Difference between revisions of "2007 AMC 10A Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | + | [[Circle]]s centered at <math>A</math> and <math>B</math> each have radius <math>2</math>, as shown. Point <math>O</math> is the [[midpoint]] of <math>\overline{AB}</math>, and <math>OA = 2\sqrt {2}</math>. Segments <math>OC</math> and <math>OD</math> are tangent to the circles centered at <math>A</math> and <math>B</math>, respectively, and <math>EF</math> is a common [[tangent (geometry)|tangent]]. What is the area of the shaded region <math>ECODF</math>? | |
− | + | <asy> | |
+ | size(5cm); | ||
+ | pair A=(-2*sqrt(2),0), B = (2*sqrt(2),0), C = A+2*dir(45), D = B+2*dir(135), E = A+2*dir(90), F = B+2*dir(90); | ||
+ | |||
+ | fill((0,0)--C--E--F--D--cycle,gray(0.6)); | ||
+ | unfill(circle(A,2)); | ||
+ | unfill(circle(B,2)); | ||
+ | draw(circle(A,2)); | ||
+ | draw(circle(B,2)); | ||
+ | draw(E--F); | ||
+ | draw(C--(0,0)--D); | ||
+ | draw(A+2*dir(300)--A--B--B+2*dir(240)); | ||
+ | |||
+ | dot((0,0)); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | dot(F); | ||
+ | |||
+ | label("$A$",A,dir(180)); | ||
+ | label("$B$",B,dir(0)); | ||
+ | label("$C$",C,dir(240)); | ||
+ | label("$D$",D,dir(300)); | ||
+ | label("$E$",E,dir(90)); | ||
+ | label("$F$",F,dir(90)); | ||
+ | label("$O$",(0,0),dir(270)); | ||
+ | label("$2$",A+dir(300),dir(210)); | ||
+ | label("$2$",B+dir(240),dir(330)); | ||
+ | </asy> | ||
<math>\text{(A)}\ \frac {8\sqrt {2}}{3} \qquad \text{(B)}\ 8\sqrt {2} - 4 - \pi \qquad \text{(C)}\ 4\sqrt {2} \qquad \text{(D)}\ 4\sqrt {2} + \frac {\pi}{8} \qquad \text{(E)}\ 8\sqrt {2} - 2 - \frac {\pi}{2}</math> | <math>\text{(A)}\ \frac {8\sqrt {2}}{3} \qquad \text{(B)}\ 8\sqrt {2} - 4 - \pi \qquad \text{(C)}\ 4\sqrt {2} \qquad \text{(D)}\ 4\sqrt {2} + \frac {\pi}{8} \qquad \text{(E)}\ 8\sqrt {2} - 2 - \frac {\pi}{2}</math> | ||
==Solution== | ==Solution== | ||
− | {{ | + | The area we are trying to find is simply <math>ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD}).\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>. | ||
+ | |||
+ | <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 degrees 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>. | |
==See also== | ==See also== | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:53, 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.