AMC 10 · 2003 · #25

Grade 10 geometry-2d
Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Volume
geometric-probabilityarc-measureequilateral-triangle caseworksymmetry-argument ↑ Prerequisites: probability-basic
📏 Long solution 💡 3 insights
Problem
Drop three points onto a circle at random, each landing anywhere with equal chance and independently. The three points give three straight-line distances. Find the probability that all three of those distances come out shorter than the circle's radius.

Pick an answer.

(A)
$\dfrac{1}{36}$
(B)
$\dfrac{1}{24}$
(C)
$\dfrac{1}{18}$
(D)
$\dfrac{1}{12}$
(E)
$\dfrac{1}{9}$
How to solve
Strategy Change Focus / Count the Complement

Distances between points on a circle are governed by the angle at the centre, so tool #1 (Draw a Diagram) plus tool #13 (Convert to Algebra) trade the three chord lengths for three angles: a chord is shorter than the radius exactly when its central angle is under 60°. The three conditions that remain are tangled, because they involve all three pairs at once. Tool #16 (Change Focus) untangles them: instead of watching three distances, watch which single point the other two follow around the circle. Tool #2 (Make a Systematic List) then splits the event into three cases — one per possible leader — that cannot happen together, so their probabilities simply add.

1STEP 1

Chord shorter than radius means under 60 degrees

A chord equals the radius exactly at a sixty degree central angle, so shorter means angle under sixty.

PQ ＜ r ⇔ θ ＜ 60°, since θ=60° makes △ OPQ equilateral
2STEP 2

Restate everything with angles

Restating in angles makes the radius disappear from the problem.

win⇔ all three pairwise separations ＜ 60°=1/6 · 360°
3STEP 3

All three must huddle in one short arc

Since the largest gap is at least a third of the circle, all three must huddle in one short arc.

g₁+g₂+g₃=360°, G ≥ 120° → 360°-G ＜ 60°
4STEP 4

Count by asking who leads

Asking which point leads that arc splits the event into three mutually exclusive cases.

win=E₁∪ E₂∪ E₃ with E₁,E₂,E₃ mutually exclusive
5STEP 5

Add the three equal cases

Each case has probability one thirty-sixth, so the total is 1/12, choice (D).

P=3(1/6)²=3/36=1/12→(D)
Answer
1/12
The same leader argument with a half-circle instead of a 60° arc gives 3(1/2)²=3/4 for the classic probability that three random points lie on some semicircle, which is the known value — good evidence the counting scheme is set up correctly. The size is sensible too: demanding that three independent points crowd into one sixth of the circle should be rare, and 1/12≈ 0.083 is rare without being extreme. A simulation of two million random triples gave a success rate of about 0.0831, matching 1/12. The near-miss choices are traceable: 1/36 forgets the three choices of leader, and 1/9 would come from allowing all three to be leaders separately with no overlap correction on a wider arc.
💡Key takeaway

A chord is shorter than the radius exactly when the two points are under 60 degrees apart, so the question is really whether all three points huddle inside one sixth of the circle.

  • Chord shorter than radius means under 60 degrees
  • Restate everything with angles
  • All three must huddle in one short arc
  • Count by asking who leads
  • Add the three equal cases

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