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Difference between revisions of "2016 AMC 10B Problems/Problem 23"

(→‎Solution)
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==Solution==
 
==Solution==
−
<math>\textbf{(C)}\ \frac{11}{27}</math>
+
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 <math>AB</math> is of length <math>1</math>.  Therefore, the area of <math>ABCDEF</math> is <math>\frac{3\sqrt 3}2</math>.  To find the area of <math>WCXYFZ</math>, we draw <math>\overline{CF}</math>, and find the area of the trapezoids <math>WCFZ</math> and <math>CXYF</math>. 
 +
 +
<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 <math>CF=2</math>.  We also know that the combined heights of the trapezoids is <math>\frac{\sqrt 3}3</math>, since <math>\overline{ZW}</math> and <math>\overline{YX}</math> are equally spaced, and the height of each of the trapezoids is <math>\frac{\sqrt 3}6</math>.  From this, we know <math>\overline{ZW}</math> and <math>\overline{YX}</math> are each <math>\frac 13</math> of the way from <math>\overline{CF}</math> to <math>\overline{DE}</math> and <math>\overline{AB}</math>, respectively.  We know that these are both equal to <math>\frac 53</math>.
 +
 +
We find the area of each of the trapezoids, which both happen to be <math>\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}</math>, and the combined area is <math>\frac{11\sqrt 3}{18}^{*}</math>.
 +
 +
We find that <math>\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>^*</math> At this point, you can answer <math>\textbf{(C)}</math> and move on with your test.
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:18, 22 February 2016

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

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.

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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

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