Question

A square with side length $8$ is unshaded except for $4$ shaded isosceles right triangular regions with legs of length $2$ in each corner of the square and a shaded diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram.

A circular coin with diameter $1$ is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the shaded region of the square can be written as $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right),$ where $a$ and $b$ are positive integers. What is $a+b?$

$64$

$66$

$68$

$70$

$72$

Solution:

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

$3+2\sqrt2+8+4\sqrt2+\frac\pi4=11+6\sqrt2+\frac\pi4.$

The probability is

$\frac{11+6\sqrt2+\frac\pi4}{49}=\frac{44+24\sqrt2+\pi}{196}.$

So $a+b=44+24=68$ .

Thus, the answer is C .

2021 AMC 10B Spring Problem 23

Solution:

Problem 23 in Other Years