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Difference between revisions of "1990 USAMO Problems/Problem 2"
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==Problem==
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== Problem ==
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A sequence of [[function]]s <math>\, \{f_n(x) \} \,</math> is defined [[recursion|recursively]] as follows:
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<cmath> \begin{align*}
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f_1(x) &= \sqrt {x^2 + 48}, \quad \text{and} \\
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f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1.
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\end{align*} </cmath>
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(Recall that <math>\sqrt {\makebox[5mm]{}}</math> is understood to represent the positive [[square root]].) For each positive integer <math>n</math>, find all real solutions of the equation <math>\, f_n(x) = 2x \,</math>.
  
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A sequence of functions <math>\, \{f_n(x) \} \,</math> is defined recursively as follows:
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== Solution ==
  
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<math>
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We define <math>f_0(x) = 8</math>.  Then the recursive relation holds for <math>n=0</math>, as well.
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f_1(x) = \sqrt {x^2 + 48}, \quad \mbox{and} \\
 
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f_{n + 1}(x) = \sqrt {x^2 + 6f_n(x)} \quad \mbox{for } n \geq 1.
 
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</math>
 
  
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(Recall that <math>\sqrt {\makebox[5mm]{}}</math> is understood to represent the positive square root.) For each positive integer <math>n</math>, find all real solutions of the equation <math>\, f_n(x) = 2x \,</math>.
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Since <math>f_n (x) \ge 0</math> for all nonnegative integers <math>n</math>, it suffices to consider nonnegative values of <math>x</math>.
  
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We claim that the following set of relations hold true for all natural numbers <math>n</math> and nonnegative reals <math>x</math>:
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<cmath> \begin{align*}
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f_n(x) &< 2x \text{ if }x>4 ; \\
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f_n(x) &= 2x \text{ if }x=4 ; \\
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f_n(x) &> 2x \text{ if }x<4 .
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\end{align*} </cmath>
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To prove this claim, we induct on <math>n</math>.  The statement evidently holds for our base case, <math>n=0</math>.
  
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==Solution==
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Now, suppose the claim holds for <math>n</math>.  Then
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<cmath> \begin{align*}
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f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2+12x} < \sqrt{4x^2} = 2x, \text{ if } x>4 ; \\
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f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} = \sqrt{x^2 + 12x} = \sqrt{4x^2} = 2x, \text{ if } x=4 ; \\
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f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} > \sqrt{x^2+12x} > \sqrt{4x^2} = 2x, \text{ if } x<4 .
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\end{align*} </cmath>
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The claim therefore holds by induction.  It then follows that for all nonnegative integers <math>n</math>, <math>x=4</math> is the unique solution to the equation <math>f_n(x) = 2x</math>.  <math>\blacksquare</math>
  
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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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{{alternate solutions}}
  
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<math>2x=\sqrt{x^2+6\sqrt{x^2+48}}</math>
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==See Also==
  
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<math>3x^2=6\sqrt{x^2+48}</math>
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{{USAMO box|year=1990|num-b=1|num-a=3}}
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356624#p356624 Discussion on AoPS/MathLinks]
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<math>x^4=4x^2+192</math>
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{{MAA Notice}}
  
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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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[[Category:Olympiad Algebra Problems]]
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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.
 
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{{solution}}
 
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==See Also==
 
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{{USAMO box|year=1990|num-b=1|num-a=3}}
 

Latest revision as of 18:14, 18 July 2016

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.

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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