Difference between revisions of "1997 JBMO Problems/Problem 2"

(Solution to Problem 2)
 
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== Problem ==
 
== Problem ==
  
''(Cyprus)'' Let <math>\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k</math>. Compute the following expression in terms of <math>k</math>:  
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Let <math>\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k</math>. Compute the following expression in terms of <math>k</math>:  
 
<cmath> E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.  </cmath>
 
<cmath> E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.  </cmath>
  
 
== Solution ==
 
== Solution ==
  
== See also ==
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To start, we add the two fractions and simplify.
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<cmath>\begin{align*}
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k &= \frac{(x^2+y^2)^2 + (x^2-y^2)^2}{x^4-y^4} \\
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&= \frac{2x^4 + 2y^4}{x^4 - y^4}.
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\end{align*}</cmath>
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Dividing both sides by two yields
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<cmath>\frac{k}{2} = \frac{x^4 + y^4}{x^4 - y^4}.</cmath>
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That means
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<cmath>\begin{align*}
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\frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4} &= \frac{k}{2} + \frac{2}{k} \\
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\frac{(x^4 + y^4)^2 + (x^4 - y^4)^2}{x^8 - y^8} &= \frac{k^2 + 4}{2k} \\
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\frac{2x^8 + 2y^8}{x^8 - y^8} &= \frac{k^2 + 4}{2k}.
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\end{align*}</cmath>
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Dividing both sides by two yields
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<cmath>\frac{x^8 + y^8}{x^8 - y^8} = \frac{k^2 + 4}{4k}.</cmath>
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That means
 +
<cmath>\begin{align*}
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\frac{x^8 + y^8}{x^8 - y^8} - \frac{x^8 - y^8}{x^8 + y^8} &= \frac{k^2 + 4}{4k} - \frac{4k}{k^2 + 4} \\
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&= \frac{k^4 + 8k^2 + 16 - 16k^2}{4k(k^2 + 4)} \\
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&= \frac{k^4 - 8k^2 + 16}{4k(k^2 + 4)} \\
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&= \boxed{\frac{(k^2 - 4)^2}{4k(k^2 + 4)}}.
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\end{align*}</cmath>
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 +
== See Also ==
  
 
{{JBMO box|year=1997|num-b=1|num-a=3}}
 
{{JBMO box|year=1997|num-b=1|num-a=3}}
  
 
  [[Category:Intermediate Algebra Problems]]
 
  [[Category:Intermediate Algebra Problems]]

Latest revision as of 13:49, 4 August 2018

Problem

Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$. Compute the following expression in terms of $k$: \[E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.\]

Solution

To start, we add the two fractions and simplify. \begin{align*} k &= \frac{(x^2+y^2)^2 + (x^2-y^2)^2}{x^4-y^4} \\ &= \frac{2x^4 + 2y^4}{x^4 - y^4}. \end{align*} Dividing both sides by two yields \[\frac{k}{2} = \frac{x^4 + y^4}{x^4 - y^4}.\] That means \begin{align*} \frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4} &= \frac{k}{2} + \frac{2}{k} \\ \frac{(x^4 + y^4)^2 + (x^4 - y^4)^2}{x^8 - y^8} &= \frac{k^2 + 4}{2k} \\ \frac{2x^8 + 2y^8}{x^8 - y^8} &= \frac{k^2 + 4}{2k}. \end{align*} Dividing both sides by two yields \[\frac{x^8 + y^8}{x^8 - y^8} = \frac{k^2 + 4}{4k}.\] That means \begin{align*} \frac{x^8 + y^8}{x^8 - y^8} - \frac{x^8 - y^8}{x^8 + y^8} &= \frac{k^2 + 4}{4k} - \frac{4k}{k^2 + 4} \\ &= \frac{k^4 + 8k^2 + 16 - 16k^2}{4k(k^2 + 4)} \\ &= \frac{k^4 - 8k^2 + 16}{4k(k^2 + 4)} \\ &= \boxed{\frac{(k^2 - 4)^2}{4k(k^2 + 4)}}. \end{align*}

See Also

1997 JBMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5
All JBMO Problems and Solutions
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