Difference between revisions of "1990 USAMO Problems/Problem 2"

Revision as of 17:33, 17 November 2007

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 nonnegative, since the natural root of any number is $\ge 0$ . 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 $n$ .

@@ Line 28: / Line 28: @@
 <cmath>f_{n+1}(x)=\sqrt{x^2+48}</cmath>
-So if 4 is a solution for <math>n=x</math>, it is a solution for <math>n=x+1</math>. From [[induction]], 4 is a solution for all n.
+So if 4 is a solution for <math>n=x</math>, it is a solution for <math>n=x+1</math>. From [[induction]], <math>4</math> is a solution for all <math>n</math>.
-{{solution}}
 ==See also==

1990 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "1990 USAMO Problems/Problem 2"

Revision as of 17:33, 17 November 2007

Problem

Solution

See also