Difference between revisions of "1987 OIM Problems/Problem 1"
Archieguan (talk | contribs) |
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== Solution == | == Solution == | ||
− | {{ | + | We have the following equations: |
+ | <cmath>(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x</cmath> | ||
+ | <cmath>(2) \left[ f\left(\frac{1-x}{1+x}\right) \right]^2f\left( x \right)=64 \cdot \frac{1-x}{1+x}</cmath> | ||
+ | Multiplying <math>(1)</math> and <math>(2)</math>, we have | ||
+ | <cmath>(3) f(x)f\left(\frac{1-x}{1+x}\right) = 16 \cdot \sqrt[3]{x\cdot \frac{1-x}{1+x}}</cmath> | ||
+ | Dividing <math>(1)</math> by <math>(3)</math> gives | ||
+ | <cmath>f(x) = 4 \cdot \sqrt[3]{x^2\cdot \frac{1+x}{1-x}}</cmath> | ||
+ | Checking to see if it works… | ||
+ | <cmath>(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right) = 64 \cdot \sqrt[3]{x^4 \cdot \left(\frac{1+x}{1-x}\right)^2 \cdot \left(\frac{1-x}{1+x}\right)^2 \cdot \frac{1}{x}} = 64x</cmath> | ||
+ | ~Archieguan | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe2.htm | https://www.oma.org.ar/enunciados/ibe2.htm |
Latest revision as of 00:25, 23 October 2024
Problem
Find all such that: for , , ,
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
We have the following equations: Multiplying and , we have Dividing by gives Checking to see if it works… ~Archieguan