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

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

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 $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$. $^*$ At this point, you can answer $\textbf{(C)}$ and move on with your test.

## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z,S,K,R,U,H,I,O,P,Q; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); S=(-1/6,sqrt(3)/6); H=(-1/6, 5sqrt(3)/6); P=(7/6, 5sqrt(3)/6); U=(7/6,sqrt(3)/6); K=(1/3, 0); R=(2/3, 0); I=(1/3,sqrt(3)); O=(2/3,sqrt(3)); Q=(1/2, sqrt(3)/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); draw(S--U); draw(K--R); draw(Z--K); draw(H--R); draw(I--U); draw(O--X); draw(H--P); draw(I--Y); draw(O--S); draw(P--K); draw(W--R); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); label("S",S,WSW); label("K",K,SSW); label("R",R,SSE); label("U",U,ESE); label("H",H,WNW); label("I",I,NNW); label("O",O,NNE); label("P",P,ENE); label("Q",Q,N); [/asy]$

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 $22$ small triangles in hexagon $ZWCXYF$, and $9 \text{ small triangles} \cdot 6 \text{ triangles}= 54$ small triangles in the whole hexagon $ABCDEF$.

Thus, the answer is $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 