AMC 10 · 2013 · #24

Grade 11 geometry-2d
polygon-inequalityprobability-basiccombinations-basictrigonometric-ratios complementary-countingextreme-principlesystematic-enumeration ↑ Prerequisites: polygon-inequalityprobability-basic
📏 Long solution 💡 4 insights
Problem
Three chords of a regular polygon are drawn at random and must form a real triangle. Find the probability.

Pick an answer.

(A)
$\frac{553}{715}$
(B)
$\frac{443}{572}$
(C)
$\frac{111}{143}$
(D)
$\frac{81}{104}$
(E)
$\frac{223}{286}$
How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Count the Complement) is primary for a countable reason: the three chosen lengths come from only six values, so there are C(8, 3) = 56 possible length patterns, of which 46 work and 10 fail. Listing ten failures is honest work; listing forty-six successes is a bookkeeping accident waiting to happen. Tool #14 (Extreme Principle) does the structural cutting twice — once to note that only the longest of the three sides can break the inequality, and once to note that any pair of lengths already past the diameter can never fail, which kills most of the case work before it starts. Tool #1 (Draw a Diagram) supplies the fact that makes six values possible at all: a chord of a circle depends only on how many steps it spans. Tool #2 (Make a Systematic List) then sweeps the surviving pairs in a fixed order and attaches the right multiplicity to each pattern. The one place none of these tools helps is the crux: two of the failures are exact equalities, not near misses, and only algebra decides those.

1STEP 1

Six lengths, fixed by the step

The step size gives only six lengths.

k = min(|i-j|, 12 - |i-j|) ∈ {1,2,…,6}, a₁ < a₂ < a₃ < a₄ < a₅ < a₆
2STEP 2

Write the six lengths exactly

Each length is written out exactly.

a_k = 2sinkπ/12: a₁ = (√(6)-√(2))/2, a₂ = 1, a₃ = √(2), a₄ = √(3), a₅ = (√(6)+√(2))/2, a₆ = 2
3STEP 3

Count the segments in each class

Counting per class gives the sample space.

n₁ = n₂ = n₃ = n₄ = n₅ = 12, n₆ = 6, Σ_k n_k = 66, C(66, 3) = 45760
4STEP 4

Only the longest side can fail

Only the longest side can fail the test.

positive area ⇔ a_x + a_y > a_z (x ≤ y ≤ z)
5STEP 5

Two exact ties, not near misses

Two exact ties appear, not near misses.

a₁ + a₃ = 2(sin 15° + sin 45°) = 4sin 30° cos 15° = 2cos 15° = 2sin 75° = a₅, 2a₂ = 2 = a₆
6STEP 6

Pairs with no shortest chord

Pairs without the shortest chord always pass.

x ≥ 2, y ≥ 3 ⟹ a_x + a_y ≥ 1 + √(2) > 2 = a₆; x = y = 2 ⟹ a_x + a_y = a₆
7STEP 7

Thresholds for the pairs starting at a₁

The remaining thresholds are checked one by one.

a₂ < 2a₁ < a₃, a₃ < a₁ + a₂ < a₄, a₁ + a₃ = a₅ < a₆, a₁ + a₄ > a₆
8STEP 8

Weight the ten failing patterns

Ten failing patterns weigh 10080 in total.

{1,1,3},{1,1,4},{1,1,5}: 3C(12, 2) · 12 = 2376; {1,1,6},{2,2,6}: 2C(12, 2) · 6 = 792; {1,2,4},{1,2,5},{1,3,5}: 3 · 12³ = 5184; {1,2,6},{1,3,6}: 2 · 12² · 6 = 1728; total 10080
9STEP 9

Take the complement and divide

The complement gives 223/286, choice (E).

1 - p = 10080/45760 = 63/286, p = 1 - 63/286 = 223/286 → (E)
Answer
223/286
The answer is 223/286 ≈ 0.7797, and the five choices run from 0.7734 to 0.7797 — a total spread of 0.0064. No estimate, no scale drawing, and no three-decimal chord table can separate them, which is exactly why the two exact ties had to be settled algebraically. Structural check: 10080/45760 must reduce to a fraction whose denominator divides 45760 = 2⁶ · 5 · 11 · 13, and 286 = 2 · 11 · 13 does. Independent recount, grouping by the longest side instead of by pattern: z = 3 contributes 792, z = 4 contributes 792 + 1728 = 2520, z = 5 contributes 792 + 1728 + 1728 = 4248, and z = 6 contributes 396 + 864 + 864 + 396 = 2520, totalling 10080 again. Boundary spot checks: three sides of the 12-gon do form a triangle, and {1,1,2} is the tightest survivor anywhere in the problem, clearing by 2a₁ - a₂ = √(6) - √(2) - 1 ≈ 0.035. Finally, the phrase "positive area" earns its keep — counting the two flat triples as triangles would leave only 7956 failures and give 727/880 ≈ 0.826, which is not among the choices at all.
💡Key takeaway

Sort the three lengths first, because only the two short ones against the long one can fail — and settle the close calls with exact radicals, since in a regular 12-gon two of those sums land dead on another length and a flat triangle has no area.

  • Six lengths, fixed by the step
  • Write the six lengths exactly
  • Count the segments in each class
  • Only the longest side can fail
  • Two exact ties, not near misses
  • Pairs with no shortest chord
  • Thresholds for the pairs starting at a₁
  • Weight the ten failing patterns
  • Take the complement and divide