AMC 10 · 2011 · #20

Grade 7 geometry-2d
Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region
geometric-probabilityarc-measureequilateral-triangle symmetry-argumenteasier-related-problem ↑ Prerequisites: probability-basic
📏 Long solution 💡 3 insights
Problem
Two points are placed on a circle of radius r, each chosen at random and independently. From each point, draw a chord of length r going clockwise. Find the probability that these two chords cross.

Pick an answer.

(A)
$\frac{1}{6}$
(B)
$\frac{1}{5}$
(C)
$\frac{1}{4}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A picture is the key: drawing the circle shows that a chord of length r cuts off a 60-degree arc, and it makes the crossing condition visible. Fixing the first point by symmetry turns a two-random-point problem into one uniform random point, and a variable for the second point's angle lets us measure exactly when the chords cross.

1STEP 1

Chord equal to radius cuts a 60 degree arc

Join a chord of length r to the center: all three sides equal r, so the triangle is equilateral and the arc it cuts is 60°.

chord=r, two radii=r → △ equilateral → central angle=arc=60°
2STEP 2

Fix the first point by symmetry

A rotation cannot change a crossing, so pin the first point down; only the second point's relative position matters, and it is uniform.

P(cross) depends only on the second point's relative angle
3STEP 3

Give the endpoints angle labels

Put the first point at 0°, so chord A ends at 300°; with the second point at angle b, chord B has ends b and b minus 60°.

A:{0°, 300°}, B:{b, b-60°}, 0° ≤ b ＜ 360°
4STEP 4

Chords cross when endpoints alternate

They cross when exactly one end of B sits in A's 60° gap: b from 300° to 360°, or b from 0° to 60° — 120° in all.

b∈(300°,360°) or b∈(0°,60°) → 60°+60°=120°
5STEP 5

Turn favorable arc into probability

The second point is uniform over the full 360°, so the probability is 120360=13\frac{120}{360}=\frac{1}{3}, choice (D).

P=120°/360°=1/3
Answer
1/3
The result 1/3 lies between 0 and 1, as any probability must. It also makes sense in scale: each chord only blocks a short 60-degree arc, so a crossing is possible but not likely, and two such 60-degree windows out of 360 degrees giving one third fits that expectation. This matches choice (D).
💡Key takeaway

A chord as long as the radius always cuts a 60-degree arc, so the two chords cross only when the second point falls in one of two 60-degree windows: 120 out of 360 degrees, or one third of the time.

  • Chord equal to radius cuts a 60 degree arc
  • Fix the first point by symmetry
  • Give the endpoints angle labels
  • Chords cross when endpoints alternate
  • Turn favorable arc into probability

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