AMC 10 · 2011 · #23

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

complex-numbersfunction-compositionabsolute-value convert-to-algebracasework ↑ Prerequisites: complex-numbers

📏 Long solution 💡 4 insights

Problem

Applying one fractional map four times returns every legal input unchanged. Find the spread of the possible sizes.

Pick an answer.

(A)

(B)

$\sqrt{2}-1$

(C)

$\sqrt{3}-1$

(D)

(E)

How to solve

Strategy Organize Information in More Ways

Writing out f(f(f(f(z)))) as one giant fraction is hopeless. Every map of the shape (pz+q)/(rz+s) is completely described by its four coefficients, and plugging one into another combines those coefficients by a fixed row-times-column rule. Storing the map as four numbers instead of as a fraction turns four nested substitutions into two rounds of the same small computation. Once the four final coefficients are in hand, the condition g(g(z))=z becomes three equations, and looking at their structure shows two of the three are automatic. That leaves a single equation, which splits into two branches; the extreme principle and the triangle inequality then bound |b|, and explicit values of a confirm both ends are actually reached.

1STEP 1

Store a map as four numbers

The map lives as four numbers.

F(G(z))=(pp'+qr')z+(pq'+qs')/(rp'+sr')z+(rq'+ss')

2STEP 2

Throw out the collapsing case

One degenerate case is thrown out first.

b=a → f≡ 1 → g(g(z)) constant ≠ z

3STEP 3

Run the rule twice

Running it twice keeps the same shape.

g(z)=(a+1)z+a(b+1)/(b+1)z+(a+b²), h(z)=(Pz+Q)/(Rz+S) ; P=(a+1)²+a(b+1)², Q=a(b+1)(b²+2a+1) ; R=(b+1)(b²+2a+1), S=a(b+1)²+(a+b²)²

4STEP 4

Turn the identity into equations

The identity condition becomes three equations.

(Pz+Q)/(Rz+S)=z ⟺ Rz²+(S-P)z-Q=0 ⟺ R=0, P=S, Q=0, P ≠ 0

5STEP 5

Two of the three come free

Two of them follow free from the first.

Q=aR, P-S=(a+1)²-(a+b²)²=(1-b)R ; → R=0 alone gives Q=0 and P=S

6STEP 6

Split the equation, keep P alive

That splits into two branches.

(b+1)(b²+2a+1)=0 ; b=-1: P=(a+1)² ≠ 0⇔ a ≠ -1 ; b²+2a+1=0: P=-((b+1)⁴)/4 ≠ 0⇔ b ≠ -1

7STEP 7

Measure b on each branch

Sizes run between 1 and the root of three.

|b|²=|b²|=|-1-2a|=|1+2a| ; |2a|-|1| ≤ |1+2a| ≤ |1|+|2a| → 1 ≤ |b|² ≤ 3 → 1 ≤ |b| ≤ √(3)

8STEP 8

Show both ends are reached

Both ends are reached, so the spread is √3-1.

a=1: b²=-3, b=± i√(3), |b|=√(3) ; a=-1: b²=1, b=1, |b|=1 ; √(3)-1

Answer

√(3)-1

The answer √(3)-1≈ 0.73 is small, which fits: both branches force |b| ≥ 1, so |b| is trapped in a narrow window and cannot be tiny or huge. Choice (A) 0 would mean |b| is pinned to a single value, but b=-1 and b=i√(3) both work and have different sizes, so (A) fails. Choice (E) 2 would need |b| to reach 3, which would demand |b²|=9 while |1+2a| can never exceed 3. Choice (D) 1 is what you get by confusing |b|² ≤ 3 with |b| ≤ 3 or by stopping at |b²|∈[1,3] and subtracting there — a real trap, since 3-1=2 and √(3)-1 are both on the list. Two independent safety checks passed: numerically substituting a=1, b=i√(3) and a=i, b=-1 into f four times returns the starting z to full precision, and the degenerate corner a=b=-1 only ever produces |b|=1, a value already inside the range, so it cannot shift either endpoint.

💡Key takeaway

A map like (z+a)/(z+b) is just its four coefficients, and composing maps multiplies those coefficients like a grid — once you see that, "apply it four times and get z back" shrinks to the single equation (b+1)(b²+2a+1)=0, and |a|=1 then squeezes |b| between 1 and √(3).

Store a map as four numbers
Throw out the collapsing case
Run the rule twice
Turn the identity into equations
Two of the three come free
Split the equation, keep P alive
Measure b on each branch
Show both ends are reached