2018 AMC 10B Problem 24

Place the hexagon with center $O$ at the origin and circumradius $1$: $A = (1, 0)$, $B = \left(\tfrac12, \tfrac{\sqrt3}{2}\right)$, $C = \left(-\tfrac12, \tfrac{\sqrt3}{2}\right)$, $D = (-1, 0)$, $E = \left(-\tfrac12, -\tfrac{\sqrt3}{2}\right)$, $F = \left(\tfrac12, -\tfrac{\sqrt3}{2}\right)$.

Triangle $ACE$ is equilateral with side $\sqrt3$, so $[ACE] = \frac{\sqrt3}{4} \cdot 3 = \frac{3\sqrt3}{4}$.

The midpoints are $X = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$, $Y = \left(-\tfrac34, \tfrac{\sqrt3}{4}\right)$, $Z = \left(0, -\tfrac{\sqrt3}{2}\right)$. Both triangles are centered at $O$, so the whole picture has $120^\circ$ rotational symmetry, and it suffices to see what one side of $XYZ$ does to $ACE$.

Side $XY$ is the horizontal line $y = \frac{\sqrt3}{4}$. Vertex $C$ (at height $\frac{\sqrt3}{2}$) is above it while $A$ and $E$ are below, so this line slices off the corner at $C$. It meets $CE$ (the vertical line $x = -\tfrac12$, from height $\frac{\sqrt3}{2}$ down to $-\frac{\sqrt3}{2}$) one quarter of the way from $C$ to $E$, and it meets $CA$ at $\left(\tfrac14, \tfrac{\sqrt3}{4}\right)$, the midpoint of $CA$.

A triangle cut from corner $C$ using $\tfrac14$ of $CE$ and $\tfrac12$ of $CA$ has area $\tfrac14 \cdot \tfrac12 = \tfrac18$ of $[ACE]$. By symmetry, sides $YZ$ and $ZX$ remove congruent corners at $E$ and $A$, and the three corners do not overlap. The hexagon that remains is
$$\left(1 - \frac{3}{8}\right)[ACE] = \frac{5}{8} \cdot \frac{3\sqrt3}{4} = \frac{15\sqrt3}{32}.$$

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

Two concentric equilateral triangles that are rotated relative to each other intersect in a hexagon made of one triangle minus three corner triangles. Coordinates on a regular hexagon are cheap, and the ratio lemma ("a triangle with two sides scaled by $p$ and $q$ from a shared vertex has $pq$ of the area") turns each corner into a fraction of the whole, so no lengths need to be computed.

Without coordinates: line $XY$ joins the midpoints of $AB$ and $CD$, so it is the midline of trapezoid $ABCD$ and bisects every segment from line $BC$ to line $AD$, in particular $CA$. It is also one quarter of the way from line $BC$ to the parallel line $EF$ ($XY$ is midway between $BC$ and $AD$, and $AD$ is midway between $BC$ and $EF$), so it cuts $CE$ at its quarter point. The rest is as above.

• Assuming the two triangles are homothetic (parallel sides) and using a simple scale factor, or assuming the intersection is a regular hexagon.
• Computing the area of $XYZ$ ($\frac{9\sqrt3}{16}$, choice E) or of $ACE$ minus $XYZ$ instead of the intersection.