2016 AMC 10B Problems/Problem 23

Revision as of 20:21, 25 February 2016 by Kyriegon (talk | contribs) (Solution 2 (30 sec))

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("<math>A<math>",A,SW); label("<math>B<math>",B,SE); label("<math>C<math>",C,ESE); label("<math>D<math>",D,NE); label("<math>E<math>",E,NW); label("<math>F<math>",F,WSW); label("<math>W<math>",W,ENE); label("<math>X<math>",X,ESE); label("<math>Y<math>",Y,WSW); label("<math>Z<math>",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("<math>A<math>",A,SW); label("<math>B<math>",B,SE); label("<math>C<math>",C,ESE); label("<math>D<math>",D,NE); label("<math>E<math>",E,NW); label("<math>F<math>",F,WSW); label("<math>W<math>",W,ENE); label("<math>X<math>",X,ESE); label("<math>Y<math>",Y,WSW); label("<math>Z<math>",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}<math> is equal to <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}<math>.


<math>^*$ (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("<math>A<math>",A,SW); label("<math>B<math>",B,SE); label("<math>C<math>",C,ESE); label("<math>D<math>",D,NE); label("<math>E<math>",E,NW); label("<math>F<math>",F,WSW); label("<math>W<math>",W,ENE); label("<math>X<math>",X,ESE); label("<math>Y<math>",Y,WSW); label("<math>Z<math>",Z,WNW); label("<math>S<math>",S,WSW); label("<math>K<math>",K,SSW); label("<math>R<math>",R,SSE); label("<math>U<math>",U,ESE); label("<math>H<math>",H,WNW); label("<math>I<math>",I,NNW); label("<math>O<math>",O,NNE); label("<math>P<math>",P,ENE); label("<math>Q<math>",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<math> small triangles in hexagon <math>ZWCXYF<math>, and <math>9 \text{ small triangles} \cdot 6 \text{ triangles}= 54<math> small triangles in the whole hexagon <math>ABCDEF<math>.

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

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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

Invalid username
Login to AoPS