Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 10A Problems/Problem 18"

(→‎Solution 2)
(→‎Solution 2)
Line 27: Line 27:
 
draw((0,0)--(0,1));
 
draw((0,0)--(0,1));
 
fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5));
 
fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5));
−
 
+
label("A",(0,0),SW);
−
 
+
label("B",(2,0),E);
 +
label("C",(1,1),N);
 
</asy>
 
</asy>
  

Revision as of 11:13, 29 December 2019

Problem 18

Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ? [asy] pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); [/asy]

$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\pi}{2}$

Solution

Not specific: Draw a rectangle with vertices at the centers of $A$ and $B$ and the intersection of $A, C$ and $B, C$. Then, we can compute the shaded area as the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$. This is $\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}$.

Solution 2

[asy] unitsize(12mm); defaultpen(linewidth(.6pt)); pair A=(0,0); pair B=(2,0); pair C=(1,1); draw(Circle((0,0),1)); draw(Circle((2,0),1)); draw(Circle((1,1),1)); dot((0,0)); dot((2,0)); dot((1,1)); draw((0,0)--(0,1)); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); label("A",(0,0),SW); label("B",(2,0),E); label("C",(1,1),N); [/asy]

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_18&oldid=113692"