2016 AMC 10B Problem 23

Let the side be $1$ and draw $AB$ at the top and $ED$ at the bottom, so the hexagon has height $\sqrt3$. The four parallel lines cut it into three strips of height $\tfrac{\sqrt3}{3}$; hexagon $WCXYFZ$ is exactly the middle strip, with $C$ and $F$ its widest points.

The hexagon's width grows linearly from $1$ at $AB$ to $2$ at the line $FC$ through the center, which is $\tfrac{\sqrt3}{2}$ below $AB$. Line $ZW$ is $\tfrac{\sqrt3}{3}$ below $AB$, that is $\tfrac{2}{3}$ of the way down to $FC$, so
$$ZW = 1 + \tfrac23(2 - 1) = \tfrac53.$$

The top strip $ABWZ$ is a trapezoid with parallel sides $1$ and $\tfrac53$ and height $\tfrac{\sqrt3}{3}$:
$$[ABWZ] = \tfrac12\left(1 + \tfrac53\right)\tfrac{\sqrt3}{3} = \tfrac{4\sqrt3}{9}.$$
By symmetry the bottom strip has the same area, so together they cover $\tfrac{8\sqrt3}{9}$.

The whole hexagon has area $\tfrac{3\sqrt3}{2}$, so the middle strip has area
$$\tfrac{3\sqrt3}{2} - \tfrac{8\sqrt3}{9} = \tfrac{27\sqrt3 - 16\sqrt3}{18} = \tfrac{11\sqrt3}{18},$$
and the requested ratio is $\dfrac{11\sqrt3/18}{27\sqrt3/18} = \dfrac{11}{27}$.

The answer is $\boxed{\textbf{(C)}\ \frac{11}{27}}$.

"Parallel and equally spaced" lines slice the hexagon into strips of equal height, so the problem reduces to how the width varies: linearly along the slanted sides. Computing the two congruent outer trapezoids and subtracting is quicker than working with the six-sided middle piece directly. Fix a convenient side length; the ratio is scale-free.

Scale to side $3$ and tile with unit equilateral triangles: the hexagon holds $6 \cdot 9 = 54$ of them. Each strip is two rows of triangles. The top strip's rows have $3 + 4 = 7$ and $4 + 5 = 9$ triangles, so the top and bottom strips hold $32$ and the middle holds $54 - 32 = 22$; ratio $\tfrac{22}{54} = \tfrac{11}{27}$.

• Assuming three equally spaced strips have equal areas and answering $\tfrac13$; the hexagon widens toward the middle, so the middle strip is larger.
• Taking $ZW$ to be $\tfrac{4}{3}$ (one third of the way from width $1$ to width $2$) instead of $\tfrac{5}{3}$; the widest line $FC$ is only $\tfrac{\sqrt3}{2}$ below $AB$, not $\sqrt3$.