Difference between revisions of "2025 AMC 8 Problems/Problem 12"
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| + | == Problem == | ||
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The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries? | The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries? | ||
<asy> | <asy> | ||
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| − | |||
size(100); | size(100); | ||
| − | |||
| − | |||
void drawSquare(pair p) { | void drawSquare(pair p) { | ||
| − | draw(box(p, p + (1,1)), | + | draw(box(p, p + (1,1)), black); |
} | } | ||
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<math>\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi</math> | <math>\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi</math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | |||
| + | The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in <math>8</math> points. By the Pythagorean Theorem, the distance from the center to one of these <math>8</math> points is <math>\sqrt{2^2 + 1^2} = \sqrt5</math>, so the area of this circle is <math>\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}</math>. | ||
| + | |||
| + | ~Soupboy0 | ||
| + | |||
| + | == Solution 2 == | ||
| + | Draw the circle in the grid and analyze the radius. It's radius is a little more than 2 but a lot less than 2.5, so the area is a little more than 4π. So, the area of the circle is <math>\boxed{\textbf{(C)} 5\pi}</math> with a radius of approximately 2.23. | ||
| + | -- leafy | ||
| + | |||
| + | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
| + | https://youtu.be/VP7g-s8akMY?si=Oe8Ka0kEZiLNXA5g&t=1096 | ||
| + | ~hsnacademy | ||
| + | ==Video Solution(Quick, fast, easy!)== | ||
| + | https://youtu.be/fdG7EDW_7xk | ||
| + | |||
| + | ~MC | ||
| + | |||
| + | == Video Solution 1 by SpreadTheMathLove == | ||
| + | |||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | == Video Solution 2 by Thinking Feet == | ||
| + | |||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC8 box|year=2025|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 19:46, 26 May 2025
Contents
Problem
The region shown below consists of 24 squares, each with side length 1 centimeter. What is the area, in square centimeters, of the largest circle that can fit inside the region, possibly touching the boundaries?
![[asy] size(100); void drawSquare(pair p) { draw(box(p, p + (1,1)), black); } int[][] grid = { {0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0}, {0, 1, 1, 1, 1, 0}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {0, 1, 1, 1, 1, 0}, {0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0} }; int rows = grid.length; int cols = grid[0].length; for (int i = 0; i < rows; ++i) { for (int j = 0; j < cols; ++j) { if (grid[i][j] == 1) { drawSquare((j, rows - i - 1)); } } } [/asy]](http://latex.artofproblemsolving.com/a/5/2/a52017dd3b0fc3a1b2e79871e510b0ca50e1e4db.png)
Solution 1
The largest circle that can fit inside the figure has its center in the middle of the figure and will be tangent to the figure in 
points. By the Pythagorean Theorem, the distance from the center to one of these points is 
, so the area of this circle is 
.
~Soupboy0
Solution 2
Draw the circle in the grid and analyze the radius. It's radius is a little more than 2 but a lot less than 2.5, so the area is a little more than 4π. So, the area of the circle is 
with a radius of approximately 2.23.
-- leafy
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=Oe8Ka0kEZiLNXA5g&t=1096 ~hsnacademy
Video Solution(Quick, fast, easy!)
~MC
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 2 by Thinking Feet
See Also
| 2025 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
