2012 AIME II Problem 11

Question

Let

f_1(x) = \frac{2}{3} - \frac{3}{3x + 1},

and for

n \ge 2,

define

f_n(x) = f_1(f_{n-1}(x)).

The value of

x

that satisfies

f_{1001}(x) = x - 3

can be expressed in the form

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Write

f_1

as a single fraction and compute

f_2

and

f_3

by hand — a pattern appears

f_3(x) = x,

so

f_{1001} = f_2;

setting

f_2(x) = x - 3

gives a quadratic with a double root

Combining fractions,

f_1(x) = \frac{2(3x + 1) - 9}{3(3x + 1)} = \frac{6x - 7}{9x + 3}.

Composing once,

f_2(x) = \frac{6 f_1(x) - 7}{9 f_1(x) + 3} = \frac{-3x - 7}{9x - 6},

and composing again gives

f_3(x) = x.

So the iteration is periodic with period

3.

Since

1001 \equiv 2 \pmod{3},

we have

f_{1001} = f_2,

and the equation becomes

\frac{-3x - 7}{9x - 6} = x - 3,

that is

9x^2 - 33x + 18 = -3x - 7,

or

9x^2 - 30x + 25 = (3x - 5)^2 = 0.

The unique solution is

x = \frac{5}{3},

so

m + n = 5 + 3 = 8.

Problem 11 in other years