2025 AMC 8 Problems/Problem 12

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?

$[asy] size(100); void drawSquare(pair p) { draw(box(p, p + (1,1)), black); } int[][] grid = { {0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0}, {0, 1, 1, 1, 1, 0}, {1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1}, {0, 1, 1, 1, 1, 0}, {0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 0, 0} }; int rows = grid.length; int cols = grid[0].length; for (int i = 0; i < rows; ++i) { for (int j = 0; j < cols; ++j) { if (grid[i][j] == 1) { drawSquare((j, rows - i - 1)); } } } [/asy]$

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

Solution 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 $8$ points. By the Pythagorean Theorem, the distance from the center to one of these $8$ points is $\sqrt{2^2 + 1^2} = \sqrt5$ , so the area of this circle is $\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}$ .

~Soupboy0

Video Solution 1 by SpreadTheMathLove

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

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2025 AMC 8 Problems/Problem 12

Contents

Problem

Solution 1

Video Solution 1 by SpreadTheMathLove

Video Solution 2 by Thinking Feet

See Also