Difference between revisions of "2025 AMC 8 Problems/Problem 12"
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| + | ==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? | 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? | ||
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==Solution== | ==Solution== | ||
| − | 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. The distance from the center to one of these <math>8</math> points can be found with the Pythagorean Theorem: <math>r^2 = 2^2 + 1^2 \rightarrow r = \sqrt5</math>. Therefore, the area of this circle = <math>\pi( | + | 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. The distance from the center to one of these <math>8</math> points can be found with the Pythagorean Theorem: <math>r^2 = 2^2 + 1^2 \rightarrow r = \sqrt5</math>. Therefore, the area of this circle = <math>\pi (\sqrt{5^2}) = \boxed{\textbf{(C)} 5\pi}</math>. |
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| + | ~Soupboy0 | ||
| + | |||
| + | ==Video Solution 1 by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:12, 30 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] import graph; size(100); pen gridPen = black; void drawSquare(pair p) { draw(box(p, p + (1,1)), gridPen); } 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/2/2/b/22b12d5f528f4fc1b25a685e25d32476c453ec95.png)
Solution
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. The distance from the center to one of these points can be found with the Pythagorean Theorem: 
. Therefore, the area of this circle = 
.
~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. 
