Difference between revisions of "2023 AMC 8 Problems/Problem 12"

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Minor edits by ~Shriyans Chowdhury
 
Minor edits by ~Shriyans Chowdhury
  

Revision as of 10:02, 1 December 2024

Problem

The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded?

[asy] // Diagram by TheMathGuyd size(6cm); draw(circle((3,3),3)); filldraw(circle((2,3),2),lightgrey); filldraw(circle((3,3),1),white); filldraw(circle((1,3),1),white); filldraw(circle((5.5,3),0.5),lightgrey); filldraw(circle((4.5,4.5),0.5),lightgrey); filldraw(circle((4.5,1.5),0.5),lightgrey); int i, j; for(i=0; i<7; i=i+1) { draw((0,i)--(6,i), dashed+grey); draw((i,0)--(i,6), dashed+grey); } [/asy]


$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{11}{36} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{19}{36} \qquad \textbf{(E)}\ \frac{5}{9}$

Solution 1

First, the total area of the radius $3$ circle is simply just $9\cdot \pi$ when using our area of a circle formula.

Now from here, we have to find our shaded area. This can be done by adding the areas of the $\frac{1}{2}$-radius circles and add; then, take the area of the $1$ radius circles and subtract that from the area of the $2$ radius circle to get our resulting complex shape area. Adding these up, we will get $3\cdot \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11\cdot \pi}{4}$.

So, our answer is $\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}$.

~apex304 Minor edits by ~NXC

Minor edits by ~Shriyans Chowdhury

Solution 2

Pretend each circle is a square. The large shaded circle is a square with area $16~\text{units}^2$, and the two white circles inside it each have areas of $4~\text{units}^2$, which adds up to $8~\text{units}^2$. The three small shaded circles become three squares with area $1~\text{units}^2$, and add up to $3~\text{units}^2$. Adding the areas of the shaded circles (19) and subtracting the areas of the white circles (8), we get $11~\text{units}^2$. Since the largest white circle in which all these other circles are becomes a square that has area $36~\text{units}^2$, our answer is $\boxed{\textbf{(B)}\ \dfrac{11}{36}}$.

-claregu LaTeX (edits -apex304, CoOlPoTaToEs)

Solution 3

after eyeballing it, it appears that 1/3 of the area is shaded. however, since it is #12 on amc8, it cannot be that simple, so choose (B) since its closest

Video Solution by Math-X (How to do this question under 30 seconds)

https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X

Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=539

~hsnacademy

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/5wpEBWZjl6o

~Education the Study of everything


Video Solution (Animated)

https://youtu.be/5RRo6pQqaUI

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4590

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=UWoUhV5T92Y

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=1137

Video Solution by harungurcan

https://www.youtube.com/watch?v=oIGy79w1H8o&t=1154s

~harungurcan

Video Solution by Dr. David

https://youtu.be/2Ih7F0XHmls

Video Solution by WhyMath

https://youtu.be/ZOi0faHzBR4

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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

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