Difference between revisions of "2007 AMC 10A Problems/Problem 24"
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Since <math>\overline{OC}</math> is tangent to circle <math>A</math>, <math>\triangle{ACO}</math> is a right triangle. We know <math>AO=2\sqrt{2}</math> and <math>AC=2</math>, so <math>\triangle{ACO}</math> is isosceles, a <math>45</math>-<math>45</math> right triangle, and has <math>\overline{CO}</math> with length <math>2</math>. The area of <math>\triangle{ACO}=\frac{1}{2}bh=2</math>. By symmetry, <math>\triangle{ACO}\cong\triangle{BDO}</math>, and so the area of <math>\triangle{BDO}</math> is also <math>2</math>. | Since <math>\overline{OC}</math> is tangent to circle <math>A</math>, <math>\triangle{ACO}</math> is a right triangle. We know <math>AO=2\sqrt{2}</math> and <math>AC=2</math>, so <math>\triangle{ACO}</math> is isosceles, a <math>45</math>-<math>45</math> right triangle, and has <math>\overline{CO}</math> with length <math>2</math>. The area of <math>\triangle{ACO}=\frac{1}{2}bh=2</math>. By symmetry, <math>\triangle{ACO}\cong\triangle{BDO}</math>, and so the area of <math>\triangle{BDO}</math> is also <math>2</math>. | ||
− | <math>\overarc{AEC}</math> (or <math>\overarc{BFD}</math>, for that matter) is <math>\frac{1}{8}</math> the area of its circle. Thus <math>\overarc{AEC}</math> and <math>\overarc{BFD}</math> both have an area of <math>\frac{\pi}{2}</math>. | + | <math>\overarc{AEC}</math> (or <math>\overarc{BFD}</math>, for that matter) is <math>\frac{1}{8}</math> the area of its circle since <math>\angle{OAC}</math> is 45 degree and <math>\angle{OAE}</math> forms a right [[triangle]]. Thus <math>\overarc{AEC}</math> and <math>\overarc{BFD}</math> both have an area of <math>\frac{\pi}{2}</math>. |
Plugging all of these areas back into the original equation yields <math>8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}</math>. | Plugging all of these areas back into the original equation yields <math>8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}</math>. |
Revision as of 19:50, 1 October 2020
Problem
Circles centered at and
each have radius
, as shown. Point
is the midpoint of
, and
. Segments
and
are tangent to the circles centered at
and
, respectively, and
is a common tangent. What is the area of the shaded region
?
Solution
The area we are trying to find is simply .
Obviously,
. Thus,
is a rectangle, and so its area is
.
Since is tangent to circle
,
is a right triangle. We know
and
, so
is isosceles, a
-
right triangle, and has
with length
. The area of
. By symmetry,
, and so the area of
is also
.
(or
, for that matter) is
the area of its circle since
is 45 degree and
forms a right triangle. Thus
and
both have an area of
.
Plugging all of these areas back into the original equation yields .
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.