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Difference between revisions of "1990 USAMO Problems/Problem 2"
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==Solution==
 
==Solution==
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x must be positive, since if x is negative, we would be taking a negative square root.
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Solving for n=1, we get x=4 and only 4. We solve for n=2:
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<math>2x=\sqrt{x^2+6\sqrt{x^2+48}}</math>
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<math>3x^2=6\sqrt{x^2+48}</math>
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<math>x^4=4x^2+192</math>
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<math>x^2=\dfrac{4+28}{2}=16</math>
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<math>x=4</math>
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We get x=4 again. We can conjecture that x=4 is the only solution.
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Plugging 2x=8 into <math>f_n(x)</math>, we get
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<math>f_{n+1}(x)=\sqrt{x^2+48}</math>
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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.
  
 
{{solution}}
 
{{solution}}
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==See Also==
 
==See Also==
  
 
{{USAMO box|year=1990|num-b=1|num-a=3}}
 
{{USAMO box|year=1990|num-b=1|num-a=3}}

Revision as of 07:58, 19 October 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 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.

This problem needs a solution. If you have a solution for it, please help us out by adding it.


See Also

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