1990 USAMO Problems/Problem 2

Problem

A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows:

$f_1(x) = \sqrt {x^2 + 48}, \quad \mbox{and} \\ f_{n + 1}(x) = \sqrt {x^2 + 6f_n(x)} \quad \mbox{for } n \geq 1.$

(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

x must be positive, since if x is negative, we would be taking a negative square root.

Solving for n=1, we get x=4 and only 4. We solve for n=2:

$2x=\sqrt{x^2+6\sqrt{x^2+48}}$

$3x^2=6\sqrt{x^2+48}$

$x^4=4x^2+192$

$x^2=\dfrac{4+28}{2}=16$

$x=4$

We get x=4 again. We can conjecture that x=4 is the only solution.

Plugging 2x=8 into $f_n(x)$ , we get

$f_{n+1}(x)=\sqrt{x^2+48}$

So if 4 is a solution for n=x, it is a solution for n=x+1. From induction, 4 is a solution for all n.

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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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