AMC 10 · 2017 · #25

Grade 12 probability

Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue

complex-polar-formroots-of-unityprobability-basicmodular-arithmetic identify-subproblemscasework ↑ Prerequisites: complex-polar-formprobability-basic

📏 Long solution 💡 4 insights

Problem

Twelve values are drawn from a small symmetric set and multiplied together. Find the chance of one exact product.

Pick an answer.

(A)

$\dfrac{5\cdot11}{3^{10}}$

(B)

$\dfrac{5^2\cdot11}{2\cdot3^{10}}$

(C)

$\dfrac{5\cdot11}{3^{9}}$

(D)

$\dfrac{5\cdot7\cdot11}{2\cdot3^{10}}$

(E)

$\dfrac{2^2\cdot5\cdot11}{3^{10}}$

How to solve

Strategy Organize Information in More Ways

In rectangular form the six vertices look like a random assortment and their 12-fold product is hopeless. Rewritten in polar form they become one of two lengths times a power of ζ = e^iπ/4, and the product splits cleanly: lengths multiply, exponents add mod 8. That single re-organization turns the problem into two small independent subproblems — a counting question about how many long vertices were drawn, and a modular arithmetic question about the exponent sum.

1STEP 1

Rewrite every vertex in polar form

Polar form makes every value a length and an angle.

V = {√(2) ζ², √(2) ζ⁶, 1/2ζ¹, 1/2ζ³, 1/2ζ⁵, 1/2ζ⁷}, ζ = e^iπ/4

2STEP 2

Split the target into length and angle

The target splits into two separate conditions.

P = (Π_j=1¹² r_j)ζ^{Σ_j=1¹² e_j} = -1 ⇔ Π_j=1¹² r_j = 1 and Σ_j=1¹² e_j ≡ 4 (mod 8)

3STEP 3

Length forces exactly eight long vertices

The length condition fixes an eight to four split.

(√(2))ⁿ(1/2)¹²⁻ⁿ = 2³n/2-12 = 1 → n = 8

4STEP 4

Probability of the eight-four split

That split's chance is a plain binomial.

Pr(|P| = 1) = C(12, 4)(1/3)⁸(2/3)⁴ = (495 · 2⁴)/3¹²

5STEP 5

The long block contributes 0 or 4

One block contributes only two possible angles.

Σ_long e_j ≡ 4b (mod 8), #{b even} = #{b odd} = 2⁷ = 128

6STEP 6

The short block is uniform mod 8

The other is spread evenly.

Σ_short e_j = 2K + 4 (mod 8), K uniform mod 4 → each of {0,2,4,6} has 64 patterns

7STEP 7

Combine the two conditions

Combining gives choice (C).

C(12, 4)(1/3)⁸(2/3)⁴·1/4 = (495 · 2²)/3¹² = (3² · 5 · 11 · 2²)/3¹² = (2² · 5 · 11)/3¹⁰

Answer

(2²·5·11)/3¹⁰

Re-running the whole thing as an exact integer count agrees. Requiring eight long draws gives C(12, 4) · 2⁸ · 4⁴ = 495 · 65536 = 32,440,320 sequences with |P| = 1, and exactly one quarter of them, 8,110,080, also have exponent sum ≡ 4 (mod 8). Against the 6¹² = 2,176,782,336 total sequences that is 8110080/2176782336 = 220/59049 = (2² · 5 · 11)/3¹⁰, which is choice (E). The magnitude is sensible: about 0.0037, since even hitting the required eight-four split is only about a 1.5% event. One more check comes free from symmetry — replacing z₁ by -z₁, which is also in V, matches every P = -1 sequence with a P = +1 sequence, so those two counts must be equal, and the count above gives 8,110,080 for each.

💡Key takeaway

In polar form a product of complex numbers is two separate problems at once — lengths multiply and angles add — so pin down the lengths first, then count the angles modulo a full turn.

Rewrite every vertex in polar form
Split the target into length and angle
Length forces exactly eight long vertices
Probability of the eight-four split
The long block contributes 0 or 4
The short block is uniform mod 8
Combine the two conditions