Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
(14 intermediate revisions by 8 users not shown)
Line 11: Line 11:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==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 <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 30-60-90 triangle by <math>2</math>: <cmath>2\cdot 2\sqrt{3} = 4\sqrt{3}.</cmath>
+
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, 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>  
 
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>  
  
 
Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>.
 
Summing the areas from the equilateral triangle and the remaining circle sections gives us: <math>\boxed{\textbf{(A)} 10\pi + 4\sqrt3}</math>.
 +
 +
==Solution 2==
 +
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.
 +
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
 +
 +
==Solution 3 (answer choices, not recommended)==
 +
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>.
 +
~chrisdiamond10
 +
 +
==Solution 4==
 +
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>
 +
 +
==Video Solution==
 +
https://youtu.be/G44CDSfgt7Y
 +
 +
~savannahsolver
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/NsQbhYfGh1Q?t=1569
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_16&oldid=231120"