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

(→‎Solution 3 (Similar to 2))
(→‎Solution 1)
 
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==Solution 1==
 
==Solution 1==
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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.  
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First, the total area of the radius <math>3</math> circle is simply just <math>9\cdot \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}{4}</math>.
+
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\cdot \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11\cdot \pi}{4}</math>.
  
 
So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>.
 
So, our answer is <math>\frac {\frac{11}{4} \pi}{9 \pi} = \boxed{\textbf{(B)}\ \frac{11}{36}}</math>.
  
 
~apex304
 
~apex304
 +
Minor edits by ~NXC
  
 
==Solution 2==
 
==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 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>.
+
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>.
  
 
-claregu
 
-claregu
−
LaTeX (edits -apex304)
+
LaTeX (edits -apex304, CoOlPoTaToEs)
  
−
==Solution 3 (Similar to 2)==
+
==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
−
Start by solving for the area of each of the smaller circles. We find 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</math>. Dividing /frac{2.75 /pi}{9 /pi}, we simplify to <math>\boxed{\textbf{(B)}\ \dfrac{11}{36}}</math>.
 
  
 
==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)==
 
https://youtu.be/Ku_c1YHnLt0?si=stUHQ9nHZZE_x-CC&t=1852 ~Math-X
 
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!!!) ==
 
==Video Solution (HOW TO THINK CREATIVELY!!!) ==
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~harungurcan
 
~harungurcan
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/2Ih7F0XHmls
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/ZOi0faHzBR4
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=11|num-a=13}}
 
{{AMC8 box|year=2023|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:19, 25 November 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 $3$ $\frac{1}{2}$-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 $3\cdot \frac{1}{4} \pi + 4 \pi -\pi - \pi = \frac{3}{4} \pi + 2 \pi = \frac{11\cdot \pi}{4}$.

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

~apex304 Minor edits by ~NXC

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 (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

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

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