AMC 10 · 2015 · #25

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

A square has side length 1. Two points are dropped independently and uniformly at random somewhere on the square's boundary (its four sides). Find the chance that the straight-line distance between the two points is at least

\frac{1}{2}

. The answer comes out as

\frac{a-bπ}{c}

in lowest terms, and you report a+b+c.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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 $\frac{1}{2}$ " 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

Fix the first point; the second lands on the same, an adjacent ( $\frac{2}{4}=\frac{1}{2}$ ), or the opposite side — handle each separately.

P(same)=

\frac{1}{4}

, P(adjacent)=

\frac{1}{2}

, P(opposite)=

\frac{1}{4}

2STEP 2

Same side: an area model

Same-side positions form a random point in the unit square, so the qualifying region's area gives probability $\frac{1}{4}$ .

P_same=P (|x-y| ≥

\frac{1}{2}

)=2·

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{4}

3STEP 3

Adjacent sides: set up the distance

Set the shared corner as the origin so each point's distance becomes a Pythagorean hypotenuse in x and y.

distance²=x²+y², distance ≥

\frac{1}{2}

⇔ x²+y² ≥

\frac{1}{4}

4STEP 4

Adjacent sides: subtract the failing quarter-disk

The failing region is a quarter-disk of radius $\frac{1}{2}$ , so subtracting its area leaves probability 1- $\frac{π}{16}$ .

P_adj=1-(

\frac{1}{4}

·π(

\frac{1}{2}

)²)/1=1-

\frac{π}{16}

5STEP 5

Opposite sides: always far enough

Opposite-side points are always at least 1 apart, well past $\frac{1}{2}$ , so this case's probability is 1.

distance ≥ 1 ≥

\frac{1}{2}

→ P_opp=1

6STEP 6

Average the cases and read off the answer

Weighting the three cases $\frac{1}{4}$ , $\frac{1}{2}$ , $\frac{1}{4}$ and summing gives a+b+c=59, answer (A).

\frac{1}{4}

\frac{1}{4}

\frac{1}{2}

(1-

\frac{π}{16}

\frac{1}{4}

·1=

\frac{26-π}{32}

→ a+b+c=59 (A)

Answer

The result

\frac{26-π}{32}

≈

\frac{26-3.14}{32}

≈

\frac{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-

\frac{π}{16}

≈0.80 (adjacent), and 1 (opposite), and it leans toward the larger ones because the adjacent and opposite cases together carry

\frac{3}{4}

of the weight. Finally gcd(26,1,32)=1, so the form

\frac{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 into an area on a unit square, then average by how often it 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