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 $\triangle XYZ?$

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

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

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

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

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

Answer: C

Concepts:regular polygon equilateral triangle area decomposition

Difficulty rating: 2470

Solution:

The triangle $XYZ$ is equilateral with side $3/2,$ so its area is $\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}.$

The triangles $ACE$ and $XYZ$ are concentric and rotated $30^\circ$ from each other. At each vertex of $XYZ,$ the sides of $ACE$ cut off a $30$ - $60$ - $90$ triangle whose hypotenuse is the half-side segment $AX=1/2.$ Its legs are $1/4$ and $\sqrt3/4,$ so each corner has area $\sqrt3/32.$

Removing the three corners gives $\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}.$ Therefore, the answer is C.

2018 AMC 10B Problem 24

Answer: C

Difficulty rating: 2470

Solution:

Problem 24 in Other Years