2002 AMC 12A Problems/Problem 5

The following problem is from both the 2002 AMC 12A #5 and 2002 AMC 10A #5, so both problems redirect to this page.

Problem

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$

$\textbf{(A)}\ \pi \qquad \textbf{(B)}\ 1.5\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ 3\pi \qquad \textbf{(E)}\ 3.5\pi$

Solution

The outer circle has radius $1+1+1=3$ , and thus area $9\pi$ . The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$ . Thus, our answer is $9\pi-7\pi=\boxed{2\pi\Rightarrow \textbf{(C)}}$ .

Video Solution by Daily Dose of Math

https://youtu.be/VSPM5_pTK34?si=EIQDi3PF6I0P7wMR

~Thesmartgreekmathdude

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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.

Page

Toolbox

Search

2002 AMC 12A Problems/Problem 5

Contents

Problem

Solution

Video Solution by Daily Dose of Math

See Also