Difference between revisions of "2011 AMC 12A Problems/Problem 11"
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\textbf{(C)}\ 2 \qquad | \textbf{(C)}\ 2 \qquad | ||
\textbf{(D)}\ \frac{3\pi}{4} \qquad | \textbf{(D)}\ \frac{3\pi}{4} \qquad | ||
− | \textbf{(E)}\ 1+\frac{\pi}{2 | + | \textbf{(E)}\ 1+\frac{\pi}{2} </math> |
− | == Solution == | + | ==Solution 1== |
<asy> | <asy> | ||
Line 44: | Line 44: | ||
Let <math>M</math> be the midpoint of <math>\overline{AB}</math> and <math>D</math> be the other intersection of circles <math>C</math> and <math>B</math>. | Let <math>M</math> be the midpoint of <math>\overline{AB}</math> and <math>D</math> be the other intersection of circles <math>C</math> and <math>B</math>. | ||
− | + | The area shared between <math>C</math>, <math>A</math> and <math>B</math> is <math>4</math> of the regions between arc <math>\widehat {MD}</math> and line <math>\overline{MD}</math>, which is (considering the arc on circle <math>B</math>) a quarter of the circle <math>B</math> minus <math>\triangle MDB</math>: | |
<math>\frac{\pi r^2}{4}-\frac{bh}{2}</math> | <math>\frac{\pi r^2}{4}-\frac{bh}{2}</math> | ||
Line 50: | Line 50: | ||
<math>b = h = r = 1</math> | <math>b = h = r = 1</math> | ||
− | (We can assume this because <math>\angle DBM</math> is 90 degrees, since <math>CDBM</math> is a square, due the application of the tangent chord theorem at point <math>M</math>) | + | (We can assume this because <math>\angle DBM</math> is 90 degrees, since <math>CDBM</math> is a square, due to the application of the tangent chord theorem at point <math>M</math>) |
So the area of the small region is | So the area of the small region is | ||
Line 58: | Line 58: | ||
The requested area is area of circle <math>C</math> minus 4 of this area: | The requested area is area of circle <math>C</math> minus 4 of this area: | ||
− | <math>\pi 1^2 - 4(\frac{\pi}{4}-\frac{1}{2}) | + | <math>\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) |
= \pi - \pi + 2 | = \pi - \pi + 2 | ||
= 2</math> | = 2</math> | ||
<math>\boxed{\textbf{C}}</math>. | <math>\boxed{\textbf{C}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | <asy> | ||
+ | unitsize(1.1cm); | ||
+ | defaultpen(linewidth(.8pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(0,0), B=(2,0), C=(1,1); | ||
+ | pair D=(2,1); | ||
+ | pair E=(0,1); | ||
+ | pair F = (1, 2); | ||
+ | pair M = (1, 0); | ||
+ | dot (A); | ||
+ | dot (B); | ||
+ | dot (C); | ||
+ | dot (D); | ||
+ | dot (E); | ||
+ | dot (F); | ||
+ | dot (M); | ||
+ | |||
+ | draw(Circle(A,1)); | ||
+ | draw(Circle(B,1)); | ||
+ | draw(Circle(C,1)); | ||
+ | |||
+ | draw (D--F--E--M--D); | ||
+ | |||
+ | label("$A$",A,W); | ||
+ | label("$B$",B,E); | ||
+ | label("$C$",C,W); | ||
+ | label("$M$",M,NE); | ||
+ | label("$D$",D,E); | ||
+ | label("$E$",E,W); | ||
+ | label("$F$",F,N); | ||
+ | </asy> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=u23iWcqbJlE | ||
+ | ~Shreyas S | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=olRZuK11mAI | ||
== 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}} | ||
+ | {{AMC10 box|year=2011|num-b=17|num-a=19|ab=A}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:56, 10 August 2023
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
Solution 1
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
.
The 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 to 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
.
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=olRZuK11mAI
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 |
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.