AMC 10 · 2010 · #23

Grade 9 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta
function-compositionpolynomial-rootsquadratic-equationsvieta-formulas symmetry-argumentconvert-to-algebra ↑ Prerequisites: quadratic-equationsfunction-composition
📏 Long solution 💡 3 insights
Problem
Two hidden upward parabolas are known only through where their composites vanish. Find the sum of their lowest values.

Pick an answer.

(A)
-100
(B)
-82
(C)
-73
(D)
-64
(E)
0
How to solve
Strategy Organize Information in More Ways

Tool #15 (Organize Information in More Ways): a quartic with four given roots looks like a lot of data, but "P(Q(x))=0" says something much simpler — Q(x) has landed on a root of P. Re-read that way, one quartic becomes two quadratic equations Q(x)=r₁ and Q(x)=r₂, and the four listed numbers are just their solutions. Tool #4 (Introduce a Variable): the question asks for the two minimum values, so name them p and q up front and write P(x)=(x-h)²+p, Q(x)=(x-k)²+q. Then every fact given becomes an equation in h,k,p,q, and the target p+q is one of the unknowns rather than something to be dug out at the end. Tool #3 (Eliminate Possibilities): the four zeros must be split into two pairs, one pair per quadratic equation, and only one of the three possible splits is compatible with both equations sharing the vertex of Q — so the split is forced, not chosen. Tool #6 (Guess and Check): forcing values only proves that nothing else can work; building the actual pair P,Q and checking all eight zeros proves that something does.

1STEP 1

Read the composite as two equations

A vanishing composite means the inner value hits a root.

P(Q(x))=(Q(x)-r₁)(Q(x)-r₂), Q(P(x))=(P(x)-s₁)(P(x)-s₂)
2STEP 2

Name the two minimum values

Vertex form names each lowest value.

P(x)=(x-h)²+p, Q(x)=(x-k)²+q; Q(x)=r ⇔ (x-k)²=r-q ⇔ x=k±√(r-q)
3STEP 3

Add all four zeros to find each vertex

Averaging the four zeros finds each vertex.

4k=-76 → k=-19; 4h=-216 → h=-54
4STEP 4

The pairing is forced by symmetry

Symmetry forces which zeros pair together.

{r₁,r₂}={q+16, q+4}, {s₁,s₂}={p+25, p+9}
5STEP 5

Close the loop with the sum of roots

The sum of roots closes the loop.

(q+16)+(q+4)=-108 → q=-64; (p+25)+(p+9)=-38 → p=-36; p+q=-100
6STEP 6

Check that such P and Q exist

Such parabolas exist, so the sum is -100.

P(x)=(x+48)(x+60), Q(x)=(x+11)(x+27); Q(x)=-48 → x∈{-23,-15}, Q(x)=-60 → x∈{-21,-17}; P(x)=-11 → x∈{-59,-49}, P(x)=-27 → x∈{-57,-51}
Answer
-100
Three internal checks agree. First, the required gaps came out as perfect squares — 16,4 on one side and 25,9 on the other — which they had to, since a distance from the vertex is √(r-q); had the zeros been spaced so that these were not squares of the observed distances, no such P,Q would exist. Second, the constant terms cross-check: P(x)=x²+108x+2880 has r₁r₂=(-48)(-60)=2880, and the roots of P predicted in Step 4 are q+16=-48 and q+4=-60; likewise Q(x)=x²+38x+297 has s₁s₂=(-11)(-27)=297, matching p+25=-11 and p+9=-27. Third, choice (E) 0 can be ruled out before any computation: P(Q(x)) has four distinct real zeros, so P must take at least two distinct real values that Q actually attains, forcing P to have two distinct real roots and hence a strictly negative minimum — and the same for Q. The distractors are all near-misses of the real work: (D) -64 is the minimum of Q alone, stopping one step early, and (C) -73 is h+k=-54+(-19), the sum of the two vertex locations rather than the two minimum values — the easiest way to answer the wrong question here.
💡Key takeaway

If P(Q(x))=0, then Q(x) must have landed on a root of P — and since a parabola is a mirror, the x values that land there always come in pairs balanced around its vertex.

  • Read the composite as two equations
  • Name the two minimum values
  • Add all four zeros to find each vertex
  • The pairing is forced by symmetry
  • Close the loop with the sum of roots
  • Check that such P and Q exist

More like this

Same archetype — closest grade level first.