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

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Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2<math>, we have that each of the smaller shaded circles have area 0.25/pi~\text{units}^2</math>, for a total area of 0.75/pi~\text{units}^2<math>. The bigger shaded circle has area 4/pi~\text{units}^2</math>, and the two medium white circles each have area /pi~\text{units}^2<math>, for a total of 2/pi~\text{units}^2</math>. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area.
 
Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2<math>, we have that each of the smaller shaded circles have area 0.25/pi~\text{units}^2</math>, for a total area of 0.75/pi~\text{units}^2<math>. The bigger shaded circle has area 4/pi~\text{units}^2</math>, and the two medium white circles each have area /pi~\text{units}^2<math>, for a total of 2/pi~\text{units}^2</math>. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area.
  
−
Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2$. Dividing /frac{2.75/pi}{9/pi}, we simplify to
+
Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2<math>. Dividing /frac{2.75/pi}{9/pi}, we simplify to </math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}$.
  
 
==Video Solution by Math-X (How to do this question under 30 seconds)==
 
==Video Solution by Math-X (How to do this question under 30 seconds)==

Revision as of 13:05, 21 January 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 $9* \pi$ 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 $3$ $\frac{1}{2}$-radius circles and add; then, take the area of the $2$ radius circle and subtract that from the area of the $2$ radius 1 circles to get our resulting complex area shape. Adding these up, we will get $3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11}{4}$.

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

~apex304

Solution 2

Pretend each circle is a square. The second largest circle is a square with area $16~\text{units}^2$ and there are two squares in that square that each has areas of $4~\text{units}^2$ which add up to $8~\text{units}^2$. Subtracting the medium-sized squares' areas from the second-largest square's area, we have $8~\text{units}^2$. The largest circle becomes a square that has area $36~\text{units}^2$, and the three smallest circles become three squares with area $8~\text{units}^2$ and add up to $3~\text{units}^2$. Adding the areas of the shaded regions, we get $11~\text{units}^2$, so our answer is $\boxed{\textbf{(B)}\ \dfrac{11}{36}}$.

-claregu LaTeX (edits -apex304)

Solution 3 (Similar to 2)

Start by finding the area of each of the smaller circles. Using the formula /pi~\text{r}^2$, we have that each of the smaller shaded circles have area 0.25/pi~\text{units}^2$, for a total area of 0.75/pi~\text{units}^2$. The bigger shaded circle has area 4/pi~\text{units}^2$, and the two medium white circles each have area /pi~\text{units}^2$, for a total of 2/pi~\text{units}^2$. Adding the shaded values (0.75/pi + 4/pi = 4.75/pi), and subtracting the white value (4.75/pi - 2/pi = 2.75/pi), we get that there is a total of 2.75/pi~\text{units}^2 shaded area.

Using the same formula, we have that the big white circle has area 9/pi~\text{units}^2$. Dividing /frac{2.75/pi}{9/pi}, we simplify to$\boxed{\textbf{(B)}\ \dfrac{11}{36}}$.

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 (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

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
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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