Difference between revisions of "2023 AMC 8 Problems/Problem 12"
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+ | ~Education the Study of everything | ||
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==Video Solution (Animated)== | ==Video Solution (Animated)== |
Revision as of 12:35, 11 June 2023
Contents
[hide]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?
Solution 1
First, the total area of the radius circle is simply just 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 -radius circles and add; then, take the area of the radius circle and subtract that from the area of the radius 1 circles to get our resulting complex area shape. Adding these up, we will get .
So, our answer is .
~apex304
Solution 2
Pretend each circle is a square. The second largest circle is a square with area and there are two squares in that square that each has areas of which add up to . Subtracting the medium-sized squares' areas from the second-largest square's area, we have . The largest circle becomes a square that has area , and the three smallest circles become three squares with area and add up to . Adding the areas of the shaded regions, we get , so our answer is .
-claregu LaTeX (edits -apex304)
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education the Study of everything
Video Solution (Animated)
~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/1bA7fD7Lg54?t=854
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.