCDEF. Each of △U2​U3​U4​,△U4​U5​U6​, and △U6​U1​U2​ is congruent to half of △U2​U4​U6​, so the total shaded area is 25​ times the area of △U2​U4​U6​ and therefore 25​⋅81​=165​ of the area of ABCDEF. The area of ABCDEF is 6⋅43​​⋅12, so the requested area is (C)3215​3​​.
