AMC 10 · 2021 · #23

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

geometric-probabilityarea-trianglesarea-circlesminkowski-sum identify-subproblemsarea-difference ↑ Prerequisites: geometric-probabilityarea-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

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 $a+b$ ?

Pick an answer.

(A)

~64

(B)

~66

(C)

~68

(D)

~70

(E)

~72

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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

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

#9 Solve an Easier Related Problem 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 $= 49$.

$$A_{\text{sample}} = 7 \times 7 = 49$$

💡 If the center is at least the radius from every edge of the big square, the coin is fully inside.

#9 Solve an Easier Related Problem 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.

$$\text{Favorable region } F = \{p : d(p,\text{black}) \le \tfrac{1}{2}\} \cap \text{sample square}$$

💡 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 Identify Subproblems 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.

$$A_{\text{diamond}} = 8 + 4\sqrt{2} + \tfrac{\pi}{4}$$

💡 Buffer = the original shape + rectangles along each edge + a full circle at the corners (the four quarter-circles glue into one).

#1 Draw a Diagram 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 $\tfrac{|2 + \tfrac{\sqrt{2}}{2} - 2|}{\sqrt{1^2 + 1^2}} = \tfrac{\tfrac{\sqrt{2}}{2}}{\sqrt{2}} = \tfrac{1}{2}$.
✓

$$\text{boundary lines: } x = \tfrac{1}{2},\; y = \tfrac{1}{2},\; x + y = 2 + \tfrac{\sqrt{2}}{2}$$

💡 Sample-square edges already give two sides of the favorable region — only the hypotenuse needs a parallel shift.

#7 Identify Subproblems 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 $4 \cdot \tfrac{3 + 2\sqrt{2}}{4} = 3 + 2\sqrt{2}$.

$$A_{\text{one corner}} = \tfrac{3 + 2\sqrt{2}}{4},\; A_{\text{4 corners}} = 3 + 2\sqrt{2}$$

💡 Sample-square clipping removes the parts of the buffer that would have been outside; left over is a clean right triangle.

#7 Identify Subproblems 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 $4$: $A_F = \tfrac{44 + 24\sqrt{2} + \pi}{4}$.

$$A_F = \tfrac{44 + 24\sqrt{2} + \pi}{4}$$

💡 Combine the two contributions by common-denominator addition — the diamond and the four corners don't overlap, so adding is legitimate.

#3 Eliminate Possibilities 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).

$$P = \tfrac{1}{196}(44 + 24\sqrt{2} + \pi);\;\; a + b = 44 + 24 = 68 \Rightarrow \textbf{(C)}$$

💡 Match coefficients of $1, \sqrt{2}, \pi$ separately — $a$ is the rational part, $b$ is the $\sqrt{2}$ coefficient.

[1] #9 7.G.B.6 Sample space. The coin (radius $\tfrac{1}{2}$) lies entirely inside the $8 \time

[2] #9 7.G.B.6 Reframe "coin overlaps black" as "center within $\tfrac{1}{2}$ of black". Call $

[3] #7 7.G.B.6 Diamond's contribution. The central diamond is a square of side $s = 2\sqrt{2}$

[4] #1 8.G.B.8 Corner triangle contribution. Take the bottom-left black triangle with vertices

[5] #7 7.G.B.6 Compute the favorable triangular region for one corner. Its three vertices are $

[6] #7 6.NS.B.3 Sum favorable area. $A_F = A_{\text{diamond}} + A_{\text{4 corners}} = \bigl(8 +

[7] #3 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.4 Fluently add and subtract multi-digit whole numbers (Final $a + b = 44 + 24 = 68$.)
6.NS.B.3 Fluently 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.6 Solve 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.8 Apply 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 $a + b = 44 + 24 = 68$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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