# 2016 AMC 10B Problems/Problem 23

## Problem

In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$

## Solution 1

We draw a diagram to make our work easier: $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); label("[itex]A[itex]",A,SW); label("[itex]B[itex]",B,SE); label("[itex]C[itex]",C,ESE); label("[itex]D[itex]",D,NE); label("[itex]E[itex]",E,NW); label("[itex]F[itex]",F,WSW); label("[itex]W[itex]",W,ENE); label("[itex]X[itex]",X,ESE); label("[itex]Y[itex]",Y,WSW); label("[itex]Z[itex]",Z,WNW); [/asy]$

Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.

$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("[itex]A[itex]",A,SW); label("[itex]B[itex]",B,SE); label("[itex]C[itex]",C,ESE); label("[itex]D[itex]",D,NE); label("[itex]E[itex]",E,NW); label("[itex]F[itex]",F,WSW); label("[itex]W[itex]",W,ENE); label("[itex]X[itex]",X,ESE); label("[itex]Y[itex]",Y,WSW); label("[itex]Z[itex]",Z,WNW); [/asy]$

From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac 53$.

We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$, and the combined area is $\frac{11\sqrt 3}{18}^{*}$.

We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}[itex] is equal to [itex]\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}[itex]. [itex]^*$ (Error compiling LaTeX. ! Missing $inserted.) At this point, you can answer $\textbf{(C)}$ and move on with your test. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z,S,K,R,U,H,I,O,P,Q; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); S=(-1/6,sqrt(3)/6); H=(-1/6, 5sqrt(3)/6); P=(7/6, 5sqrt(3)/6); U=(7/6,sqrt(3)/6); K=(1/3, 0); R=(2/3, 0); I=(1/3,sqrt(3)); O=(2/3,sqrt(3)); Q=(1/2, sqrt(3)/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); draw(S--U); draw(K--R); draw(Z--K); draw(H--R); draw(I--U); draw(O--X); draw(H--P); draw(I--Y); draw(O--S); draw(P--K); draw(W--R); label("[itex]A[itex]",A,SW); label("[itex]B[itex]",B,SE); label("[itex]C[itex]",C,ESE); label("[itex]D[itex]",D,NE); label("[itex]E[itex]",E,NW); label("[itex]F[itex]",F,WSW); label("[itex]W[itex]",W,ENE); label("[itex]X[itex]",X,ESE); label("[itex]Y[itex]",Y,WSW); label("[itex]Z[itex]",Z,WNW); label("[itex]S[itex]",S,WSW); label("[itex]K[itex]",K,SSW); label("[itex]R[itex]",R,SSE); label("[itex]U[itex]",U,ESE); label("[itex]H[itex]",H,WNW); label("[itex]I[itex]",I,NNW); label("[itex]O[itex]",O,NNE); label("[itex]P[itex]",P,ENE); label("[itex]Q[itex]",Q,N); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once u have draw these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \cdot 6 = 54$ small triangles in the whole hexagon. There are$22[itex] small triangles in hexagon [itex]ZWCXYF[itex], and [itex]9 \text{ small triangles} \cdot 6 \text{ triangles}= 54[itex] small triangles in the whole hexagon [itex]ABCDEF[itex].

Thus, the answer is [itex]\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$(Error compiling LaTeX. ! Missing$ inserted.).