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Difference between revisions of "1997 JBMO Problems/Problem 2"
(Solution to Problem 2)
 
Line 1: Line 1:
 
== 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>:  
+
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 ==
+
To start, we add the two fractions and simplify.
 +
<cmath>\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*}</cmath>
 +
Dividing both sides by two yields
 +
<cmath>\frac{k}{2} = \frac{x^4 + y^4}{x^4 - y^4}.</cmath>
 +
That means
 +
<cmath>\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*}</cmath>
 +
Dividing both sides by two yields
 +
<cmath>\frac{x^8 + y^8}{x^8 - y^8} = \frac{k^2 + 4}{4k}.</cmath>
 +
That means
 +
<cmath>\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*}</cmath>
 +
 
 +
== 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 (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All JBMO Problems and Solutions
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