Difference between revisions of "2016 AMC 10B Problems/Problem 23"
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==Solution== | ==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 <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 22:18, 22 February 2016
Problem
In regular hexagon , points , , , and are chosen on sides , , , and respectively, so lines , , , and are parallel and equally spaced. What is the ratio of the area of hexagon to the area of hexagon ?
Solution
We draw a diagram to make our work easier:
Assume that is of length . Therefore, the area of is . To find the area of , we draw , and find the area of the trapezoids and .
From this, we know that . We also know that the combined heights of the trapezoids is , since and are equally spaced, and the height of each of the trapezoids is . From this, we know and are each of the way from to and , respectively. We know that these are both equal to .
We find the area of each of the trapezoids, which both happen to be , and the combined area is .
We find that is equal to .
At this point, you can answer 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.