Question

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 $\triangle ACE$ and $\triangle XYZ?$

$\dfrac{3}{8}\sqrt{3}$

$\dfrac{7}{16}\sqrt{3}$

$\dfrac{15}{32}\sqrt{3}$

$\dfrac{1}{2}\sqrt{3}$

$\dfrac{9}{16}\sqrt{3}$

Solution:

Center the hexagon at the origin. Then $\triangle ACE$ is equilateral with side $AC = \sqrt{3},$ so its area is $\tfrac{\sqrt3}{4}(\sqrt3)^2 = \tfrac{3\sqrt3}{4}.$ And $\triangle XYZ$ is equilateral with side $\tfrac32,$ area $\tfrac{\sqrt3}{4}\left(\tfrac32\right)^2 = \tfrac{9\sqrt3}{16}.$ The two are concentric and rotated $30^\circ$ apart, so their intersection is $\triangle XYZ$ with three congruent corners (each of area $\tfrac{\sqrt3}{32}$ ) cut off: $\tfrac{9\sqrt3}{16} - 3\cdot\tfrac{\sqrt3}{32} = \tfrac{15}{32}\sqrt{3}.$ Therefore, the answer is C.

2018 AMC 10B Problem 24

Solution:

Problem 24 in Other Years