2021 AMC 10B Problem 23

The coin has radius $\frac12,$ so its center is uniformly distributed over a $7\times7$ square of area $49.$ A shaded corner triangle contributes the set of center positions within distance $\frac12$ of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is $\frac{1+\sqrt2}{2},$ so its area is $$\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.$$ All four corners contribute $3+2\sqrt2.$ The center shaded diamond is a square of side $2\sqrt2.$ Expanding it by distance $\frac12$ adds four rectangles of total area $4\sqrt2$ and four quarter-circles of total area $\frac\pi4,$ in addition to the diamond’s area $8.$ Thus the center contribution is $$8+4\sqrt2+\frac\pi4.$$ The favorable area is $$\begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}$$ The probability is $$\begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}$$ So $a+b=44+24=68.$ Thus, the answer is C .