2024 AMC 10B Problem 14

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 $\boxed{\textbf{(B)}\ 71}$.

Geometric probability is a ratio of areas, so the task is to recognize each region: an absolute-value inequality gives a diamond, and a squared inequality $(u - 25)^2 \le 49$ unpacks to a band $18 \le u \le 32$, which for $u = x^2 + y^2$ is an annulus. The one thing that can go wrong is containment, so always compare the target's extent with the board's inradius.

• Forgetting to check that the outer circle of radius sqrt32 lies inside the board; here it is exactly inscribed, so no part of the ring is cut off.
• Using $8^2 = 64$ for the area of the board (as if it were an axis-aligned square of side $8$), which gives $\frac{7}{32}\pi$ and $m + n = 39$.