2021 AMC 10B Problem 23

Step 1: follow the center of the coin.

1. The coin has radius $\frac12$. It stays inside the square exactly when its center is at least $\frac12$ from every side. So the center lands anywhere in a smaller $7 \times 7$ square, with area $49$. This is our "all possible outcomes" area.
2. The coin touches a black shape exactly when its center is within $\frac12$ of that shape. So we need the area of all such center points (call it the bad area), and the probability is bad area $\div\ 49$.
3. Place the square with corners $(0, 0)$ and $(8, 8)$, so the center can be anywhere with $x$ and $y$ from $\frac12$ to $\frac{15}{2}$.

Step 2: the diamond.

4. The diamond has side $2\sqrt2$, so its diagonals are $4$ and its area is $\frac12 \cdot 4 \cdot 4 = 8$. Its perimeter is $4 \times 2\sqrt2 = 8\sqrt2$.
5. The points within $\frac12$ of the diamond are made of three parts:
- the diamond itself: $8$;
- a strip of width $\frac12$ along each of the four sides: $\frac12 \times 8\sqrt2 = 4\sqrt2$;
- rounded pieces at the four corners. Together these four pieces make one whole circle of radius $\frac12$: area $\frac{\pi}{4}$.
6. Bad area near the diamond: $8 + 4\sqrt2 + \frac{\pi}{4}$.

Step 3: one corner triangle.

7. Look at the triangle at the corner $(0, 0)$, with vertices $(0,0)$, $(2,0)$, $(0,2)$. Its slanted side lies on the line $x + y = 2$.
8. The center always has $x \ge \frac12$ and $y \ge \frac12$, so it is never too close to the two short sides of the triangle. Only the slanted side matters.
9. Moving from the line $x + y = 2$ straight away from it by a distance of $\frac12$ reaches the line $x + y = 2 + \frac{\sqrt2}{2}$. (Moving by $\frac12$ in the slanted direction adds $\frac{1}{2\sqrt2}$ to each of $x$ and $y$, so $x + y$ grows by $\frac{1}{\sqrt2} = \frac{\sqrt2}{2}$.)
10. So the bad points here are those with $x, y \ge \frac12$ and $x + y \le 2 + \frac{\sqrt2}{2}$. That is a right triangle with its corner at $(\frac12, \frac12)$ and two equal legs of length $2 + \frac{\sqrt2}{2} - 1 = 1 + \frac{\sqrt2}{2}$.
11. Its area is
$$\tfrac12\left(1 + \tfrac{\sqrt2}{2}\right)^2 = \tfrac12\left(1 + \sqrt2 + \tfrac12\right) = \tfrac34 + \tfrac{\sqrt2}{2}.$$
12. Four corners: $4 \times \left(\frac34 + \frac{\sqrt2}{2}\right) = 3 + 2\sqrt2$.

Step 4: put it together.

13. These regions do not overlap (the diamond's region stays near the middle). Total bad area:
$$8 + 4\sqrt2 + \tfrac{\pi}{4} + 3 + 2\sqrt2 = 11 + 6\sqrt2 + \tfrac{\pi}{4} = \frac{44 + 24\sqrt2 + \pi}{4}.$$
14. Divide by $49$:
$$\frac{44 + 24\sqrt2 + \pi}{196} = \frac{1}{196}\left(44 + 24\sqrt2 + \pi\right).$$
15. So $a = 44$, $b = 24$, and $a + b = 68$.

The answer is $\boxed{\textbf{(C)}\ 68}$.

A coin is hard to track, but its center is a single point. "The coin stays inside" shrinks the square, and "the coin touches black" grows each black shape by the radius. Growing a shape adds strips along its sides plus rounded corners that make one full circle, which is where the $\pi$ comes from. The corner triangles are different because the center cannot get close to the square's edges, so only their slanted sides count.