Difference between revisions of "2011 AMC 12A Problems/Problem 11"
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The requested area is area of circle <math>C</math> minus 4 of this area: | The requested area is area of circle <math>C</math> minus 4 of this area: | ||
− | <math>\pi 1^2 - 4(\frac{\pi}{4}-\frac{1}{2}) | + | <math>\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) |
= \pi - \pi + 2 | = \pi - \pi + 2 | ||
= 2</math> | = 2</math> |
Revision as of 18:56, 10 August 2023
Contents
[hide]Problem
Circles and each have radius 1. Circles and share one point of tangency. Circle has a point of tangency with the midpoint of What is the area inside circle but outside circle and circle
Solution 1
The requested area is the area of minus the area shared between circles , and .
Let be the midpoint of and be the other intersection of circles and .
The area shared between , and is of the regions between arc and line , which is (considering the arc on circle ) a quarter of the circle minus :
(We can assume this because is 90 degrees, since is a square, due to the application of the tangent chord theorem at point )
So the area of the small region is
The requested area is area of circle minus 4 of this area:
.
Solution 2
We can move the area above the part of the circle above the segment down, and similarly for the other side. Then, we have a square, whose diagonal is , so the area is then just .
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=olRZuK11mAI
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.