AMC 10 · 2012 · #25
Grade 11 algebraPick an answer.
Nothing here can be solved by algebra: f is defined by absolute values of fractional parts, so nf(xf(x)) = x is really thousands of little quadratic equations glued together. The move that makes it finite is Tool #1 (Draw a Diagram): divide by n to get f(xf(x)) = x/n, which asks where a fast zigzag meets a nearly flat line. Tool #14 (Extreme Principle) uses 0 ≤ f ≤ 1 to trap every solution inside [0, n], so the picture only has to be drawn once, over a finite window. Tool #4 (Introduce a Variable) names the inner function g(x) = xf(x), and Tool #7 (Identify Subproblems) chops [0,n] into half-unit windows on which g is a single monotone piece of a parabola — that is the fact that lets each window be counted independently. Inside a window the counting is a clean one-crossing-per-swing argument, and it must be proved, not assumed: the zigzag rises and falls with slope at least 2 while the line has slope 1/n ≤ 1, so crossings cannot be missed or doubled. Tool #5 (Look for a Pattern) sums the per-window counts 2, 2, 4, 4, 6, … into a closed form, and Tool #3 (Eliminate Possibilities) reads off the smallest n clearing 2012.
Trap every solution inside [0,n]
Every solution is trapped inside a bounded range.
Dividing by n turns an unbounded-looking equation into a line that rises from 0 to 1 over exactly the stretch where the other side is allowed to live.
9.A-CED.A.3Extreme PrincipleRead the shape of f: a row of V's
The function is a row of identical V shapes.
Every half-unit of input to f buys exactly one full sweep of the output from floor to ceiling, or ceiling to floor.
9.F-IF.B.4Draw A DiagramName the inner function g(x) = xf(x)
Naming the inner function makes the nesting readable.
Multiplying the V-wave by x pins its zeros at the half-odd-integers but lets the peaks grow taller and taller.
9.F-BF.A.1Introduce A Variableg is monotone on each half-unit window
It runs one way on each half-unit window.
Both parabola pieces have their turning point at or before the start of their window, so inside the window each one only goes one way.
9.A-REI.B.4Identify SubproblemsExactly one crossing per swing
Each sweep gives exactly one crossing.
The zigzag climbs at least twice as fast as the line and covers the line's whole range each swing, so it must cut across once and cannot double back for a second cut.
11.A-REI.D.11Draw A DiagramThe first half-unit contributes nothing
The first half-unit contributes nothing.
Near the origin the inner value is squashed so small that the outer V never gets down off its peak, and the line is still far too low to reach it.
9.F-IF.B.4Extreme PrincipleAdd up the swings: 2n²
Adding the sweeps gives a clean square formula.
Each new unit of x adds two more windows whose swing counts grow steadily, and the running total of an arithmetic list like that is a perfect square times two.
Each new unit adds two more windows whose swing counts grow steadily, and such a running total is a square.
▸ Why?
The counts climb by the same fixed step from window to window, so they form an evenly spaced list.
▸ Why?
Pairing the first with the last gives the same total as pairing inward, so the sum is a single product.
Pick the smallest n
The smallest multiplier is 32, choice (C).
The count grows like a square, so finding the threshold is just squeezing 1006 between two consecutive perfect squares.
9.A-CED.A.1Eliminate PossibilitiesRewrite it as one fast zigzag against one slow straight line: every time the zigzag makes a full trip from 0 up to 1 it has to cross the line exactly once, so counting solutions is just counting trips.
- Trap every solution inside [0,n]
- Read the shape of f: a row of V's
- Name the inner function g(x) = xf(x)
- g is monotone on each half-unit window
- Exactly one crossing per swing
- The first half-unit contributes nothing
- Add up the swings: 2n²
- Pick the smallest n