AMC 10 · 2021 · #23
Grade 8 geometry-2dProblem
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 22 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 1961(a+b2+π), where a and b are positive integers. What is a+b?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: An $8 \times 8$ white square has $5$ black regions: a central diamond (a square of side $2\sqrt{2}$, rotated $45^\circ$) and four isosceles right triangles of leg $2$ in the four corners. A coin of diameter $1$ is dropped uniformly at random in any position where it lies entirely inside the $8 \times 8$ square. Find the probability that the coin overlaps any black region. The probability has form $\tfrac{1}{196}(a + b\sqrt{2} + \pi)$ for positive integers $a, b$; report $a + b$.
Givens: $8 \times 8$ white square; coordinates $0 \le x, y \le 8$; $4$ corner triangles: isosceles right, leg $2$, at each corner of the big square; $1$ central diamond: square of side $2\sqrt{2}$ rotated $45^\circ$, vertices at $(4, 2), (6, 4), (4, 6), (2, 4)$; Coin radius $r = \tfrac{1}{2}$ (diameter $1$); Coin lands uniformly at random subject to being entirely inside the $8 \times 8$ square; Probability form: $\tfrac{1}{196}(a + b\sqrt{2} + \pi)$, $a, b$ positive integers; Answer choices: (A) $64$, (B) $66$, (C) $68$, (D) $70$, (E) $72$
Unknowns: $a + b$
Understand
Restated: An $8 \times 8$ white square has $5$ black regions: a central diamond (a square of side $2\sqrt{2}$, rotated $45^\circ$) and four isosceles right triangles of leg $2$ in the four corners. A coin of diameter $1$ is dropped uniformly at random in any position where it lies entirely inside the $8 \times 8$ square. Find the probability that the coin overlaps any black region. The probability has form $\tfrac{1}{196}(a + b\sqrt{2} + \pi)$ for positive integers $a, b$; report $a + b$.
Givens: $8 \times 8$ white square; coordinates $0 \le x, y \le 8$; $4$ corner triangles: isosceles right, leg $2$, at each corner of the big square; $1$ central diamond: square of side $2\sqrt{2}$ rotated $45^\circ$, vertices at $(4, 2), (6, 4), (4, 6), (2, 4)$; Coin radius $r = \tfrac{1}{2}$ (diameter $1$); Coin lands uniformly at random subject to being entirely inside the $8 \times 8$ square; Probability form: $\tfrac{1}{196}(a + b\sqrt{2} + \pi)$, $a, b$ positive integers; Answer choices: (A) $64$, (B) $66$, (C) $68$, (D) $70$, (E) $72$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #10 Create a Physical Representation, #3 Eliminate Possibilities
Tool #1 (Diagram) — draw the $8 \times 8$ square with the five black regions and shade where the coin's *center* can sit. Tool #9 (Easier) — switch from "the coin overlaps black" to "the center lies within $\tfrac{1}{2}$ of black" (geometric-probability standard move). Tool #7 (Subproblems) — split the favorable area into (a) diamond + buffer, (b) four corner triangles + buffer; the regions don't overlap. Tool #10 (Physical) — try with paper cutouts to feel why the diamond's buffer adds rectangles + a full disc, and the corner triangle's buffer is clipped to a single right triangle. Tool #3 (Eliminate) — final $a + b$ must match one of five integers.
Execute — Answer: C
7.G.B.6 Step 1 - Sample space.
- The coin (radius $\tfrac{1}{2}$) lies entirely inside the $8 \times 8$ square iff its center is in $[\tfrac{1}{2}, \tfrac{15}{2}] \times [\tfrac{1}{2}, \tfrac{15}{2}]$, a $7 \times 7$ square of area $49$.
- Sample-space area $= 49$.
💡 If the center is at least the radius from every edge of the big square, the coin is fully inside.
7.G.B.6 Step 2 - Reframe "coin overlaps black" as "center within $\tfrac{1}{2}$ of black".
- Call $F$ the favorable region (the set of allowed centers that produce an overlap with black).
- $F$ is the Minkowski sum of each black region with a disk of radius $\tfrac{1}{2}$, intersected with the $7 \times 7$ sample square.
- The diamond's buffer and the corner triangles' buffers are well separated (the diamond's nearest edge to a corner triangle is several units away), so $F$'s area is the sum of each piece's contribution.
- Compute the diamond and corner contributions separately.
💡 Coin overlaps a region exactly when its center is within radius distance of the region — translate physical contact into a single center-distance condition.
7.G.B.6 Step 3 - Diamond's contribution.
- The central diamond is a square of side $s = 2\sqrt{2}$ rotated $45^\circ$.
- Its area is $s^2 = (2\sqrt{2})^2 = 8$.
- The buffer (Minkowski with a $\tfrac{1}{2}$-disk) adds: (i) four rectangles, one along each side of the diamond, each of dimensions $2\sqrt{2} \times \tfrac{1}{2}$, total area $4 \cdot (2\sqrt{2})(\tfrac{1}{2}) = 4\sqrt{2}$; (ii) four quarter-disks at the four diamond vertices, together forming a full disk of radius $\tfrac{1}{2}$, area $\pi (\tfrac{1}{2})^2 = \tfrac{\pi}{4}$.
- So $A_{\text{diamond}} = 8 + 4\sqrt{2} + \tfrac{\pi}{4}$.
- The diamond is centered at $(4, 4)$ and extends out to distance $2$, plus another $\tfrac{1}{2}$ for the buffer, so the buffered diamond stays well inside the $7 \times 7$ sample square.
💡 Buffer = the original shape + rectangles along each edge + a full circle at the corners (the four quarter-circles glue into one).
8.G.B.8 Step 4 - Corner triangle contribution.
- Take the bottom-left black triangle with vertices $(0,0), (2,0), (0,2)$; the other three are mirror images.
- The favorable region for *this* triangle (set of centers within $\tfrac{1}{2}$ of it, lying inside the sample square $\{x, y \ge \tfrac{1}{2}\}$) is bounded by: (a) the sample-square edges $x = \tfrac{1}{2}$ and $y = \tfrac{1}{2}$, and (b) the line parallel to the hypotenuse $x + y = 2$ at distance $\tfrac{1}{2}$ above it.
- The line $x + y = 2 + \tfrac{\sqrt{2}}{2}$ has distance from $x + y = 2$ equal to $\tfrac{|2 + \tfrac{\sqrt{2}}{2} - 2|}{\sqrt{1^2 + 1^2}} = \tfrac{\tfrac{\sqrt{2}}{2}}{\sqrt{2}} = \tfrac{1}{2}$.
- ✓
💡 Sample-square edges already give two sides of the favorable region — only the hypotenuse needs a parallel shift.
7.G.B.6 Step 5 - Compute the favorable triangular region for one corner.
- Its three vertices are $(\tfrac{1}{2}, \tfrac{1}{2})$, $(\tfrac{1}{2}, \tfrac{3}{2} + \tfrac{\sqrt{2}}{2})$, and $(\tfrac{3}{2} + \tfrac{\sqrt{2}}{2}, \tfrac{1}{2})$ — solve $x + y = 2 + \tfrac{\sqrt{2}}{2}$ with $x = \tfrac{1}{2}$ to get $y = \tfrac{3}{2} + \tfrac{\sqrt{2}}{2}$.
- This is an isosceles right triangle with leg $\ell = (\tfrac{3}{2} + \tfrac{\sqrt{2}}{2}) - \tfrac{1}{2} = 1 + \tfrac{\sqrt{2}}{2}$.
- Area: $\tfrac{1}{2}\ell^2 = \tfrac{1}{2}(1 + \tfrac{\sqrt{2}}{2})^2 = \tfrac{1}{2}\bigl(1 + \sqrt{2} + \tfrac{1}{2}\bigr) = \tfrac{1}{2} \cdot \tfrac{3 + 2\sqrt{2}}{2} = \tfrac{3 + 2\sqrt{2}}{4}$.
- (Note: $(1 + \tfrac{\sqrt{2}}{2})^2 = 1 + 2 \cdot \tfrac{\sqrt{2}}{2} + \tfrac{2}{4} = \tfrac{3}{2} + \sqrt{2}$.) The four corners are symmetric, so total corner-contribution area is $4 \cdot \tfrac{3 + 2\sqrt{2}}{4} = 3 + 2\sqrt{2}$.
💡 Sample-square clipping removes the parts of the buffer that would have been outside; left over is a clean right triangle.
6.NS.B.3 Step 6 - Sum favorable area.
- $A_F = A_{\text{diamond}} + A_{\text{4 corners}} = \bigl(8 + 4\sqrt{2} + \tfrac{\pi}{4}\bigr) + (3 + 2\sqrt{2}) = 11 + 6\sqrt{2} + \tfrac{\pi}{4}$.
- Put on common denominator $4$: $A_F = \tfrac{44 + 24\sqrt{2} + \pi}{4}$.
💡 Combine the two contributions by common-denominator addition — the diamond and the four corners don't overlap, so adding is legitimate.
4.NBT.B.4 Step 7 - Compute the probability.
- $P = \dfrac{A_F}{A_{\text{sample}}} = \dfrac{(44 + 24\sqrt{2} + \pi)/4}{49} = \dfrac{44 + 24\sqrt{2} + \pi}{196} = \dfrac{1}{196}(44 + 24\sqrt{2} + \pi)$.
- Comparing to $\tfrac{1}{196}(a + b\sqrt{2} + \pi)$: $a = 44, b = 24$.
- So $a + b = 44 + 24 = 68$, matching choice (C).
💡 Match coefficients of $1, \sqrt{2}, \pi$ separately — $a$ is the rational part, $b$ is the $\sqrt{2}$ coefficient.
7.G.B.6 Sample space. The coin (radius $\tfrac{1}{2}$) lies entirely inside the $8 \time 7.G.B.6 Reframe "coin overlaps black" as "center within $\tfrac{1}{2}$ of black". Call $ 7.G.B.6 Diamond's contribution. The central diamond is a square of side $s = 2\sqrt{2}$ 8.G.B.8 Corner triangle contribution. Take the bottom-left black triangle with vertices 7.G.B.6 Compute the favorable triangular region for one corner. Its three vertices are $ 6.NS.B.3 Sum favorable area. $A_F = A_{\text{diamond}} + A_{\text{4 corners}} = \bigl(8 + 4.NBT.B.4 Compute the probability. $P = \dfrac{A_F}{A_{\text{sample}}} = \dfrac{(44 + 24\s Review
Reasonableness: Sanity. $P = \dfrac{44 + 24\sqrt{2} + \pi}{196} \approx \dfrac{44 + 24 \cdot 1.414 + 3.14}{196} \approx \dfrac{44 + 33.9 + 3.14}{196} \approx \dfrac{81.0}{196} \approx 0.413$. That's about $41\%$ — the black regions take up area $4 \cdot \tfrac{1}{2} \cdot 2 \cdot 2 + 8 = 8 + 8 = 16$ out of $64$ ($25\%$), and the buffer around each black region adds significant probability, so $41\%$ is reasonable. Coefficient match: $a + b = 68$ is one of the listed integers — and the only "middle" answer choice, neither too small ($64$, $66$) nor too large ($70$, $72$).
Alternative: Tool #10 (Physical) — for each black region, cut out a paper copy and roll a coin of diameter $1$ around its perimeter, tracing the path of the *center*; the path encloses exactly the favorable region. This visually confirms the diamond's contribution is (interior + perimeter $\times \tfrac{1}{2}$ + a full small circle) and the corner triangle's contribution is clipped to a right triangle. Alternatively, Tool #16 (Complement): compute area NOT covered (coin is fully on white) and subtract — but the buffer-around-black direct approach is cleaner.
CCSS standards used (min grade 8)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Final $a + b = 44 + 24 = 68$.)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Combining $A_F = 11 + 6\sqrt{2} + \tfrac{\pi}{4} = \tfrac{44 + 24\sqrt{2} + \pi}{4}$ over a common denominator.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Computing areas of the buffered diamond ($8 + 4\sqrt{2} + \tfrac{\pi}{4}$) and the four corner triangular regions ($3 + 2\sqrt{2}$), and translating coin-overlap into a center-distance condition (geometric probability sample-space area).)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Using the formula for distance from a point to the line $x + y = 2$ to derive the parallel line $x + y = 2 + \tfrac{\sqrt{2}}{2}$.)
⭐ This hard AMC 10 problem only needs Grade 7-8 area-and-distance you already know — once you switch from "coin overlaps black" to "coin center within $\tfrac{1}{2}$ of black", the favorable region splits into the central buffered diamond ($8 + 4\sqrt{2} + \tfrac{\pi}{4}$) and four small corner triangles ($\tfrac{3 + 2\sqrt{2}}{4}$ each), giving probability $\tfrac{44 + 24\sqrt{2} + \pi}{196}$ and $a + b = 44 + 24 = 68$.
⭐ This hard AMC 10 problem only needs Grade 7-8 area-and-distance you already know — once you switch from "coin overlaps black" to "coin center within $\tfrac{1}{2}$ of black", the favorable region splits into the central buffered diamond ($8 + 4\sqrt{2} + \tfrac{\pi}{4}$) and four small corner triangles ($\tfrac{3 + 2\sqrt{2}}{4}$ each), giving probability $\tfrac{44 + 24\sqrt{2} + \pi}{196}$ and $a + b = 44 + 24 = 68$.
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