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2021 AMC 10B Problem 23

2021 AMC 10B Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Geometric Probability · difficulty 4 of 5 · about 8 min · position P21-25

A square with side length $8$ is colored white except for $4$ black isosceles right triangular regions with legs of length $2$ in each corner of the square and a black 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 black 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 $a+b$ ?

A)
$64$
B)
$66$
C)
$68$
D)
$70$
E)
$72$

Answer

Key insight

The center lives in a 7×7 square; it hits black iff within 1/2 of a black shape, so expand the diamond by 1/2 and clip each corner triangle.

Solution

The coin has radius $\frac12$ , so its center is uniformly distributed over the $7 \times 7$ square of points at least $\frac12$ from every side (area $49$ ). The coin touches black exactly when its center is within $\frac12$ of a black region. We find the area of that "bad" set.

Diamond. The diamond is a square with diagonals $4$ and $4$ , area $8$ , perimeter $4 \cdot 2\sqrt{2} = 8\sqrt{2}$ . Points within $\frac12$ of a convex polygon form the polygon, plus a strip of width $\frac12$ along each side, plus quarter-discs at the vertices that together make one full disc of radius $\frac12$ :
$8 + \tfrac12 \cdot 8\sqrt{2} + \pi\left(\tfrac12\right)^2 = 8 + 4\sqrt{2} + \tfrac{\pi}{4}.$

Corner triangles. Take the corner at the origin: the triangle with vertices $(0,0), (2,0), (0,2)$ . Allowed centers have $x, y \ge \frac12$ , so they are already at least $\frac12$ from the two legs; only the hypotenuse $x + y = 2$ matters. A center is within $\frac12$ of it when $x + y \le 2 + \frac{\sqrt2}{2}$ . The bad set is the right isosceles triangle with vertex $(\frac12, \frac12)$ and legs $2 + \frac{\sqrt2}{2} - 1 = 1 + \frac{\sqrt2}{2}$ :
$\tfrac12\left(1 + \tfrac{\sqrt2}{2}\right)^2 = \tfrac12\left(\tfrac32 + \sqrt2\right) = \tfrac34 + \tfrac{\sqrt2}{2}.$
Four corners: $3 + 2\sqrt2$ .

Total. Bad area $= 11 + 6\sqrt2 + \frac{\pi}{4} = \frac{44 + 24\sqrt2 + \pi}{4}$ , so the probability is
$\frac{44 + 24\sqrt2 + \pi}{4 \cdot 49} = \frac{1}{196}\left(44 + 24\sqrt2 + \pi\right).$
Hence $a + b = 44 + 24 = 68$ .

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

Why this works

Replace the coin by its center: the sample space is a shrunken square, and "coin touches region" becomes "center lies in the region grown by the radius." Growing a convex polygon adds strips along edges and a full disc's worth of corner sectors, which is where the $\pi$ comes from. The corner triangles are different only because part of their grown region is cut off by where the center is allowed to be; always intersect with the sample space.

The trap

Treating the corner triangles like the diamond, adding quarter-circles or strips along their legs, which lie outside the region the center can occupy.

Common mistakes

Treating the corner triangles like the diamond, adding quarter-circles or strips along their legs, which lie outside the region the center can occupy.
Dividing by $64$ (the full square) instead of $49$ , or using diameter $1$ as the growth distance instead of the radius $\frac12$ .

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)