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AMC 10 · 2023B · #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

Use symmetry to pick one direction

By 90° symmetry all four directions share the same p, so the law of total probability gives P(out) = 14\frac{1}{4}(p + p + p + p) = p.

P(out) = Σ_dir 1/4 P(out ∣ dir) = p
2STEP 2

Turn it into a 2D problem

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

Draw the sample rectangle

On the 6 × 1 (Y, D) rectangle, the region Y + D ＞ 6 is a right triangle with legs 1 and area 12\frac{1}{2}.

triangle area = 1/2 · 1 · 1 = 1/2
4STEP 4

Compare the areas

Probability = area ratio = (12\frac{1}{2})/6 = 112\frac{1}{12}, so by step 1 P(out) = 112\frac{1}{12}.

P(out) = 1/2/6 = 1/12 → (B) 1/12
Answer
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 112\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 12\frac{1}{2}. Ratio (12\frac{1}{2})/6 = 112\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 12\frac{1}{2}. Divide by the total area 6 to get (B) 112\frac{1}{12}.

  • Use symmetry to pick one direction
  • Turn it into a 2D problem
  • Draw the sample rectangle
  • Compare the areas

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