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Difference between revisions of "2025 AMC 8 Problems/Problem 12"
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==Video Solution by Thinking Feet==
 
==Video Solution by Thinking Feet==
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 19:05, 30 January 2025

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]

$\textbf{(A)}\ 3\pi\qquad \textbf{(B)}\ 4\pi\qquad \textbf{(C)}\ 5\pi\qquad \textbf{(D)}\ 6\pi\qquad \textbf{(E)}\ 8\pi$


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 $8$ points. The distance from the center to one of these $8$ points can be found with the Pythagorean Theorem: $r^2 = 2^2 + 1^2 \rightarrow r = \sqrt5$. Therefore, the area of this circle = $\pi (\sqrt{5^2}) = \boxed{\textbf{(C)} 5\pi}$.

~Soupboy0

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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

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