AMC 10 · 2023 · #19

Grade 7 geometry-2d

geometric-probabilityarea-rectanglessymmetry-argumentprobability-basic easier-related-problemidentify-subproblemssymmetry-argument ↑ Prerequisites: geometric-probabilityarea-rectangles

📏 Medium solution 💡 3 insights

Problem

Sonya the frog lands at a uniformly random point (X, Y) in the square [0, 6] × [0, 6], then picks a uniformly random distance D in [0, 1] and a uniformly random cardinal direction. She hops that distance in that direction. Find the probability that her landing spot is outside the square.

Pick an answer.

(A)

$frac{1}{6}$

(B)

$frac{1}{12}$

(C)

$frac{1}{4}$

(D)

$frac{1}{10}$

(E)

$frac{1}{9}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Direct computation across all four directions is messy, but symmetry collapses it. Tool #9 (Easier Problem): solve the conditional probability for ONE direction — say north — then use Tool #7 (Identify Subproblems) and the law of total probability with symmetry to get the full answer. Tool #1 (Diagram) does the heavy lifting: draw the (Y, D) rectangle and shade Y + D ＞ 6 — the favorable region is a small right triangle whose area is read off by inspection.

1STEP 1

By 90° symmetry all four directions share the same p, so the law of total probability gives P(out) = $\frac{1}{4}$ (p + p + p + p) = p.

P(out) = Σ_dir

\frac{1}{4}

P(out ∣ dir) = p

2STEP 2

Hop north → (X, Y + D); X stays in [0, 6], so escaping means Y + D ＞ 6 with Y ∼ U[0, 6] and D ∼ U[0, 1].

p = P(Y + D ＞ 6), Y ∼ U[0, 6], D ∼ U[0, 1]

3STEP 3

On the 6 × 1 (Y, D) rectangle, the region Y + D ＞ 6 is a right triangle with legs 1 and area $\frac{1}{2}$ .

triangle area =

\frac{1}{2}

· 1 · 1 =

\frac{1}{2}

4STEP 4

Probability = area ratio = ( $\frac{1}{2}$ )/6 = $\frac{1}{12}$ , so by step 1 P(out) = $\frac{1}{12}$ .

P(out) = (

\frac{1}{2}

)/6 =

\frac{1}{12}

→ (B)

\frac{1}{12}

Answer

\frac{1}{12}

Sanity check the magnitude: the dangerous strip near each edge is only 1 unit wide (since D ≤ 1), so the dangerous strip area is at most 4 · (6 · 1) = 24 out of 36 — but escape requires both being in the strip AND drawing a D large enough, which trims it heavily. The answer

\frac{1}{12}

≈ 0.083 feels right — small but not tiny. Cross-check the triangle: vertices (5,1), (6,1), (6,0), legs of length 1, area

\frac{1}{2}

. Ratio (

\frac{1}{2}

)/6 =

\frac{1}{12}

. Matches choice (B).

💡Key takeaway

By rotational symmetry, the overall escape probability equals the probability of escaping when hopping in any one chosen direction (say north). For north, only the y-coordinate matters: shade the rectangle [0,6] × [0,1] of (Y, D) pairs, and the escape region Y + D ＞ 6 is a right triangle with legs 1 and area $\frac{1}{2}$ . Divide by the total area 6 to get (B) $\frac{1}{12}$ .