AMC 10 · 2021 · #23

Grade 11 algebra

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

quadratic-equationsfunction-compositionvieta-formulascompleting-the-square identify-subproblemsextreme-principle ↑ Prerequisites: quadratic-equationsfunction-composition

📏 Long solution 💡 4 insights

Problem

Consider the monic quadratics with real coefficients whose composition with themselves has exactly three real solutions. Exactly one of them has the largest possible sum of its own two roots. Find that quadratic's value at one.

Pick an answer.

(A)

$\dfrac{5}{16}$

(B)

$\dfrac{1}{2}$

(C)

$\dfrac{5}{8}$

(D)

(E)

$\dfrac{9}{8}$

How to solve

Strategy Extreme Principle

The word disrespectful hides a plain structural condition, so my first job is to unpack it. A composition p(p(x)) = 0 always splits: the inner value p(x) must land on a root of p, which breaks one quartic into two quadratics. Counting how three real solutions can come out of two quadratics leaves exactly one shape, and that shape has a clean picture: a horizontal line touching the parabola at its lowest point. That picture turns the whole condition into one equation linking the two roots. Then I name the gap between the roots with a single letter, which reduces the quantity I want to maximize to one quadratic in one variable. A quadratic's maximum is found by completing the square, so the extreme case falls out with no calculus and no guessing.

1STEP 1

Split the composition into two quadratics

Split it into two quadratics.

p(p(x)) = (p(x) - r)(p(x) - s) = 0

2STEP 2

Three real solutions forces one double root

Three solutions force a double root.

#{p(x) = r} + #{p(x) = s} = 3 ⟹ {1, 2}

3STEP 3

The double root sits at the vertex

The double root sits at the vertex.

p(x) = (x - (r+s)/2)² - ((s-r)²)/4, r = -((s-r)²)/4

4STEP 4

Name the gap between the roots

Name the gap between the roots.

d = s - r ＞ 0, r = -d²/4, s = d - d²/4, r + s = d - d²/2

5STEP 5

Maximize and evaluate at x = 1

Maximizing and evaluating at one gives five sixteenths.

r + s = 1/2 - ((d-1)²)/2 ≤ 1/2, d = 1 → p(x) = (x + 1/4)(x - 3/4), p(1) = 5/4·1/4 = 5/16

Answer

5/16

Expanding gives p(x) = x² - 1/2x - 3/16, and p(1) = 1 - 1/2 - 3/16 = 5/16, matching the factored evaluation. The disrespectful property checks out directly. The minimum of p is at x = 1/4 with value -1/4, which is exactly the root r, so p(x) = -1/4 becomes (x - 1/4)² = 0 and has the single solution x = 1/4. Meanwhile p(x) = 3/4 becomes x² - 1/2x - 15/16 = 0, with discriminant 1/4 + 15/4 = 4 ＞ 0 and solutions x = 5/4 and x = -3/4. Together that is exactly three real solutions: -3/4, 1/4, 5/4. The size is also sensible: the roots sum to 1/2, so p(1) should be small and positive since 1 sits just outside the larger root 3/4, and 5/16 fits that.

💡Key takeaway

When an equation stacks a function on itself, ask what the inside has to equal, and when a quadratic must have a repeated root, that root is sitting right at the bottom of the parabola.

Split the composition into two quadratics
Three real solutions forces one double root
The double root sits at the vertex
Name the gap between the roots
Maximize and evaluate at x = 1

Split the composition into two quadratics

Three real solutions forces one double root

The double root sits at the vertex

Name the gap between the roots

Maximize and evaluate at x = 1

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