AMC 10 · 2019 · #17

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta

polynomial-rootsrecursive-sequencesymmetric-polynomialspolynomial-substitution easier-related-problempattern-recognitionpolynomial-substitution ↑ Prerequisites: polynomial-rootsrecursive-sequence

📏 Medium solution 💡 3 insights

Problem

A cubic has three roots. For each index, let s be the sum of the roots raised to that power. Three fixed real numbers make the next s equal to a fixed combination of the previous three, and this holds for every index from two upward. Find the sum of those three numbers.

Pick an answer.

(A)

-6

(B)

(C)

(D)

(E)

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier related problem): chasing the sum s_k directly is hard, but a single root is easy — a root plugged into the polynomial gives 0, which is one clean equation. Solve the one-root version, then add the three copies back. Tool #4 (Introduce a Variable): name the roots r₁, r₂, r₃ so the sum s_k can be written and manipulated. Tool #5 (Pattern): multiplying the one-root equation by a power of that root shifts every exponent up by the same amount, so the same three coefficients reappear at every level — that is exactly what a recurrence is. Tool #3 (Eliminate Possibilities): the choices are separated by sign patterns (5 - 8 + 13 versus 5 + 8 + 13), so the signs must be tracked, not guessed.

1STEP 1

Name the roots

Name the power sums.

s_k = r₁^k + r₂^k + r₃^k

2STEP 2

One root, one equation

Each root gives one equation.

r_i³ = 5r_i² - 8r_i + 13 (i = 1, 2, 3)

3STEP 3

Push the identity up to exponent k

Multiplying pushes the exponent up.

r_i^ k+1 = 5 r_i^ k - 8 r_i^ k-1 + 13 r_i^ k-2

4STEP 4

Add the three copies

Adding them hands over the coefficients.

s_k+1 = 5 s_k - 8 s_k-1 + 13 s_k-2

5STEP 5

Match coefficients

Match the coefficients.

a = 5, b = -8, c = 13

6STEP 6

Add and pick the choice

Adding gives 10.

a + b + c = 5 + (-8) + 13 = 10 → (D)

Answer

Numerical test at k = 2: the recurrence predicts s₃ = 5(9) - 8(5) + 13(3) = 45 - 40 + 39 = 44. Newton's identities give the same value independently — with e₁ = 5, e₂ = 8, e₃ = 13 from Vieta, p₃ = e₁ p₂ - e₂ p₁ + 3e₃ = 45 - 40 + 39 = 44. One more level agrees too: s₄ = 5(44) - 8(9) + 13(5) = 213, matching p₄ = e₁ p₃ - e₂ p₂ + e₃ p₁ = 220 - 72 + 65 = 213. There is also a shortcut check on the sum itself: for a monic cubic x³ + Ax² + Bx + C the same derivation gives a + b + c = -(A + B + C) = 1 - p(1), and here p(1) = 1 - 5 + 8 - 13 = -9, so a + b + c = 1 - (-9) = 10. Two routes, same 10, so (D) stands; (E) 26 is the sign-dropping trap and -6 or 0 do not arise from any consistent sign pattern.

💡Key takeaway

A root turns the polynomial into a trade: r³ = 5r² - 8r + 13. Multiply that by r^k-2, add the three roots' copies, and the recurrence appears with the polynomial's own coefficients sign-flipped — a = 5, b = -8, c = 13, so a + b + c = 10, choice (D).

Name the roots
One root, one equation
Push the identity up to exponent k
Add the three copies
Match coefficients
Add and pick the choice