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

Revision as of 02:30, 16 February 2023

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)

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

@@ Line 35: / Line 35: @@
 ==Solution 2==
-Pretend each circle is a square. The second largest circle is a square with area <math>16~\text{units}^2</math> and there are two squares in that square that each have area <math>4~\text{units}^2</math> which add up to <math>8~\text{units}^2</math>. Subtracting the medium-sized squares' areas from the second-largest square's area, we have <math>8~\text{units}^2</math>. The largest circle becomes a square that has area <math>36~\text{units}^2</math>, and the three smallest circles become three squares with area <math>8~\text{units}^2</math> and add up to <math>3~\text{units}^2</math>. Adding the areas of the shaded regions we get <math>11~\text{units}^2</math>, so our answer is <math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}</math>.
+Pretend each circle is a square. The second largest circle is a square with area <math>16~\text{units}^2</math> and there are two squares in that square that each has areas of <math>4~\text{units}^2</math> which add up to <math>8~\text{units}^2</math>. Subtracting the medium-sized squares' areas from the second-largest square's area, we have <math>8~\text{units}^2</math>. The largest circle becomes a square that has area <math>36~\text{units}^2</math>, and the three smallest circles become three squares with area <math>8~\text{units}^2</math> and add up to <math>3~\text{units}^2</math>. Adding the areas of the shaded regions, we get <math>11~\text{units}^2</math>, so our answer is <math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}</math>.
 -claregu
-LaTeX edits -apex304
+LaTeX (edits -apex304)
 ==Video Solution (Animated)==

2023 AMC 8 (Problems • Answer Key • Resources)
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