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

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First the total area of the <math>3</math> radius circle is simply just <math>9* \pi</math>. Using our area of a circle formula.
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==Problem==
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The figure below shows a large white circle with a number of smaller white and shaded circles in its
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interior. What fraction of the interior of the large white circle is shaded?
  
Now from here we have to find our shaded area. This can be done by adding the areas of the <math>3</math> <math>\frac{1}{2}</math> radius circles and add then take the area of the <math>2</math> radius circle and subtracting that from the area of the <math>2</math>, 1 radius circles to get our resulting complex area shape. Adding these up we will get <math>3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11}{4}</math>
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<asy>
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// Diagram by TheMathGuyd
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size(6cm);
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draw(circle((3,3),3));
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filldraw(circle((2,3),2),lightgrey);
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filldraw(circle((3,3),1),white);
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filldraw(circle((1,3),1),white);
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filldraw(circle((5.5,3),0.5),lightgrey);
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filldraw(circle((4.5,4.5),0.5),lightgrey);
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filldraw(circle((4.5,1.5),0.5),lightgrey);
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int i, j;
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for(i=0; i<7; i=i+1)
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{
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draw((0,i)--(6,i), dashed+grey);
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draw((i,0)--(i,6), dashed+grey);
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}
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</asy>
  
Our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\text{(B)}\frac{11}{36}}</math>
 
  
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<math>\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}</math>
  
==Animated Video Solution==
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==Solution 1==
 +
First, the total area of the radius <math>3</math> circle is simply just <math>9* \pi</math> 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 <math>3</math> <math>\frac{1}{2}</math>-radius circles and add; then, take the area of the <math>2</math> radius circle and subtract that from the area of the <math>2</math> radius 1 circles to get our resulting complex area shape. Adding these up, we will get <math>3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11 * \pi}{4}</math>.
 +
 
 +
So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>.
 +
 
 +
~apex304
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==Solution 2==
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Pretend each circle is a square. The large shaded circle is a square with area <math>16~\text{units}^2</math>, and the two white circles inside it each have areas of <math>4~\text{units}^2</math>, which adds up to <math>8~\text{units}^2</math>. The three small shaded circles become three squares with area <math>1~\text{units}^2</math>, and add up to <math>3~\text{units}^2</math>. Adding the areas of the shaded circles (19) and subtracting the areas of the white circles (8), we get <math>11~\text{units}^2</math>. Since the largest white circle in which all these other circles are becomes a square that has area <math>36~\text{units}^2</math>, our answer is <math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}</math>.
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-claregu
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LaTeX (edits -apex304, CoOlPoTaToEs)
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==Video Solution by Math-X (How to do this question under 30 seconds)==
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https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X
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 +
 
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==Video Solution (HOW TO THINK CREATIVELY!!!) ==
 +
https://youtu.be/5wpEBWZjl6o
 +
 
 +
~Education the Study of everything
 +
 
 +
 
 +
==Video Solution (Animated)==
 
https://youtu.be/5RRo6pQqaUI
 
https://youtu.be/5RRo6pQqaUI
  
 
~Star League (https://starleague.us)
 
~Star League (https://starleague.us)
 +
 +
==Video Solution by Magic Square==
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https://youtu.be/-N46BeEKaCQ?t=4590
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 +
==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=UWoUhV5T92Y
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=1137
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 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=oIGy79w1H8o&t=1154s
 +
 +
~harungurcan
 +
 +
==See Also==
 +
{{AMC8 box|year=2023|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 16:31, 18 May 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 $3 * \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11 * \pi}{4}$.

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 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)

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png