Difference between revisions of "2016 AMC 10B Problems/Problem 23"
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− | 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>. | + | 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> | <asy> | ||
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<math>^*</math> At this point, you can answer <math>\textbf{(C)}</math> and move on with your test. | <math>^*</math> At this point, you can answer <math>\textbf{(C)}</math> and move on with your test. | ||
− | ==Solution 2== | + | ==Solution 2 (a lot faster than Solution 1)== |
<asy> | <asy> | ||
− | pair A,B,C,D,E,F,W,X,Y,Z | + | pair A,B,C,D,E,F,W,X,Y,Z; |
A=(0,0); | A=(0,0); | ||
B=(1,0); | B=(1,0); | ||
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Y=(-1/3,sqrt(3)/3); | Y=(-1/3,sqrt(3)/3); | ||
Z=(-1/3,2sqrt(3)/3); | Z=(-1/3,2sqrt(3)/3); | ||
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draw(A--B--C--D--E--F--cycle); | draw(A--B--C--D--E--F--cycle); | ||
draw(W--Z); | draw(W--Z); | ||
draw(X--Y); | draw(X--Y); | ||
draw(F--C--B--E--D--A); | draw(F--C--B--E--D--A); | ||
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label("$A$",A,SW); | label("$A$",A,SW); | ||
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label("$Y$",Y,WSW); | label("$Y$",Y,WSW); | ||
label("$Z$",Z,WNW); | label("$Z$",Z,WNW); | ||
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</asy> | </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 | + | 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 <math>22</math> small triangles in hexagon <math>ZWCXYF</math>, and <math>9 \cdot 6 = 54</math> small triangles in the whole hexagon. |
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>. |
Revision as of 23:31, 21 April 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 1
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.
Solution 2 (a lot faster than Solution 1)
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 small triangles in hexagon , and small triangles in the whole hexagon.
Thus, the answer is .
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.