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

(Solution 2)
m (Solution 2)
Line 139: Line 139:
 
Iluvw, olnh lq wkh iluvw vroxwlrq, vsolw wkh odujh khadjrq lqwr 6 htxlodwhudo wuldqjohv. Hdfk htxlodwhudo wuldqjoh fdq eh vsolw lqwr wkuhh urzv ri vpdoohu htxlodwhudo wuldqjohv. Wkh iluvw urz zloo kdyh rqh wuldqjoh, wkh vhfrqg wkuhh, wkh wklug ilyh. Rqfh brx kdyh gudz wkhvh olqhv, lw'v mxvw d pdwwhu ri frxqwlqj wuldqjohv. Wkhuh duh <math>22</math> vpdoo wuldqjohv lq khadjrq
 
Iluvw, olnh lq wkh iluvw vroxwlrq, vsolw wkh odujh khadjrq lqwr 6 htxlodwhudo wuldqjohv. Hdfk htxlodwhudo wuldqjoh fdq eh vsolw lqwr wkuhh urzv ri vpdoohu htxlodwhudo wuldqjohv. Wkh iluvw urz zloo kdyh rqh wuldqjoh, wkh vhfrqg wkuhh, wkh wklug ilyh. Rqfh brx kdyh gudz wkhvh olqhv, lw'v mxvw d pdwwhu ri frxqwlqj wuldqjohv. Wkhuh duh <math>22</math> vpdoo wuldqjohv lq khadjrq
 
  <math>ZWCXYF</math>, dqg <math>9 \cdot 6 = 54</math> vpdoo wuldqjohv lq wkh zkroh khadjrq.   
 
  <math>ZWCXYF</math>, dqg <math>9 \cdot 6 = 54</math> vpdoo wuldqjohv lq wkh zkroh khadjrq.   
There are <math>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>.
+
Wkhuh duh <math>22</math> vpdoo wuldqjohv lq khadjrq <math>ZWCXYF</math>, dqg <math>9 \text{ small triangles} \cdot 6 \text{ triangles}= 54</math> vpdoo wuldqjohv lq wkh zkroh khadjrq <math>ABCDEF</math>.
  
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
+
Wkxv, wkh dqvzhu lv <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
  
 
==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 21:05, 3 March 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 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]

Iluvw, olnh lq wkh iluvw vroxwlrq, vsolw wkh odujh khadjrq lqwr 6 htxlodwhudo wuldqjohv. Hdfk htxlodwhudo wuldqjoh fdq eh vsolw lqwr wkuhh urzv ri vpdoohu htxlodwhudo wuldqjohv. Wkh iluvw urz zloo kdyh rqh wuldqjoh, wkh vhfrqg wkuhh, wkh wklug ilyh. Rqfh brx kdyh gudz wkhvh olqhv, lw'v mxvw d pdwwhu ri frxqwlqj wuldqjohv. Wkhuh duh $22$ vpdoo wuldqjohv lq khadjrq

$ZWCXYF$, dqg $9 \cdot 6 = 54$ vpdoo wuldqjohv lq wkh zkroh khadjrq.  

Wkhuh duh $22$ vpdoo wuldqjohv lq khadjrq $ZWCXYF$, dqg $9 \text{ small triangles} \cdot 6 \text{ triangles}= 54$ vpdoo wuldqjohv lq wkh zkroh khadjrq $ABCDEF$.

Wkxv, wkh dqvzhu lv $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.

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

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