1990 USAMO Problems/Problem 2

Problem

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 $x$ : $\begin{align*}f_n(x) &< 2x \text{ if }x>4 ; \\f_n(x) &= 2x \text{ if }x=4 ; \\f_n(x) &> 2x \text{ if }x&...$ 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 $\begin{align*}f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2+12x} < \sqrt{4x^2} = 2x, \text{ if } x>4 ; \\f_{n+1...$ 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)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5	Followed by Problem 3
All USAMO Problems and Solutions