Difference between revisions of "2007 AMC 10A Problems/Problem 24"
(→Solution) |
m (edited tone of writing) |
||
Line 38: | Line 38: | ||
==Solution== | ==Solution== | ||
The area we are trying to find is simply <math>ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD})</math>. | The area we are trying to find is simply <math>ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD})</math>. | ||
− | + | <math>\overline{EF}\parallel\overline{AB}</math>. Thus, <math>ABFE</math> is a [[rectangle]], and so its area is <math>b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{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 an isosceles 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 an isosceles 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>. |
Revision as of 12:52, 6 August 2024
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 .
. Thus,
is a rectangle, and so its area is
.
Since is tangent to circle
,
is a right triangle. We know
and
, so
is an isosceles 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 degrees 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.