AMC 10 · 2021 · #23

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

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

📏 Long solution 💡 4 insights 📊 Diagram

Problem

An 8 × 8 white square has 5 black regions: a central diamond (a square of side 2√(2), rotated 45°) 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 × 8 square. Find the probability that the coin overlaps any black region. The probability has form

\frac{1}{196}

(a + b√(2) + π) for positive integers a, b; report a + b.

Pick an answer.

(A)

~64

(B)

~66

(C)

~68

(D)

~70

(E)

~72

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — draw the 8 × 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 $\frac{1}{2}$ 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.

1STEP 1

The coin (radius $\frac{1}{2}$ ) fits inside iff its center stays in a 7 × 7 square, so the sample-space area is 49.

A_sample = 7 × 7 = 49

2STEP 2

Recast "coin overlaps black" as "center within $\frac{1}{2}$ of black"; the buffers of the diamond and the corners stay separate, so add their areas.

Favorable region F = {p : d(p,black) ≤

\frac{1}{2}

} ∩ sample square

3STEP 3

Buffering the side-2√(2) diamond (area 8) adds four 2√(2) × $\frac{1}{2}$ rectangles (4√(2)) and a full $\frac{1}{2}$ -disk ( $\frac{π}{4}$ ), giving 8 + 4√(2) + $\frac{π}{4}$ .

A_diamond = 8 + 4√(2) +

\frac{π}{4}

4STEP 4

For the bottom-left triangle the favorable region is bounded by x = $\frac{1}{2}$ , y = $\frac{1}{2}$ , and the hypotenuse pushed out $\frac{1}{2}$ to x + y = 2 + $\frac{√(2)}{2}$ .

boundary lines: x =

\frac{1}{2}

, y =

\frac{1}{2}

, x + y = 2 +

\frac{√(2)}{2}

5STEP 5

That corner is a right triangle of leg 1 + $\frac{√(2)}{2}$ , area $\frac{<span class="hl-ask">3 + 2√(2)</span>}{4}$ ; the four symmetric corners together give 3 + 2√(2).

A_one corner =

\frac{3 + 2√(2)}{4}

, A₄ corners = 3 + 2√(2)

6STEP 6

Add the two separate contributions: (8 + 4√(2) + $\frac{π}{4}$ ) + (3 + 2√(2)) = 11 + 6√(2) + $\frac{π}{4}$ = $\frac{44 + 24√(2) + π}{4}$ .

A_F =

\frac{44 + 24√(2) + π}{4}

7STEP 7

Divide by 49: P = $\frac{1}{196}$ (44 + 24√(2) + π), so a = 44, b = 24 and a + b = 68, choice (C).

P =

\frac{1}{196}

(44 + 24√(2) + π); a + b = 44 + 24 = 68 → (C)

Answer

~68

Sanity. P =

\frac{44 + 24√(2) + π}{196}

≈

\frac{44 + 24 · 1.414 + 3.14}{196}

≈

\frac{44 + 33.9 + 3.14}{196}

≈

\frac{81.0}{196}

≈ 0.413. That's about 41% — the black regions take up area 4 ·

\frac{1}{2}

· 2 · 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).

💡Key takeaway

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 $\frac{1}{2}$ of black", the favorable region splits into the central buffered diamond (8 + 4√(2) + $\frac{π}{4}$ ) and four small corner triangles ( $\frac{3 + 2√(2)}{4}$ each), giving probability $\frac{44 + 24√(2) + π}{196}$ and a + b = 44 + 24 = 68.

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