Difference between revisions of "2012 AMC 10B Problems/Problem 16"
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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. | 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. | ||
(check this link for a better view: https://drive.google.com/drive/u/0/my-drive) | (check this link for a better view: https://drive.google.com/drive/u/0/my-drive) | ||
− | 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 | + | 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 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2012|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:56, 10 January 2021
Contents
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?
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 . 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: The height is and the base is . Multiplying the height and base together with , we get . Since there are two 30-60-90 triangles in the equilateral triangle, we multiply the area of the triangle by :
To find the area of the remaining sectors, which are of the original circles once we remove the triangle, we know that the sectors have a central angle of since the equilateral triangle already covered that area. Since there are pieces gone from the equilateral triangle, we have, in total, of a circle (with radius ) gone. Each circle has an area of , so three circles gives a total area of . Subtracting the half circle, we have:
Summing the areas from the equilateral triangle and the remaining circle sections gives us: .
Alternate Solution
First, the area of the 3 circles is simply . 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 and the area of the 90 degrees partial circles are . Therefore, the area of the shape in between the three circles is a little less than . Summing up the 3 circles we got and the approximate area of the middle shape, we get , which is a little more than what we want. We see that all answer choices except is greater than , 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 sectors of of a circle which makes a total of circles. Calculating this area with , we get . We also know that the equilateral triangle will give us some constant multiplied by . The only answer choice with and is . ~chrisdiamond10
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.