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:

Both $\triangle ACE$ and $\triangle XYZ$ are equilateral, and $\triangle ACE$ has half the area of the hexagon. The vertices where the two triangles cut each other let the shaded hexagon be measured against the midpoint triangle $U$ of $\triangle ACE.$

That midpoint triangle has $\tfrac14$ the area of $\triangle ACE,$ hence $\tfrac18$ of the hexagon. The shaded region equals $\tfrac52$ of the midpoint triangle, so it is $\tfrac52\cdot\tfrac18=\tfrac{5}{16}$ of the hexagon.

The hexagon has area $6\cdot\tfrac{\sqrt3}{4}\cdot1^2=\tfrac{3\sqrt3}{2},$ so the shaded area is $\tfrac{5}{16}\cdot\tfrac{3\sqrt3}{2}=\tfrac{15}{32}\sqrt3.$

Thus, the correct answer is C.

2018 AMC 12B Problem 20

Solution:

Problem 20 in Other Years