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>. | |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=10|num-a=12|ab=A}} | {{AMC12 box|year=2011|num-b=10|num-a=12|ab=A}} |
Revision as of 22:47, 24 February 2011
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
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{3\pi}{4} \qquad \textbf{(E)}\ 1+\frac{\pi}{2}}$ (Error compiling LaTeX. Unknown error_msg)
Solution
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
.
Then 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 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
.
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 |