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

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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>.
 
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>.
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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>.
 
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>.
  
 
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==Solution 2==
<math>^*</math> At this point, you can answer <math>\textbf{(C)}</math> and move on with your test.
 
 
 
==Solution 2 (cheap, but ingenious)==
 
  
 
<asy>
 
<asy>
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Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
 
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
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 +
==Solution 3 (Similar Triangles)==
 +
<asy>
 +
pair A,B,C,D,E,F,W,X,Y,Z;
 +
A=(0,0);
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B=(1,0);
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C=(3/2,sqrt(3)/2);
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D=(1,sqrt(3));
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E=(0,sqrt(3));
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F=(-1/2,sqrt(3)/2);
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W=(4/3,2sqrt(3)/3);
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X=(4/3,sqrt(3)/3);
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Y=(-1/3,sqrt(3)/3);
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Z=(-1/3,2sqrt(3)/3);
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pair G = (0.5, sqrt(3)*3/2);
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draw(A--B--C--D--E--F--cycle);
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draw(W--Z);
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draw(X--Y);
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draw(E--G--D);
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draw(F--C);
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,ESE);
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label("$D$",D,NE);
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label("$E$",E,NW);
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label("$F$",F,WSW);
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label("$W$",W,ENE);
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label("$X$",X,ESE);
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label("$Y$",Y,WSW);
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label("$Z$",Z,WNW);
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label("$G$",G,N);
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</asy>
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Extend <math>\overline{EF}</math> and <math>\overline{CD}</math> to meet at point <math>G</math>, as shown in the diagram. Then <math>\triangle GZW \sim \triangle GFC</math>. Then <math>[GZW] = \left(\dfrac53\right)^2[GED]</math> and <math>[GFC] = 2^2[GED]</math>. Subtracting <math>[GED]</math>, we find that <math>[EDWZ] = \dfrac{16}{9}[GED]</math> and <math>[EDCF] = 3[GED]</math>. Subtracting again, we find that <cmath>[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].</cmath>Finally, <cmath>\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.</cmath>
  
 
==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}}

Latest revision as of 02:38, 9 August 2020

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 $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.

Solution 2

[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]

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 you have drawn 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.

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

Solution 3 (Similar Triangles)

[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); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C);  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("$G$",G,N); [/asy] Extend $\overline{EF}$ and $\overline{CD}$ to meet at point $G$, as shown in the diagram. Then $\triangle GZW \sim \triangle GFC$. Then $[GZW] = \left(\dfrac53\right)^2[GED]$ and $[GFC] = 2^2[GED]$. Subtracting $[GED]$, we find that $[EDWZ] = \dfrac{16}{9}[GED]$ and $[EDCF] = 3[GED]$. Subtracting again, we find that \[[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].\]Finally, \[\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.\]

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

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