Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$? $\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad$ $ \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad$ $ \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad$ $ \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad$ $ \textbf{(E)}\ \frac {9}{16}\sqrt{3}$
Solution
The hexagon is $\triangle ACE$ with three congruent corner triangles cut off by the sides of $\triangle XYZ$. At the corner $E$, the cut-off triangle $EHJ$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle with $EH=\frac14 EA=\frac{\sqrt3}4$ (its legs are $EH$ and $HJ=\sqrt3\,EH=\frac34$).
So the area $[EHJ]=\frac12\cdot \frac{\sqrt3}4\cdot \frac34=\frac{3\sqrt3}{32}$.
$AC=\sqrt3$, so $[ACE]=\frac{3\sqrt3}4$.
The answer is $\frac{3\sqrt3}4-3\cdot \frac{3\sqrt3}{32}=\frac{15\sqrt3}{32}$.
A regular octagon $ABCDEFGH$ has side length $1$. Points $X,Y,Z,W$ are the midpoints of $AB,CD,EF,GH$, respectively. What is the area of the intersection of squares $ACEG$ and $XYZW$?
