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

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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?
  
To determine the area of the figure, you can connect the centers of the circles to form an equilateral triangle of length <math>4</math>. Find the area of this triangle to include the figure formed in between the circles. This area is <math>4\sqrt{3}</math>.
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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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<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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[[Category: Introductory Geometry Problems]]
  
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==Solution==
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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>
  
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 <math>2^2 \pi * 5/6 = 10\pi/3</math>. Then this area is multiplied by three to find the total area of the sectors (10 pi).
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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>  
This result is added to area of the equilateral triangle to get a final answer of <math>10\pi + 4\sqrt3</math>.
 
  
This means (A) is the right answer.
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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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==Alternate Solution==
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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 are 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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(check this link for a better view: https://drive.google.com/drive/u/0/my-drive)
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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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==See Also==
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{{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 14:06, 1 November 2020

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

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: $\boxed{\textbf{(A)} 10\pi + 4\sqrt3}$.

Alternate Solution

First, the area of the 3 circles is simply $3*\pi*2^2 = 12 \pi$. Notice that the middle are 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. (check this link for a better view: https://drive.google.com/drive/u/0/my-drive) 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

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
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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