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Difference between revisions of "2012 AMC 10B Problems/Problem 16"

(Solution to AMC 10B 2012 #16)
 
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To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle of length 4. Find the area of this triangle to include the figure formed in between the circles. This area is 4sqrt3.
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==Problem==
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Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?
  
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<asy>
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filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray);
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filldraw(circle((1,sqrt(3)),1),gray);
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filldraw(circle((0,0),1),gray);
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filldraw(circle((2,0),1),grey);</asy>
  
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To find the area of the remaining sectors, notice that the sectors have a central angle of 300 because 60 degrees were "used up" for the triangle. The area of one sector is 2^2 * pi * 5/6 = 10pi/3. Then this area is multiplied by three to find the total area of the sectors (10 pi).
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<math>\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi</math>
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This result is added to area of the equilateral triangle to get a final answer of 10pi + 4sqrt3.
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[[Category: Introductory Geometry Problems]]
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--[[User:Ramen|Ramen]] 18:15, 13 September 2012 (EDT)Justin
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==Solution 1==
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To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle with a side of length <math>4</math>. We must find the area of this triangle to include the figure formed in between the circles. Since the equilateral triangle has two 30-60-90 triangles inside, we can find the height and the base of each 30-60-90 triangle from the ratios: <math>1: \sqrt{3}: 2.</math> The height is <math>2\sqrt{3}</math> and the base is <math>2</math>. Multiplying the height and base together with <math>\dfrac{1}{2}</math>, we get <math>2\sqrt{3}</math>. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the <math>30-60-90</math> triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath>
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To find the area of the remaining sectors, which are <math>\dfrac{5}{6}</math> of the original circles once we remove the triangle, we know that the sectors have a central angle of <math>300^\circ</math> since the equilateral triangle already covered that area. Since there are <math>3</math> <math>\dfrac{1}{6}</math> pieces gone from the equilateral triangle, we have, in total, <math>\dfrac{1}{2}</math> of a circle (with radius <math>2</math>) gone. Each circle has an area of <math>\pi r^2 = 4\pi</math>, so three circles gives a total area of <math>12\pi</math>. Subtracting the half circle, we have: <cmath>12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.</cmath>
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Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>.
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==Solution 2==
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First, the area of the 3 circles is simply <math>3*\pi*2^2 = 12 \pi</math>. Notice that the middle area is a little more than a rectangle formed by completely filling the rectangle formed by connecting two 90 degrees partial circles and then subtracting the two 90 degrees partial circles.  
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The area of the rectangle is <math>4*2=8</math> and the area of the 90 degrees partial circles are <math>2*(1/4)*\pi*2^2 = 2\pi</math>. Therefore, the area of the shape in between the three circles is a little less than <math>8 - 2\pi</math>. Summing up the 3 circles we got and the approximate area of the middle shape, we get <math>10\pi+8</math>, which is a little more than what we want. We see that all answer choices except <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math> is greater than <math>10\pi+8</math>, therefore it's the only answer. -dchang0524
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==Solution 3 (answer choices, not recommended)==
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Notice as in solution 1 that the figure is simply an equilateral triangle and <math>3</math> sectors of <math>\frac{5}{6}</math> of a circle which makes a total of <math>\frac{5}{2}</math> circles. Calculating this area with <math>\pi r^2</math>, we get <math>10\pi</math>. We also know that the equilateral triangle will give us some constant multiplied by <math>\sqrt3</math>. The only answer choice with <math>10\pi</math> and <math>\sqrt3</math> is <math>\boxed{\textbf{(A)}}</math>.
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~chrisdiamond10
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==Solution 4==
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Notice that we can connect the centers of the 3 circles to make an equilateral triangle with length <math>4</math>. The equilateral triangle with length <math>4</math> has area <math>4</math><math>\sqrt{3}</math>. Because the triangle took <math>1/6</math> th of all the 3 circles away, we have the area of the circles as <math>(5/6)*(4\pi)*3 = 10\pi</math>. We add the area of the triangle to get <math>10 \pi</math> + <math>4</math><math>\sqrt{3}</math> = <math>\boxed {A}</math>
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==Video Solution==
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https://youtu.be/G44CDSfgt7Y
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~savannahsolver
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== Video Solution by OmegaLearn ==
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https://youtu.be/NsQbhYfGh1Q?t=1569
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~ pi_is_3.14
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==See Also==
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{{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 00:29, 1 November 2024

Problem

Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?

[asy] filldraw((0,0)--(2,0)--(1,sqrt(3))--cycle,gray,gray); filldraw(circle((1,sqrt(3)),1),gray); filldraw(circle((0,0),1),gray); filldraw(circle((2,0),1),grey);[/asy]

$\textbf{(A)}\ 10\pi+4\sqrt{3}\qquad\textbf{(B)}\ 13\pi-\sqrt{3}\qquad\textbf{(C)}\ 12\pi+\sqrt{3}\qquad\textbf{(D)}\ 10\pi+9\qquad\textbf{(E)}\ 13\pi$

Solution 1

To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle with a side of length $4$. We must find the area of this triangle to include the figure formed in between the circles. Since the equilateral triangle has two 30-60-90 triangles inside, we can find the height and the base of each 30-60-90 triangle from the ratios: $1: \sqrt{3}: 2.$ The height is $2\sqrt{3}$ and the base is $2$. Multiplying the height and base together with $\dfrac{1}{2}$, we get $2\sqrt{3}$. Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the $30-60-90$ triangle by $2$: \[2\cdot 2\sqrt{3} = 4\sqrt{3}.\]

To find the area of the remaining sectors, which are $\dfrac{5}{6}$ of the original circles once we remove the triangle, we know that the sectors have a central angle of $300^\circ$ since the equilateral triangle already covered that area. Since there are $3$ $\dfrac{1}{6}$ pieces gone from the equilateral triangle, we have, in total, $\dfrac{1}{2}$ of a circle (with radius $2$) gone. Each circle has an area of $\pi r^2 = 4\pi$, so three circles gives a total area of $12\pi$. Subtracting the half circle, we have: \[12\pi - \dfrac{4\pi}{2} = 12\pi - 2\pi = 10\pi.\]

Summing the areas from the equilateral triangle and the remaining circle sections gives us: $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$.

Solution 2

First, the area of the 3 circles is simply $3*\pi*2^2 = 12 \pi$. Notice that the middle area is a little more than a rectangle formed by completely filling the rectangle formed by connecting two 90 degrees partial circles and then subtracting the two 90 degrees partial circles. The area of the rectangle is $4*2=8$ and the area of the 90 degrees partial circles are $2*(1/4)*\pi*2^2 = 2\pi$. Therefore, the area of the shape in between the three circles is a little less than $8 - 2\pi$. Summing up the 3 circles we got and the approximate area of the middle shape, we get $10\pi+8$, which is a little more than what we want. We see that all answer choices except $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$ is greater than $10\pi+8$, therefore it's the only answer. -dchang0524

Solution 3 (answer choices, not recommended)

Notice as in solution 1 that the figure is simply an equilateral triangle and $3$ sectors of $\frac{5}{6}$ of a circle which makes a total of $\frac{5}{2}$ circles. Calculating this area with $\pi r^2$, we get $10\pi$. We also know that the equilateral triangle will give us some constant multiplied by $\sqrt3$. The only answer choice with $10\pi$ and $\sqrt3$ is $\boxed{\textbf{(A)}}$. ~chrisdiamond10

Solution 4

Notice that we can connect the centers of the 3 circles to make an equilateral triangle with length $4$. The equilateral triangle with length $4$ has area $4$$\sqrt{3}$. Because the triangle took $1/6$ th of all the 3 circles away, we have the area of the circles as $(5/6)*(4\pi)*3 = 10\pi$. We add the area of the triangle to get $10 \pi$ + $4$$\sqrt{3}$ = $\boxed {A}$

Video Solution

https://youtu.be/G44CDSfgt7Y

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/NsQbhYfGh1Q?t=1569

~ pi_is_3.14

See Also

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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