1997 JBMO Problems/Problem 2: Difference between revisions
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Rockmanex3 (talk | contribs) Solution to Problem 2 |
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== Problem == | == Problem == | ||
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> | |||
== Solution == | == Solution == | ||
== See | 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]] | |||
Latest revision as of 12:49, 4 August 2018
Problem
Let 
. Compute the following expression in terms of 
:
![\[E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.\]](http://latex.artofproblemsolving.com/c/9/9/c99e935bc847cdcaee76b4a8af84db262067eca7.png)
Solution
To start, we add the two fractions and simplify.
Dividing both sides by two yields
![\[\frac{k}{2} = \frac{x^4 + y^4}{x^4 - y^4}.\]](http://latex.artofproblemsolving.com/b/9/e/b9e114b045541e6bad06a83b2a0d20d1197f576f.png)
That means
Dividing both sides by two yields
![\[\frac{x^8 + y^8}{x^8 - y^8} = \frac{k^2 + 4}{4k}.\]](http://latex.artofproblemsolving.com/f/1/4/f1430cd7306cc8d7ff463ec4bcc051ec7d5dcf31.png)
That means
See Also
| 1997 JBMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
