Difference between revisions of "2025 AMC 8 Problems/Problem 12"
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| − | ==Problem== | + | == 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); | ||
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| − | |||
void drawSquare(pair p) { | void drawSquare(pair p) { | ||
| − | draw(box(p, p + (1,1)), | + | draw(box(p, p + (1,1)), black); |
} | } | ||
| Line 38: | Line 35: | ||
<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 | |
| − | + | == Video Solution 1 by SpreadTheMathLove == | |
| − | |||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | ==Video Solution by Thinking Feet== | + | == Video Solution by Thinking Feet == |
| + | |||
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==See Also== | + | == See Also == |
| + | |||
{{AMC8 box|year=2025|num-b=11|num-a=13}} | {{AMC8 box|year=2025|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 16:34, 31 January 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
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
