AMC 10 · 2016 · #24

Grade 11 algebra
vieta-formulasam-gm-inequalitypolynomial-rootsoptimization vieta-formulasextreme-principle ↑ Prerequisites: vieta-formulasam-gm-inequality
📏 Medium solution 💡 3 insights
Problem
One coefficient appears twice in a cubic whose roots must all be real. Find the other coefficient at the smallest case.

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12
How to solve
Strategy Extreme Principle

The question is a minimum question, so the Extreme Principle sets the shape of the work: find a lower bound for a, then show some polynomial actually reaches it. To get a bound, stop looking at the polynomial and look at its roots instead. Naming the roots r, s, t turns the coefficients into Vieta's relations, and the problem's quirk (same a in two places) says sum of roots = product of roots. That is exactly the setup where the AM-GM inequality collapses everything to a single inequality in a. Three subproblems fall out: prove the roots are positive so AM-GM is legal, get the bound, and check the bound is attained.

1STEP 1

Turn coefficients into root facts

The coefficients become facts about the roots.

r+s+t = a, rs+rt+st = b, rst = a
2STEP 2

Prove all three roots are positive

All three roots must be positive.

r = a+u+v > u+v → r(u+v) > (u+v)² ≥ 4uv → b = uv - r(u+v) < 0
3STEP 3

Squeeze a with AM-GM

A mean inequality then bounds the coefficient.

a/3 ≥ ∛(a) → a³/27 ≥ a → a² ≥ 27 → a ≥ 3√(3)
4STEP 4

Find what equality forces

Equality forces all three roots equal.

r = s = t, r³ = 3√(3) = (√(3))³ → r = s = t = √(3)
5STEP 5

Check that the bound is reached

That cubic really exists.

(x-√(3))³ = x³ - 3√(3) x² + 9x - 3√(3)
6STEP 6

Read off the forced b

Reading the other coefficient gives 9, choice (C).

b = rs + rt + st = 3(√(3))² = 9
Answer
9
Substitute back: with (a = 3√(3) ≈ 5.196) the polynomial is (x³ - 5.196x² + 9x - 5.196), whose only root is (√(3) ≈ 1.732), repeated three times. Its root sum (3(1.732) ≈ 5.196) and root product ((1.732)³ ≈ 5.196) really are equal, which is exactly what the shared a requires. A smaller a fails for a concrete reason: at (a = 5) we would need three positive numbers with sum 5 and product 5, but AM-GM caps that product at ((5/3)³ ≈ 4.63 < 5). And 9 lands where a quantity built from (3 · 3) should land, in the middle of the answer list.
💡Key takeaway

When the roots are forced to have the same sum as product, AM-GM squeezes that number from below, and the tightest case is always the one where all the roots are equal.

  • Turn coefficients into root facts
  • Prove all three roots are positive
  • Squeeze a with AM-GM
  • Find what equality forces
  • Check that the bound is reached
  • Read off the forced b