AMC 10 · 2012 · #23

Grade 12 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

complex-numbersroots-of-unitypolynomial-rootsgcd convert-to-algebraextreme-principlesystematic-enumeration ↑ Prerequisites: complex-numberspolynomial-roots

📏 Long solution 💡 4 insights

Problem

Integer coefficients form a decreasing chain and the polynomial must have a root of size one. Add one value over all survivors.

Pick an answer.

(A)

(B)

(C)

100

(D)

108

(E)

120

How to solve

Strategy Extreme Principle

The condition |z₀|=1 is a size condition, so the argument should be run on sizes. The trick that makes the chain 4 ≥ a ≥ b ≥ c ≥ d ≥ 0 speak is to multiply P by z-1: the coefficients collapse into the five consecutive gaps of the chain, all non-negative, and they add to exactly 4 (Tool #15, Organize Information in More Ways; Tool #4, Introduce a Variable). Then |z₀|=1 makes one side of the identity have size exactly 4 while the other side is a sum of five pieces of total size 4 -- the modulus is maxed out, so nothing may point in a wrong direction. That equality case of the triangle inequality is the whole problem (Tool #14, Extreme Principle), and it converts the analytic condition into the arithmetic condition z₀^k=1 for every gap that is actually present. Running the same chain backwards shows the condition is not merely necessary but sufficient (Tool #11, Work Backwards), which is what licenses a finite list (Tool #2, Make a Systematic List; Tool #3, Eliminate Possibilities).

1STEP 1

Name the five gaps in the chain

The chain becomes five nonnegative gaps.

e₁=4-a, e₂=a-b, e₃=b-c, e₄=c-d, e₅=d, e₁+e₂+e₃+e₄+e₅=4

2STEP 2

Multiply by z-1

One clever factor makes the sum telescope.

(z-1)P(z)=4z⁵-(e₁z⁴+e₂z³+e₃z²+e₄z+e₅)

3STEP 3

Rule out z₀=1, then trade equations

The value one is never a root here.

P(1)=4+a+b+c+d ≥ 4 ＞ 0 → z₀ ≠ 1, P(z₀)=0⇔ 4z₀⁵=Σ_k=1⁵e_kz₀^ 5-k

4STEP 4

The size is maxed out

A size argument shows the bound is tight.

4=|4z₀⁵|=|Σ_k=1⁵e_kz₀^ 5-k| ≤ Σ_k=1⁵e_k|z₀|⁵-k=Σ_k=1⁵e_k=4

5STEP 5

Equality forces z₀^k=1

Equality forces each used power to be one.

Σ_k=1⁵e_kz₀^-k=4=Σ_k=1⁵e_k, Re (z₀^-k) ≤ 1 → e_k ＞ 0→ z₀^-k=1⇔ z₀^ k=1

6STEP 6

The condition is also enough

That condition is also enough.

S={k:e_k ＞ 0}, P has a root with |z|=1⇔ gcd(S) ≥ 2

7STEP 7

List the allowed gap patterns

Only seven polynomials survive.

(a,b,c,d)∈{(4,0,0,0),(4,1,1,0),(4,2,2,0),(4,3,3,0),(4,4,4,0)}∪{(4,4,0,0)}∪{(4,4,4,4)}

8STEP 8

Add the seven values of P(1)

Their values add to 92, choice (B).

Σ_t=0⁴(8+2t)+12+20=60+12+20=92 → (B)

Answer

Each of the seven polynomials can be checked by hand, independently of the gap argument. For (4,t,t,0), P(z)=4z⁴+4z³+tz²+tz=z(z+1)(4z²+t), which has the unit-circle root z=-1 for every t. For (4,4,0,0), P(z)=4z⁴+4z³+4z²=4z²(z²+z+1), whose roots include the two non-real cube roots of unity. For (4,4,4,4), P(z)=4(z⁴+z³+z²+z+1)=4·(z⁵-1)/(z-1), whose roots are the four fifth roots of unity other than 1. All seven really do qualify, and the argument also shows nothing else can: any quadruple with a ≤ 3 has e₁ ＞ 0, which would demand z₀¹=1, i.e. the excluded z₀=1, so a=4 is genuinely forced rather than merely observed. The classic trap is dropping (4,0,0,0) as degenerate -- P(z)=4z³(z+1) does have the triple root 0, but z=-1 is still a root with |{-1}|=1, so it counts; discarding it would give 92-8=84, which is exactly the decoy (A). Missing the fifth-root polynomial instead gives 72, which is not even on the list -- a hint that the exotic case is required. The total 92 matches choice (B).

💡Key takeaway

Multiply by z-1 so the coefficient chain becomes five gaps adding to 4; on the unit circle the total size is maxed out, which forces z^k=1 at every present gap, so a root exists exactly when those positions share a common factor -- seven polynomials survive and their P(1) values add to 92.

Name the five gaps in the chain
Multiply by z-1
Rule out z₀=1, then trade equations
The size is maxed out
Equality forces z₀^k=1
The condition is also enough
List the allowed gap patterns
Add the seven values of P(1)