Difference between revisions of "2008 AMC 8 Problems/Problem 25"
Mathking999 (talk | contribs) (→Solution) |
Pi is 3.14 (talk | contribs) |
||
| (13 intermediate revisions by 6 users not shown) | |||
| Line 19: | Line 19: | ||
filldraw(circle(A5, 1), black, black); | filldraw(circle(A5, 1), black, black); | ||
</asy> | </asy> | ||
| − | <math> \textbf{(A)}\ | + | |
| + | <math> \textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad</math> | ||
==Solution== | ==Solution== | ||
| + | Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc. | ||
<cmath>\begin{array}{c|cc} | <cmath>\begin{array}{c|cc} | ||
\text{circle \#} & \text{radius} & \text{area} \\ \hline | \text{circle \#} & \text{radius} & \text{area} \\ \hline | ||
| Line 32: | Line 34: | ||
\end{array}</cmath> | \end{array}</cmath> | ||
| − | The entire circle's area is <math>144\pi</math>. The area of the black | + | The entire circle's area is <math>144\pi</math>. The area of the black regions is <math>(100-64)\pi + (36-16)\pi + 4\pi = 60\pi</math>. The percentage of the design that is black is <math>\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}</math>. |
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/j3QSD5eDpzU?t=3363 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=24|after=Last Problem}} | {{AMC8 box|year=2008|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:38, 2 January 2023
Problem
Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Which of the following is closest to the percent of the design that is black?
![[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy]](http://latex.artofproblemsolving.com/3/4/4/34420ee5e67e87663563a3047fde204ba36cff7f.png)
Solution
Let the smallest circle be 1, the second smallest circle be 2, the third smallest circle be 3, etc.
![\[\begin{array}{c|cc} \text{circle \#} & \text{radius} & \text{area} \\ \hline 1 & 2 & 4\pi \\ 2 & 4 & 16\pi \\ 3 & 6 & 36\pi \\ 4 & 8 & 64\pi \\ 5 & 10 & 100\pi \\ 6 & 12 & 144\pi \end{array}\]](http://latex.artofproblemsolving.com/8/1/4/8146393d4a39389bc89d9f904b246e939af97714.png)
The entire circle's area is 
. The area of the black regions is 
. The percentage of the design that is black is 
.
Video Solution by OmegaLearn
https://youtu.be/j3QSD5eDpzU?t=3363
~ pi_is_3.14
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
