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 △ACE and △XYZ ?
Answer Choices:
A. 833
B. 1673
C. 32153
D. 213
E. 1693 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 △U2XU1 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 △U2XU1. The hypotenuse of △U2XU1 is XU2=AX=21, so the area of △U2XU1 is
21⋅43⋅(21)2=3213
The area of the equilateral triangle XYZ with side length 23 is equal to 413⋅(23)2=1693. Hence the area of the shaded hexagon is
1693−3⋅3213=33(163−321)=32153
OR
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 U2U4U6 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 △U2U3U4,△U4U5U6, and △U6U1U2 is congruent to half of △U2U4U6, so the total shaded area is 25 times the area of △U2U4U6 and therefore 25⋅81=165 of the area of ABCDEF. The area of ABCDEF is 6⋅43⋅12, so the requested area is 32153.
