AMC 10 · 2019 · #21

Grade 11 algebra
Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role
vieta-formulasquadratic-equationspolynomial-rootscaseworksign-analysis caseworksystematic-enumerationconvert-to-algebra ↑ Prerequisites: vieta-formulasquadratic-equations
📏 Long solution 💡 4 insights
Problem
Count the quadratic polynomials with real coefficients and nonzero leading coefficient whose set of roots is exactly the same set as the set of its coefficients.

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
infinitely many
How to solve
Strategy Make a Systematic List

Tool #2 (Systematic List) is the spine: three coefficient slots have to be filled from a root set of size at most two, so a repeat is forced and there are only a few patterns of which coefficients coincide — list them all and nothing can be missed. Tool #4 (Introduce a Variable) supplies Vieta's formulas, which convert the set condition into equations in a, b, c. Tool #13 (Algebra) solves each case's small system. Tool #3 (Eliminate Possibilities) kills the branches that produce contradictions or duplicates. Tool #1 (Diagram) settles how many real roots the leftover cubic a³+a+1=0 has by looking at its graph.

1STEP 1

Write the root relations

Sum and product come from the coefficients.

r+s = -b/a, rs = c/a
2STEP 2

Two coefficients must coincide

Matching set sizes forces two to coincide.

|{a,b,c}| = |{r,s}| ≤ 2
3STEP 3

Rule out a repeated root

A repeated root makes the discriminant negative.

x²+x+1=0, Δ = 1-4 = -3 ＜ 0
4STEP 4

Set up the first case

Write the first case's equations.

a+c = -1, c(a²-1) = 0
5STEP 5

Solve the first case

It yields two polynomials.

-x²-x (roots 0, -1), x²+x-2 (roots -2, 1)
6STEP 6

Set up the second case

It reduces to a cubic.

ab = 1, a + 1/a = -1/a² → a³+a+1 = 0
7STEP 7

Count the cubic's real roots

Being increasing, it has one real root.

f(a) = a³+a+1 strictly increasing, f(-1) ＜ 0 ＜ f(0)
8STEP 8

Set up the third case

Write the third case's equations.

ab = b/a → b(a - 1/a) = 0
9STEP 9

Solve the third case

One more polynomial appears.

x² - 1/2x - 1/2 = 1/2(2x+1)(x-1) (roots 1, -1/2)
10STEP 10

Add the cases

Adding them gives 4.

2+1+1 = 4
Answer
4
Check all four polynomials against the original condition, one at a time. (1) -x²-x: coefficients {-1,-1,0} = {-1,0}; factoring gives -x(x+1), roots {0,-1} — equal. (2) x²+x-2: coefficients {1,1,-2} = {1,-2}; factoring gives (x+2)(x-1), roots {-2,1} — equal. (3) x² - 1/2x - 1/2: coefficients {1,-1/2,-1/2} = {1,-1/2}; doubling gives 2x²-x-1 = (2x+1)(x-1), roots {-1/2,1} — equal. (4) α x² + 1/αx + α where α³+α+1 = 0, so α ≈ -0.682 and 1/α ≈ -1.466: root sum should be -b/a = -1/α² ≈ -2.148, and α + 1/α ≈ -2.148 — equal; root product should be c/a = 1, and α · 1/α = 1 — equal. All four survive, and no case can yield a fifth: each case ended in either a linear equation or a cubic with a single real root, never a free parameter, so (E) infinitely many is impossible and the count is exactly 4.
💡Key takeaway

Three coefficients can only fill two root slots, so two of them must be equal — check the three ways that can happen, solve the tiny system each time, and only four quadratics survive.

  • Translate roots into Vieta equations
  • Two coefficients must coincide
  • Rule out a repeated root
  • Case a=b: set up the equations
  • Case a=b: two polynomials
  • Case a=c: reduce to a cubic
  • The cubic has one real root
  • Case b=c: set up the equations
  • Case b=c: one polynomial
  • Add the cases

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