AMC 10 · 2023 · #22
Grade 11 algebraPick an answer.
An identity that holds for all a and b is an infinite pile of equations, and nothing can be done with the pile as a whole. Tool #9 (Solve an Easier Related Problem) is the way in: replace the universal statement by two or three concrete instances that are easy enough to solve. The instance a=b=0 pins down f(0), the instance b=0 finishes one branch outright, and the instance a=b=x produces a doubling rule f(2x) = 2f(x)²-1. Tool #14 (Extreme Principle) then reads that rule at its floor: a square cannot be negative, so the right side never drops below -1, which is exactly the bound the question is testing. That kills one choice. But a "cannot be" question has a second half, because the four survivors must be shown genuinely reachable. Tool #6 (Guess and Check) supplies them: two constants, a cosine, and an exponential average, each checked against the original identity. Tool #3 (Eliminate Possibilities) is the bookkeeping that pairs the bound with the four witnesses so that exactly one choice is left standing.
Substitute zero
The value at zero has two options.
Feeding the rule its easiest input turns an equation about an unknown function into an ordinary quadratic about one unknown number.
9.A-REI.B.4Solve An Easier Related ProblemOne world collapses to zero
In one case the function is identically zero.
A zero at the origin multiplies the whole right side away, forcing every other value to be zero as well.
9.A-REI.B.3Solve An Easier Related ProblemSet the variables equal
Setting them equal gives a doubling rule.
Choosing the two free inputs to be equal is what makes a square appear, and squares are the only thing in this problem that carry a built-in inequality.
9.F-BF.A.1Solve An Easier Related ProblemThe floor from a square
The square creates a floor.
The half-input is the lever: f(1) is a square doubled and shifted down by one, and a square has nowhere below zero to go.
A square has nowhere below zero to go, so a doubled square shifted down has a hard floor.
▸ Why?
A number and its opposite have the same square, so no square can ever be negative.
▸ Why?
A quantity bounded below at zero stays bounded below after doubling and shifting.
Both worlds share the floor
Both cases share the same floor.
One bound covers both branches, so the whole set of achievable values sits above a single line at -1.
9.A-CED.A.3Eliminate PossibilitiesConstants deliver two values
Constant functions deliver two of them.
Constant functions are the free test cases, and here two of the five choices fall out of them immediately.
9.F-IF.A.2Guess And CheckCosine delivers another
A cosine delivers another.
The problem's equation is the product-to-sum identity wearing a disguise, so cosine was always a solution waiting to be named.
11.F-TF.C.9Guess And CheckAn exponential average finishes
What remains, below the floor, is negative two.
Averaging t^x with its reciprocal makes the cross terms pair up perfectly, which is the same algebra that made cosine work.
9.A-SSE.A.2Guess And CheckWhen a rule must hold for every a and b, you get to pick them: setting a=b turns this rule into f(2x) = 2f(x)²-1, and a square can never drag that below -1.
- Substitute zero to pin down f(0)
- The f(0)=0 world collapses to zero
- Set a equal to b to get a doubling rule
- A square cannot push below negative one
- Both worlds share the same floor
- Two constants deliver 0 and 1
- Cosine delivers negative one
- An exponential average delivers two