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

Revision as of 10:22, 2 February 2025

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

Solution 2 (very unrecommended)

Draw a circle in the grid. It's radius is a little more than 2 but a lot less than 2.5, so the area is a little more than 4π. So, the area of the circle is $\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}$ with a radius of approximately 2.23.

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=Oe8Ka0kEZiLNXA5g&t=1096 ~hsnacademy

Video Solution 1 by SpreadTheMathLove

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

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

@@ Line 40: / Line 40: @@
 ~Soupboy0
+== Solution 2 (very unrecommended) ==
+Draw a circle in the grid. It's radius is a little more than 2 but a lot less than 2.5, so the area is a little more than 4π.  So, the area of the circle is <math>\pi \sqrt{5}^2 = \boxed{\textbf{(C)} 5\pi}</math> with a radius of approximately 2.23.
 ==Video Solution (A Clever Explanation You’ll Get Instantly)==

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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