Difference between revisions of "2016 AMC 10B Problems/Problem 23"
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<math>\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}</math> | <math>\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}</math> | ||
− | |||
==Solution 1== | ==Solution 1== | ||
Line 16: | Line 15: | ||
E=(0,sqrt(3)); | E=(0,sqrt(3)); | ||
F=(-1/2,sqrt(3)/2); | F=(-1/2,sqrt(3)/2); | ||
− | + | X=(4/3,2sqrt(3)/3); | |
− | + | W=(4/3,sqrt(3)/3); | |
− | + | Z=(-1/3,sqrt(3)/3); | |
− | + | Y=(-1/3,2sqrt(3)/3); | |
draw(A--B--C--D--E--F--cycle); | draw(A--B--C--D--E--F--cycle); | ||
draw(W--Z); | draw(W--Z); | ||
Line 73: | Line 72: | ||
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>. | ||
− | ==Solution 2 | + | ==Solution 2== |
<asy> | <asy> | ||
Line 107: | Line 106: | ||
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>. | ||
+ | |||
+ | ==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 <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> | ||
+ | |||
+ | |||
+ | ==Solution 4 (Extending Lines)== | ||
+ | |||
+ | Refer to the diagram from Solution 1. | ||
+ | |||
+ | Let us start by connecting points <math>F</math> to <math>C</math> to create a new line segment <math>FC</math>. We drop a perpendicular line segment from <math>E</math> to side <math>FC</math> at point <math>P</math>. Since <math>\angle FED = 120</math>, and <math>\angle PED = 90</math>, we know that <math>\angle FEP = 30.</math> | ||
+ | |||
+ | Therefore, <math>\triangle EFP</math> is a 30-60-90 triangle. Assume that <math>EP = \sqrt{3}</math>. Therefore, we know that <math>FP = 1</math>, <math>EF = 2</math>. | ||
+ | |||
+ | Let us draw <math>PQ</math> such that <math>PQ</math> is perpendicular to <math>ZW</math>. Since the distance between the parallel lines are equal, we know that <math>PQ</math> is half of the distance between the parallel lines. Hence, <math>PQ</math> will be one-third the length of <math>EP</math>. | ||
+ | |||
+ | From this, we also know that <math>EQ = EP - PQ = EP - \frac{1}{3} EP = \frac{2}{3} EP</math>. | ||
+ | |||
+ | Hence, we know that <math>EQ = \frac{2}{3} EP = \frac{2\sqrt{3}}{3}.</math> Since <math>\angle EQZ = 90, FEP = 30</math> we know that <math>\triangle ZEQ</math> is also a 30-60-90 triangle. From this, we know that <math>ZQ = \frac{2}{3}</math>. | ||
+ | |||
+ | Let us drop another perpendicular line segment from <math>D</math> to point <math>T</math> such that <math>DT</math> is perpendicular to <math>ZW</math>. | ||
+ | |||
+ | Since <math>ED = QT</math> and since <math>ED</math> is the side length of the regular hexagon, we know that <math>ED = EF = QT = 2.</math> | ||
+ | |||
+ | By symmetry, we also know that <math>WT = \frac{2}{3}</math>. | ||
+ | |||
+ | Therefore, we can find the length of <math>ZW = ZQ + QT + QW = \frac{2}{3} + 2 + \frac{2}{3}</math>. Hence, we know that <math>ZW = \frac{10}{3}.</math> | ||
+ | |||
+ | Now, we can find the area of trapezoid <math>EDZW = \frac{ED + ZW}{2} \cdot EQ = \frac{2 + \frac{10}{3}}{2} \cdot \frac{2\sqrt{3}}{3} = \frac{16\sqrt{3}}{9}.</math> | ||
+ | |||
+ | By symmetry, we know that the area of trapezoid <math>YZAB = \frac{16\sqrt{3}}{9}.</math> | ||
+ | |||
+ | Using the area of a hexagon formula, we get that the area of regular hexagon <math>ABCDEF = \frac{3\sqrt{3}}{2} \cdot 2^2 = 6\sqrt{3}.</math> | ||
+ | |||
+ | Hence, the area of hexagon <math>WCXYFZ = 6\sqrt{3} - 2\left(\frac{16\sqrt{3}}{9}\right) = \frac{22\sqrt{3}}{9}.</math> | ||
+ | |||
+ | Hence, the ratio of the area of hexagon <math>WCXYFZ</math> to the area of hexagon <math>ABCDEF</math> is <math>\dfrac{\frac{22\sqrt{3}}{9}}{6\sqrt{3}} = \boxed{\frac{11}{27} \implies C}.</math> | ||
+ | |||
+ | ~yk2007 :D | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We will do this by area subtraction. Drawing in the required lines, we drop an altitude from A to WZ and an altitude from E to XY. You get 2 30-60-90 triangles, and given that the side length is <math>x</math>, AZ is <math>2x/3</math>, which means that WZ is <math>5x/3</math>, adding in AB. Using the 30-60-90 triangle again, you get that the area of trapezoid <math>ABWZ</math> is <math>\frac{4x}{3} \cdot</math> <math>\frac{x\sqrt3}{3}</math>. Repeat on the other side to get that the area of both of these trapezoids combined are <math>\frac{8x^2\sqrt3}{9}</math>. Finding the area of the hexagon, dividing, and subtracting, gets you C. | ||
+ | |||
+ | -dragoon (LATEX help) | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/GrCtzL0S-Uo?t=638 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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 16:43, 9 August 2023
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 is equal to
.
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 you have drawn 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 .
Solution 3 (Similar Triangles)
Extend
and
to meet at point
, as shown in the diagram. Then
. Then
and
. Subtracting
, we find that
and
. Subtracting again, we find that
Finally,
Solution 4 (Extending Lines)
Refer to the diagram from Solution 1.
Let us start by connecting points to
to create a new line segment
. We drop a perpendicular line segment from
to side
at point
. Since
, and
, we know that
Therefore, is a 30-60-90 triangle. Assume that
. Therefore, we know that
,
.
Let us draw such that
is perpendicular to
. Since the distance between the parallel lines are equal, we know that
is half of the distance between the parallel lines. Hence,
will be one-third the length of
.
From this, we also know that .
Hence, we know that Since
we know that
is also a 30-60-90 triangle. From this, we know that
.
Let us drop another perpendicular line segment from to point
such that
is perpendicular to
.
Since and since
is the side length of the regular hexagon, we know that
By symmetry, we also know that .
Therefore, we can find the length of . Hence, we know that
Now, we can find the area of trapezoid
By symmetry, we know that the area of trapezoid
Using the area of a hexagon formula, we get that the area of regular hexagon
Hence, the area of hexagon
Hence, the ratio of the area of hexagon to the area of hexagon
is
~yk2007 :D
Solution 5
We will do this by area subtraction. Drawing in the required lines, we drop an altitude from A to WZ and an altitude from E to XY. You get 2 30-60-90 triangles, and given that the side length is , AZ is
, which means that WZ is
, adding in AB. Using the 30-60-90 triangle again, you get that the area of trapezoid
is
. Repeat on the other side to get that the area of both of these trapezoids combined are
. Finding the area of the hexagon, dividing, and subtracting, gets you C.
-dragoon (LATEX help)
Video Solution by OmegaLearn
https://youtu.be/GrCtzL0S-Uo?t=638
~ pi_is_3.14
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.