2021 AMC 10 B Problem 23

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To help you master 2021 AMC 10 B Problem 23, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 B Problem 23

Solution

Solution 1

Geometric Expansion (Minkowski Sum) Method

Find the sample space for the coin’s center, then calculate the area of the expanded black regions (buffer zones of radius $1/2$ ) restricted to this sample space.

The area of a convex polygon expanded by a distance $r$ (its Minkowski sum with a disk of radius $r$ ) is given by $Area + Perimeter \cdot r + π r^{2}$ .

Step 1: Determine the sample space

Let the center of the square be $(0, 0)$ . The square has side length $8$ , so it covers $[- 4, 4] \times [- 4, 4]$ . The coin has diameter $1$ , so its radius is $r = 1/2$ . For the coin to be completely contained within the square, its center must be at least $1/2$ unit away from all boundaries. Thus, the center must lie in the square $[- 3.5, 3.5] \times [- 3.5, 3.5]$ . Let this region be our sample space $S$ .

Area (S) = (3.5 - (- 3.5))^{2} = 7^{2} = 49

Step 2: Calculate the expanded area of the central diamond

The black diamond $D$ in the center is a square with side length $22$ . Its area is $(22)^{2} = 8$ , and its perimeter is $4 \times 22 = 82$ . The coin intersects the diamond if its center is within $1/2$ unit of $D$ . Because $D$ is convex, the area of this expanded region (the Minkowski sum) is the original area, plus the perimeter multiplied by $r$ , plus a full circle of radius $r$ formed by the four corner sectors. The furthest point of this expanded diamond is at distance $2 + 1/2 = 2.5$ from the center, which is well within the boundary of $S$ (max coordinate $3.5$ ).

Area (D^{'}) = Area (D) + Perimeter (D) \cdot r + π r^{2}

Area (D^{'}) = 8 + 82 (\frac{1}{2}) + π (\frac{1}{2})^{2} = 8 + 42 + \frac{π}{4}

Step 3: Analyze the expanded corner triangles

Consider the top-right corner. The black triangle $T$ has legs of length $2$ , with vertices $(2, 4)$ , $(4, 4)$ , and $(4, 2)$ . Its hypotenuse lies on the line $x + y = 6$ . The locus of points within distance $1/2$ of $T$ expands its boundary. The shifted hypotenuse moves $1/2$ unit closer to the origin. The distance from the origin to $x + y = 6$ is $6/ 2 = 32$ . The shifted line has a distance of $32 - 1/2$ from the origin.

x + y = 2 (32 - \frac{1}{2}) = 6 - \frac{2}{2}

Step 4: Calculate the area of the valid corner regions

We must restrict the expanded corner to our sample space $S$ , which has its top-right corner at $(3.5, 3.5)$ . The expanded circular sectors around the triangle’s vertices fall outside $S$ . The only part inside $S$ is the isosceles right triangle cut off by the shifted hypotenuse. The sum of coordinates at the corner of $S$ is $3.5 + 3.5 = 7$ . The altitude $h$ from $(3.5, 3.5)$ to the line $x + y = 6 - \frac{2}{2}$ is the difference in their distances from the origin along the diagonal. The area of a right isosceles triangle with altitude $h$ to its hypotenuse is $h^{2}$ . There are $4$ such corners.

h = \frac{7 - ( 6 - \frac{2}{2} )}{2} = \frac{1 + \frac{2}{2}}{2} = \frac{1}{2} + \frac{1}{2}

Area of one corner = h^{2} = (\frac{1}{2} + \frac{1}{2})^{2} = \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = \frac{3}{4} + \frac{2}{2}

Area of 4 corners = 4 (\frac{3}{4} + \frac{2}{2}) = 3 + 22

Step 5: Combine areas and calculate the probability

The expanded diamond and the expanded corners are widely separated and do not overlap. The total favorable area is the sum of their valid areas. We divide this by the total area of the sample space $S$ ( $49$ ) to find the probability $P$ .

Total Favorable Area = (8 + 42 + \frac{π}{4}) + (3 + 22) = 11 + 62 + \frac{π}{4}

P = \frac{11 + 6 2 + \frac{π}{4}}{49} = \frac{44 + 24 2 + π}{196}

Step 6: Extract the requested constants

The problem states the probability can be written as $\frac{1}{196} (a + b 2 + π)$ . Matching our expression to this format, we identify $a$ and $b$ , and find their sum.

a = 44

b = 24

a + b = 44 + 24 = 68

Correct answer: C. $68$
The sum

a + b

is 68, which corresponds to choice C. A quick check of the format confirms our derivation: the fact that

π

has a coefficient of

1

inside the parenthesis means the total favorable area must contain exactly

\frac{π}{4}

before dividing by

49

. This perfectly matches our calculation, proving that the circular sectors at the square’s corners must lie entirely outside the sample space.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Minkowski Sum of a Convex Polygon:

Area (P \oplus B_{r}) = Area (P) + Perimeter (P) \cdot r + π r^{2}

$P$ is a convex polygon, $B_{r}$ is a disk of radius $r$ , and $\oplus$ denotes the set of all points within distance $r$ from $P$ . The area where the coin can be placed to intersect the central diamond is exactly the region of points within distance $r = 1/2$ from the diamond.

Distance from a Point to a Line:

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

$d$ is the perpendicular distance from point $(x_{0}, y_{0})$ to the line $A x + B y + C = 0$ . Used to find the altitude from the corner of the sample space $(3.5, 3.5)$ to the shifted boundary line of the corner triangle.

Area of a Right Isosceles Triangle via Altitude:

Area = h^{2}

$h$ is the altitude drawn to the hypotenuse. The region formed by the shifted hypotenuse at the corner of the sample space is a right isosceles triangle. Finding its area using its altitude $h$ avoids calculating the lengths of the legs.

Key Takeaway: Remember This Strategy for Next Time

A circular object is dropped into a bounded region and must be 'completely contained'.

Try: First, shrink the entire bounding region inward by the radius of the circle to create the sample space for the center of the circle.

Why: The boundary of the circle hits the outer wall exactly when its center is one radius away. The center cannot cross this inner boundary.

A problem asks for the probability that a random circle overlaps with a specific shape.

Try: Expand the target shape by the radius of the circle (Minkowski Sum) to find the 'target zone' for the circle's center. For convex polygons, the expanded area is $Area + Perimeter \cdot r + π r^{2}$ .

Why: The shortest distance between the circle and the shape is $\leq r$ if and only if the center of the circle is within distance $r$ from the shape.

A line $x + y = c$ is shifted inward or outward.

Try: Recognize that the distance from the origin to $x + y = c$ is $c / 2$ . Adjust this distance by the shift amount, then multiply back by $2$ to find the new constant $c^{'}$ .

Why: Lines of the form $x + y = c$ are perpendicular to the vector $(1, 1)$ , making their perpendicular distance to the origin purely a function of $c$ divided by $2$ .

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Categories: Math CompetitionsTags: AMC 10, AMC 10 2021, AMC 10 B 2021

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2021 AMC 10 B Problem 23

In This Page:

2021 AMC10 B Problem 23