Difference between revisions of "2016 AMC 10B Problems/Problem 23"
Math101010 (talk | contribs) m (→Solution 2 (30 sec)) |
(→Solution 2 (30 sec)) |
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Line 24: | Line 24: | ||
draw(X--Y); | draw(X--Y); | ||
− | label(" | + | label("<math>A<math>",A,SW); |
− | label(" | + | label("<math>B<math>",B,SE); |
− | label(" | + | label("<math>C<math>",C,ESE); |
− | label(" | + | label("<math>D<math>",D,NE); |
− | label(" | + | label("<math>E<math>",E,NW); |
− | label(" | + | label("<math>F<math>",F,WSW); |
− | label(" | + | label("<math>W<math>",W,ENE); |
− | label(" | + | label("<math>X<math>",X,ESE); |
− | label(" | + | label("<math>Y<math>",Y,WSW); |
− | label(" | + | label("<math>Z<math>",Z,WNW); |
</asy> | </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>. | + | 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> | ||
Line 55: | Line 55: | ||
draw(F--C--B--E--D--A); | draw(F--C--B--E--D--A); | ||
− | label(" | + | label("<math>A<math>",A,SW); |
− | label(" | + | label("<math>B<math>",B,SE); |
− | label(" | + | label("<math>C<math>",C,ESE); |
− | label(" | + | label("<math>D<math>",D,NE); |
− | label(" | + | label("<math>E<math>",E,NW); |
− | label(" | + | label("<math>F<math>",F,WSW); |
− | label(" | + | label("<math>W<math>",W,ENE); |
− | label(" | + | label("<math>X<math>",X,ESE); |
− | label(" | + | label("<math>Y<math>",Y,WSW); |
− | label(" | + | label("<math>Z<math>",Z,WNW); |
</asy> | </asy> | ||
Line 71: | Line 71: | ||
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 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}< | + | 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>. |
Line 79: | Line 79: | ||
<asy> | <asy> | ||
− | pair A,B,C,D,E,F,W,X,Y,Z; | + | pair A,B,C,D,E,F,W,X,Y,Z,S,K,R,U,H,I,O,P,Q; |
A=(0,0); | A=(0,0); | ||
B=(1,0); | B=(1,0); | ||
Line 90: | Line 90: | ||
Y=(-1/3,sqrt(3)/3); | Y=(-1/3,sqrt(3)/3); | ||
Z=(-1/3,2sqrt(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(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); | ||
+ | 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(" | + | label("<math>A<math>",A,SW); |
− | label(" | + | label("<math>B<math>",B,SE); |
− | label(" | + | label("<math>C<math>",C,ESE); |
− | label(" | + | label("<math>D<math>",D,NE); |
− | label(" | + | label("<math>E<math>",E,NW); |
− | label(" | + | label("<math>F<math>",F,WSW); |
− | label(" | + | label("<math>W<math>",W,ENE); |
− | label(" | + | label("<math>X<math>",X,ESE); |
− | label(" | + | label("<math>Y<math>",Y,WSW); |
− | label(" | + | label("<math>Z<math>",Z,WNW); |
+ | label("<math>S<math>",S,WSW); | ||
+ | label("<math>K<math>",K,SSW); | ||
+ | label("<math>R<math>",R,SSE); | ||
+ | label("<math>U<math>",U,ESE); | ||
+ | label("<math>H<math>",H,WNW); | ||
+ | label("<math>I<math>",I,NNW); | ||
+ | label("<math>O<math>",O,NNE); | ||
+ | label("<math>P<math>",P,ENE); | ||
+ | label("<math>Q<math>",Q,N); | ||
</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 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. | + | 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. |
+ | |||
+ | 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>. | ||
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 20:21, 25 February 2016
Contents
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 $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}<math> is equal to <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}<math>.
<math>^*$ (Error compiling LaTeX. ! Missing $ inserted.) At this point, you can answer and move on with your test.
Solution 2
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.
There are $22<math> small triangles in hexagon <math>ZWCXYF<math>, and <math>9 \text{ small triangles} \cdot 6 \text{ triangles}= 54<math> small triangles in the whole hexagon <math>ABCDEF<math>.
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$ (Error compiling LaTeX. ! Missing $ inserted.).
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.