2024 AMC 10B Problem 14

Probability here means (area of target) divided by (area of board). So find each area.

The board.

1. Points with $|x| + |y| \le 8$: on the axes this reaches $(8, 0)$, $(0, 8)$, $(-8, 0)$, $(0, -8)$. In the first quarter the edge is the line $x + y = 8$, and the other quarters are mirror images. So the board is a square tilted on its corner.
2. Its two diagonals each have length $16$. A square's area is half the product of its diagonals: $\frac12 \cdot 16 \cdot 16 = 128$.

The target.

3. $(x^2 + y^2 - 25)^2 \le 49$ means the number $x^2 + y^2 - 25$ is between $-7$ and $7$. Add $25$:
$$18 \le x^2 + y^2 \le 32.$$
4. $x^2 + y^2$ is the squared distance from the center $(0,0)$. So the target is the ring between the circle of radius $\sqrt{18}$ and the circle of radius $\sqrt{32}$.
5. Area of the ring $=$ big circle minus small circle $= 32\pi - 18\pi = 14\pi$.

Does the ring fit on the board?

1. The closest point of the board's edge to the center is the middle of an edge. For the edge from $(8,0)$ to $(0,8)$ that is $(4, 4)$, at distance $\sqrt{16 + 16} = \sqrt{32}$.
2. The outer circle has radius $\sqrt{32}$ as well, so it just touches the edges and the whole ring is on the board.

The probability.

1. $\dfrac{14\pi}{128} = \dfrac{7}{64}\pi$. Since $7$ and $64$ share no factor, $m = 7$, $n = 64$, and $m + n = 71$.

The answer is $\boxed{\textbf{(B)}\ 71}$.

A "random point" probability is a ratio of areas, so the real job is to recognize each shape. An equation with $|x| + |y|$ makes a tilted square, and one with $x^2 + y^2$ makes circles. Always check that the target region fits inside the board. If part of it hung over the edge, you would have to cut that part off.

The board $|x| + |y| \le 8$ is a square (a diamond) with vertices $(\pm 8, 0)$ and $(0, \pm 8)$. Its diagonals both have length $16$, so its area is $\frac12 \cdot 16 \cdot 16 = 128$.

The target condition $(x^2 + y^2 - 25)^2 \le 49$ means $-7 \le x^2 + y^2 - 25 \le 7$, i.e.
$$18 \le x^2 + y^2 \le 32.$$
This is the ring between circles of radii $\sqrt{18}$ and $\sqrt{32}$ centered at the origin, with area $\pi(32 - 18) = 14\pi$.

Does the ring lie inside the board? The nearest points of the board's boundary to the origin are on the lines $\pm x \pm y = 8$, at distance $\frac{8}{\sqrt2} = \sqrt{32}$. So the outer circle is exactly inscribed in the diamond and the whole ring is inside.

The probability is $\frac{14\pi}{128} = \frac{7}{64}\pi$, so $m + n = 7 + 64 = 71$.

The answer is $\boxed{\textbf{(B)}\ 71}$.