2018 AMC 10B Problem 24

The two triangles are both centered at the hexagon's center, and turning the picture by a third of a full turn leaves it the same. So we only need to understand what one side of $XYZ$ does to $ACE$, then repeat three times.

1. Coordinates. Put the center at $(0,0)$. A regular hexagon with side $1$ has its corners $1$ unit from the center:
$$A = (1, 0),\ B = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),\ C = \left(-\tfrac12, \tfrac{\sqrt3}{2}\right),\ D = (-1, 0),\ E = \left(-\tfrac12, -\tfrac{\sqrt3}{2}\right),\ F = \left(\tfrac12, -\tfrac{\sqrt3}{2}\right).$$
2. Area of $ACE$. It is an equilateral triangle. Its side $CE$ runs from height $\frac{\sqrt3}{2}$ to $-\frac{\sqrt3}{2}$, so it has length $\sqrt3$. The area is
$$\frac{\sqrt3}{4} \times (\sqrt3)^2 = \frac{3\sqrt3}{4}.$$
3. The side $XY$. $X$ is the midpoint of $AB$ and $Y$ is the midpoint of $CD$:
$$X = \left(\tfrac34, \tfrac{\sqrt3}{4}\right),\qquad Y = \left(-\tfrac34, \tfrac{\sqrt3}{4}\right).$$
Both have height $\frac{\sqrt3}{4}$, so $XY$ is a flat line at that height. Corner $C$ is above it, while $A$ and $E$ are below. So this line cuts the corner at $C$ off triangle $ACE$.
4. Where the cut hits the sides of $ACE$.
- Side $CE$ goes straight down from height $\frac{\sqrt3}{2}$ to $-\frac{\sqrt3}{2}$. The line at height $\frac{\sqrt3}{4}$ is only $\frac{\sqrt3}{4}$ below $C$, which is $\frac14$ of the way down $CE$.
- Side $CA$ goes from height $\frac{\sqrt3}{2}$ to height $0$. Height $\frac{\sqrt3}{4}$ is halfway, so the cut hits the midpoint of $CA$.
5. Size of the cut-off corner. It is a triangle at $C$ with one side $\frac14$ of $CE$ and the other $\frac12$ of $CA$. Use $CE$ as the base: the small triangle's base is $\frac14$ as long, and its height (from the cut point on $CA$ to line $CE$) is $\frac12$ as big as the height from $A$. So its area is
$$\tfrac14 \times \tfrac12 = \tfrac18 \text{ of } ACE.$$
6. Three corners. By the turning symmetry, sides $YZ$ and $ZX$ cut matching corners off at $E$ and $A$, and the three corners do not overlap. What remains is the hexagon:
$$\left(1 - \tfrac38\right) \times \frac{3\sqrt3}{4} = \frac58 \times \frac{3\sqrt3}{4} = \frac{15\sqrt3}{32}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{15}{32}\sqrt{3}}$.

Two triangles with the same center, one turned relative to the other, overlap in a hexagon: one triangle with three corners cut off. A corner triangle that uses fractions $p$ and $q$ of the two sides at a vertex has $p \times q$ of the full area, so no actual lengths are needed once you know where the cuts land.

Find the cut points without coordinates. Line $XY$ joins the midpoints of $AB$ and $CD$, the two slanted sides of the trapezoid $ABCD$, so it runs halfway between $BC$ and $AD$. That means it cuts every segment from $BC$ to $AD$ in half, including $CA$. Also, $AD$ runs halfway between $BC$ and $EF$, so $XY$ is a quarter of the way from $BC$ to $EF$, which puts its crossing with $CE$ a quarter of the way from $C$. The rest is the same.