AMC 10 · 2015 · #23

Grade 8 geometry-2d

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

geometric-probabilityprobability-basicarea-circles casework ↑ Prerequisites: geometric-probability

📏 Long solution 💡 4 insights

Problem

Two random points on a square's boundary must be far enough apart. Report the reduced expression's parts.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

The two points can land on the same side, on two sides that share a corner, or on two sides that face each other, and the distance behaves completely differently in each situation. Tool #7 (Identify Subproblems) splits the problem along exactly those three cases and lets symmetry fix the chance of each case. Inside a case, Tool #4 (Introduce a Variable) names each point's position by how far it sits from a corner, turning "distance at least 1/2" into an inequality in two coordinates. Tool #1 (Draw a Diagram) plots those two coordinates on a unit square so a probability becomes an area. In the corner-sharing case the qualifying region is awkward, so Tool #16 (Count the Complement) measures the small quarter-circle that fails instead and subtracts it. Weighting the three case-probabilities by how often each case happens assembles the final answer.

1STEP 1

Split by where the points land

Where the points land splits it into three cases.

P(same)=1/4, P(adjacent)=1/2, P(opposite)=1/4

2STEP 2

Same side: an area model

Same side is a plain area model.

P_same=P (|x-y| ≥ 1/2)=2·1/2·1/2·1/2=1/4

3STEP 3

Adjacent sides: set up the distance

Adjacent sides turn distance into a circle.

distance²=x²+y², distance ≥ 1/2 ⇔ x²+y² ≥ 1/4

4STEP 4

Adjacent sides: subtract the failing quarter-disk

The failing region is a quarter disk.

P_adj=1-(1/4·π(1/2)²)/1=1-π/16

5STEP 5

Opposite sides: always far enough

Opposite sides always succeed.

distance ≥ 1 ≥ 1/2 → P_opp=1

6STEP 6

Average the cases and read off the answer

Weighting the cases gives 59, choice (A).

P=1/4·1/4+1/2 (1-π/16)+1/4·1=(26-π)/32 → a+b+c=59 (A)

Answer

The result (26-π)/32≈(26-3.14)/32≈22.86/32≈ 0.71 is a sensible probability strictly between 0 and 1. It also sits correctly between the three case values it averages: 0.25 (same side), 1-π/16≈0.80 (adjacent), and 1 (opposite), and it leans toward the larger ones because the adjacent and opposite cases together carry 3/4 of the weight. Finally gcd(26,1,32)=1, so the form (a-bπ)/c is genuinely reduced and a+b+c=59 is forced, matching (A).

💡Key takeaway

Sort the two points by which sides they land on, turn each case's distance rule into an area on a unit square, then average the cases by how often each happens.

Split by where the points land
Same side: an area model
Adjacent sides: set up the distance
Adjacent sides: subtract the failing quarter-disk
Opposite sides: always far enough
Average the cases and read off the answer