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Difference between revisions of "2025 AMC 8 Problems/Problem 12"

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~Soupboy0
 
~Soupboy0
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==Solution 2==
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The largest circle that can fit in the figure is the inscribed circle of the <math>4</math> by <math>4</math> square, which has radius <math>2</math>. The answer is <math>\boxed{\text{(B) }4\pi}</math> ~Tacos_are_yummy_1
  
 
==Vide Solution 1 by SpreadTheMathLove==
 
==Vide Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI

Revision as of 10:39, 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

Solution 2

The largest circle that can fit in the figure is the inscribed circle of the $4$ by $4$ square, which has radius $2$. The answer is $\boxed{\text{(B) }4\pi}$ ~Tacos_are_yummy_1

Vide Solution 1 by SpreadTheMathLove

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

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