Difference between revisions of "2011 AMC 12A Problems/Problem 11"
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− | We can move the area above the part of the circle above the segment <math>EF</math> down, and similarly for the other side. Then, we have a square, whose diagonal is <math>2</math>, so the area is then just <math>\left(\frac{2}{\sqrt{2}}\right)^2 = 2</math>. | + | We can move the area above the part of the circle above the segment <math>EF</math> down, and similarly for the other side. Then, we have a square, whose diagonal is <math>2</math>, so the area is then just <math>\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 21:43, 2 September 2024
Contents
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.