1990 USAMO Problems/Problem 2

Problem

A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows: $\begin{align*} f_1(x) &= \sqrt {x^2 + 48}, \quad \text{and} \\ f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1. \end{align*}$ (Recall that $\sqrt {\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$ , find all real solutions of the equation $\, f_n(x) = 2x \,$ .

Solution

We define $f_0(x) = 8$ . Then the recursive relation holds for $n=0$ , as well.

Since $f_n (x) \ge 0$ for all nonnegative integers $n$ , it suffices to consider nonnegative values of $x$ .

We claim that the following set of relations hold true for all natural numbers $n$ and nonnegative reals $x$ : $\begin{align*} f_n(x) &< 2x \text{ if }x>4 ; \\ f_n(x) &= 2x \text{ if }x=4 ; \\ f_n(x) &> 2x \text{ if }x<4 . \end{align*}$ To prove this claim, we induct on $n$ . The statement evidently holds for our base case, $n=0$ .

Now, suppose the claim holds for $n$ . Then $\begin{align*} f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2+12x} < \sqrt{4x^2} = 2x, \text{ if } x>4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} = \sqrt{x^2 + 12x} = \sqrt{4x^2} = 2x, \text{ if } x=4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} > \sqrt{x^2+12x} > \sqrt{4x^2} = 2x, \text{ if } x<4 . \end{align*}$ The claim therefore holds by induction. It then follows that for all nonnegative integers $n$ , $x=4$ is the unique solution to the equation $f_n(x) = 2x$ . $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1990 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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1990 USAMO Problems/Problem 2

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