AMC 10 · 2002 · #25

Grade 9 algebra
mean-median-mode-rangegraph-readingsign-analysis invariant-monovariant ↑ Prerequisites: mean-median-mode-rangegraph-reading
📏 Long solution 💡 5 insights 📊 Diagram
Problem
Start with a polynomial P that has real coefficients. Take all of its nonzero coefficients, compute their mean, and build a new polynomial Q by putting that one mean value in place of every nonzero coefficient. Decide which of the five pictures could be a picture of y = P(x) and y = Q(x) drawn together over -4 ≤ x ≤ 4.

Pick an answer.

(A)
(graph A)
(B)
(graph B)
(C)
(graph C)
(D)
(graph D)
(E)
(graph E)
How to solve
Strategy Eliminate Possibilities

There is no formula to compute here, and P is never given, so the only way in is to find something that is true of P and Q for every possible P and then check the five pictures against it. Averaging destroys almost everything about a list of numbers, so the search is for what it protects. It protects the total: a list of k numbers with mean m adds up to km, which is the same total as before. A coefficient total is not obviously a picture fact, but substituting x = 1 turns any polynomial into the sum of its coefficients, so the protected total becomes a single shared point on both graphs. That is the whole problem: the two curves are forced to meet at x = 1. A second, independent test comes from writing Q out. Every nonzero coefficient of Q is the same number m, so Q is m times a bare sum of powers, and for positive x that sum is positive; this pins down the sign and the direction of Q on the whole right half of every legal picture. Finally, since the question asks what could be, elimination alone is not a finish: the surviving panel needs an explicit P built for it.

1STEP 1

Write down what Q actually is

Q keeps P's exponents with one shared coefficient: Q(x) = m(x^e₁ + … + x^e_k), so both share a degree.

P(x)=Σ_j=1^k c_j x^e_j, m=(c₁+…+c_k)/k, Q(x)=mΣ_j=1^k x^e_j
2STEP 2

The mean protects the total

A mean preserves exactly one thing, the total: k copies of m add to the same sum as P's coefficients.

m+m+…+m_k times = km = c₁+c₂+…+c_k
3STEP 3

Turn the total into a point on both graphs

Substituting x = 1 turns that total into a shared point, forcing P(1) = Q(1) for every P.

P(1)=Σ_j=1^kc_j, Q(1)=km=Σ_j=1^kc_j ⟹ P(1)=Q(1)
4STEP 4

Test the panels at x = 1

Checking where the curves meet, four panels fail to cross at x = 1 and are impossible.

(A): x ≈ ± 3.5, (C): x ≈ -2.7, x ≈ 0^-, (D): x ≈ -1.7, 2.6, (E): x ≈ -1.8
5STEP 5

Confirm with a second test that never looks at x = 1

A second test confirms it: for positive x, Q keeps m's sign and moves in one steady direction.

x > 0 → x^e₁+…+x^e_k > 0 → signQ(x)=signm, Q'(x)=mΣ_j e_j x^e_j-1
6STEP 6

Build a polynomial for the panel that is left

The witness P(x) = 2x⁴-3x²-3x-4 gives m = -2 and reproduces the survivor, so the answer is (B).

P(x)=2x⁴-3x²-3x-4, m=(2-3-3-4)/4=-2, Q(x)=-2(x⁴+x²+x+1), P(1)=Q(1)=-8
Answer
(graph B)
The witness reproduces the panel in detail, not just in outline. Subtracting, P(x) - Q(x) = 4x⁴ - x² - x - 2 = (x-1)(4x³+4x²+3x+2), and the cubic factor has derivative 12x²+8x+3, whose discriminant 64 - 144 is negative, so the cubic is strictly increasing and has exactly one real root, near x ≈ -0.82. The two curves therefore cross exactly twice, at x ≈ -0.82 and at x = 1 — one crossing left of the origin and one at the marked tick, which is what the panel shows. The shapes match too: P'(x) = 8x³ - 6x - 3 has a single real root near x ≈ 1.05, so P has exactly one minimum, of value about -8.03, sitting essentially at the crossing point, while Q rises to a maximum of about -1.57 near x ≈ -0.39 and falls away on both sides. Every one of these features is visible in the panel. The two tests also agree with each other rather than merely coexisting: the x = 1 test rules out (A), (C), (D) and (E), and the sign-and-direction test independently rules out (A), (D) and (E), so the only panel that either test could have spared and the other one condemned is (C), which fails the first test clearly and by a wide margin.
💡Key takeaway

Averaging a list of numbers never changes their total, and plugging in x=1 turns a polynomial into the total of its coefficients — so the two curves have no choice but to meet at x=1.

  • Write down what Q actually is
  • The mean protects the total
  • Turn the total into a point on both graphs
  • Test the panels at x = 1
  • Confirm with a second test that never looks at x = 1
  • Build a polynomial for the panel that is left