Difference between revisions of "2004 AMC 10B Problems/Problem 25"
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Hence the area of the new shaded region is <math>4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( \frac 12 \cdot 2 \cdot \sqrt{3} \right) = \frac 83 \pi - 2\sqrt 3</math>, and the area of the original shared region is | Hence the area of the new shaded region is <math>4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( \frac 12 \cdot 2 \cdot \sqrt{3} \right) = \frac 83 \pi - 2\sqrt 3</math>, and the area of the original shared region is | ||
− | <math>\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{ \frac 53 \pi - 2\sqrt 3 }</math>. | + | <math>\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }</math>. |
== See also == | == See also == |
Revision as of 19:24, 31 January 2015
Problem
A circle of radius is internally tangent to two circles of radius
at points
and
, where
is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?
Solution
The area of the small circle is . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.
Let and
be the intersections of the two large circles. Connect them to
and
to get the picture below:
Now obviously the triangles and
are equilateral with side
.
Take a look at the bottom circle. The angle is
, hence the sector
is
of the circle. The same is true for the sector
of the bottom circle, and sectors
and
of the top circle.
If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice. These triangles have a base of and a height of
.
Hence the area of the new shaded region is , and the area of the original shared region is
.
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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.