AMC 10 · 2024 · #14

Grade 8 probability

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

probability-basicarea-circlescoordinate-geometryfraction-arithmetic identify-subproblemsarea-differenceconvert-to-algebra ↑ Prerequisites: probability-basicarea-circlescoordinate-geometry

📏 Medium solution 💡 3 insights

Problem

Region B in the plane consists of all (x, y) with |x| + |y| ≤ 8. Region T consists of all (x, y) with (x² + y² - 25)² ≤ 49. A dart lands uniformly at random in B. The probability that it lands in T is

\frac{m}{n}

π in lowest terms. Find m + n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

135

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Both regions live in the plane and the answer is a ratio of areas, so the picture is everything. Tool #1 (Draw a Diagram) reveals what each inequality really is: |x| + |y| ≤ 8 is a square (rotated 45°) with diagonals on the axes, and (x² + y² - 25)² ≤ 49 is an annulus centered at the origin. Tool #7 (Identify Subproblems) splits the calculation in three clean parts — area of B (rotated square), area of T (annulus), and the ratio. Tool #13 (Convert to Algebra) tidies the target inequality: let r² = x² + y², then (r² - 25)² ≤ 49 becomes |r² - 25| ≤ 7, i.e. 18 ≤ r² ≤ 32 — two concentric circles.

1STEP 1

|x| + |y| ≤ 8 is a square tilted 45° with diagonals of length 16 on the axes, so Area(B) = $\frac{1}{2}$ ·16·16 = 128.

Area(B) =

\frac{1}{2}

(16)(16) = 128

2STEP 2

Set r² = x² + y²; then (r² - 25)² ≤ 49 means |r² - 25| ≤ 7, i.e. 18 ≤ r² ≤ 32.

(r² - 25)² ≤ 49 ⇔ 18 ≤ r² ≤ 32

3STEP 3

The ring 18 ≤ r² ≤ 32 is the r²=32 disk minus the r²=18 disk, so Area(T) = π(32) - π(18) = 14π (no square roots needed).

Area(T) = π(32) - π(18) = 14π

4STEP 4

T's outer radius √(32) = 4√(2) equals the origin-to-side distance $\frac{8}{√(2)}$ , so the ring is tangent inside the square: T ⊆ B.

√(32) = 4√(2) = dist(origin, side of B) ⟹ T ⊆ B

5STEP 5

The probability is $\frac{14π}{128}$ = $\frac{7π}{64}$ , so m = 7 and n = 64 are coprime and m + n = 71.

\frac{m}{n}

π =

\frac{14π}{128}

\frac{7π}{64}

⟹ m + n = 7 + 64 = 71 → (B)

Answer

The annulus has area 14π ≈ 44, and the square has area 128, so the probability is about

\frac{44}{128}

≈ 0.34. That fits a thick-ish ring covering a sizable chunk of the diamond — visually plausible. As a second check, the numbers 7 and 64 should be relatively prime, and they are. The other answer choices match other plausible-looking simplifications: (A) 39 = 7 + 32 would come from forgetting to subtract the inner disk, (E) 135 = 14 + 121 would come from squaring √(32) incorrectly. Our (B) = 71 is the clean derivation.

💡Key takeaway

This AMC 10 problem only needs Grade 8 square-root reasoning plus the Grade 7 circle-area formula that you already know!

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