AMC 10 · 2021 · #16

Grade 11 algebra

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

polynomial-rootsvieta-formulasfunction-evaluationpolynomial-factoring convert-to-algebrasymmetry-argumentguess-and-check ↑ Prerequisites: polynomial-rootsvieta-formulas

📏 Long solution 💡 3 insights

Problem

A cubic has three roots. Another polynomial has leading coefficient one and its three roots are exactly the reciprocals of the first cubic's roots. Express the value of the second polynomial at one using only the first cubic's coefficients.

Pick an answer.

(A)

$\frac{1+a+b+c}c$

(B)

1+a+b+c

(C)

$\frac{1+a+b+c}{c^2}$

(D)

$\frac{a+b+c}{c^2}$

(E)

$\frac{1+a+b+c}{a+b+c}$

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable) — the problem talks about roots but never names them, so name them: let r, s, t be the roots of f. Once they have names, g can be written in factored form and g(1) becomes a product of three simple factors. Tool #16 (Change Focus) — instead of hunting for the roots themselves (hopeless, they are not computable from a, b, c), focus on the two symmetric combinations that Vieta's formulas hand over for free: the product rst and the value f(1) = (1-r)(1-s)(1-t). Everything g(1) needs is built from those two. Tool #6 (Guess and Check) and Tool #3 (Eliminate Possibilities) are the safety net: build one concrete cubic satisfying 1 ＜ a ＜ b ＜ c, compute g(1) numerically, and see which of the five expressions survives.

1STEP 1

Name the roots

Give the three roots names.

f(x) = x³ + ax² + bx + c = (x-r)(x-s)(x-t), with r, s, t ≠ 0 because c ≠ 0.

2STEP 2

Evaluate the second at one

It becomes a product of three factors.

g(1) = (1 - 1/r)(1 - 1/s)(1 - 1/t)

3STEP 3

Clear the reciprocals

Split into a numerator and denominator.

g(1) = (r-1)/r · (s-1)/s · (t-1)/t = (r-1)(s-1)(t-1)/rst

4STEP 4

Recognize the numerator

The numerator is the first polynomial's value.

(r-1)(s-1)(t-1) = (-1)³ (1-r)(1-s)(1-t) = -f(1) = -(1 + a + b + c)

5STEP 5

Get the denominator

The root relations give the denominator.

-rst = c → rst = -c

6STEP 6

Combine and read

The result is the coefficient sum over the constant term.

g(1) = (-(1+a+b+c))/-c = (1+a+b+c)/c → (A)

Answer

1+a+b+cc

Test with a concrete cubic. Take roots -3, -4, -5: then f(x) = (x+3)(x+4)(x+5) = x³ + 12x² + 47x + 60, so a = 12, b = 47, c = 60 and indeed 1 ＜ 12 ＜ 47 ＜ 60. The reciprocal roots are -1/3, -1/4, -1/5, so g(1) = (1 + 1/3)(1 + 1/4)(1 + 1/5) = 4/3 · 5/4 · 6/5 = 2. Now score the choices: (A) gives (1+12+47+60)/60 = 120/60 = 2; (B) gives 120; (C) gives 120/3600 = 1/30; (D) gives 119/3600; (E) gives 120/119. Only (A) survives. A units-style check agrees: the reciprocal roots are each about 1/root, so g(1) should be a modest number, not something as large as 1+a+b+c or as tiny as a division by c² — exactly one factor of c belongs in the denominator.

💡Key takeaway

You never need the roots themselves. Name them r, s, t, write g(1) = (r-1)(s-1)(t-1)/rst, then notice the top is -f(1) = -(1+a+b+c) and the bottom is -c — so g(1) = (1+a+b+c)/c. Shortcut worth keeping: flipping every root of a polynomial just reverses its coefficient list.

Name the roots of f
Build g and evaluate at 1
Clear the reciprocals
Recognize the numerator as -f(1)
Get the denominator from Vieta
Combine and read off the choice