Difference between revisions of "1987 OIM Problems/Problem 1"
(Created page with "== Problem == Find all <math>f(x)</math> such that: <cmath>\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x</cmath> for <math>x \ne 0</math>, <math>x \ne 1</math>, <ma...") |
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<cmath>\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x</cmath> | <cmath>\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x</cmath> | ||
for <math>x \ne 0</math>, <math>x \ne 1</math>, <math>x \ne -1</math>, | for <math>x \ne 0</math>, <math>x \ne 1</math>, <math>x \ne -1</math>, | ||
| + | |||
| + | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
== 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 == | ||
| + | https://www.oma.org.ar/enunciados/ibe2.htm | ||
Latest revision as of 00:25, 23 October 2024
Problem
Find all 
such that:
![\[\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x\]](http://latex.artofproblemsolving.com/0/c/7/0c7c2664c02b3db74a07274075a0eed0992d5513.png)
for 
, 
, 
,
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
We have the following equations:
![\[(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x\]](http://latex.artofproblemsolving.com/b/d/0/bd0001ecf77a4c09dda2b2f0c75e30519a83449a.png)
![\[(2) \left[ f\left(\frac{1-x}{1+x}\right) \right]^2f\left( x \right)=64 \cdot \frac{1-x}{1+x}\]](http://latex.artofproblemsolving.com/5/d/d/5ddba62ea83aa13266cb4500010ecd3d9db41cd7.png)
Multiplying 
and 
, we have
![\[(3) f(x)f\left(\frac{1-x}{1+x}\right) = 16 \cdot \sqrt[3]{x\cdot \frac{1-x}{1+x}}\]](http://latex.artofproblemsolving.com/3/d/2/3d2600a4034fcfea75d589013d308c6d5b15ed10.png)
Dividing by 
gives
![\[f(x) = 4 \cdot \sqrt[3]{x^2\cdot \frac{1+x}{1-x}}\]](http://latex.artofproblemsolving.com/5/f/3/5f3bde7e0fd1dc427623a70556b6e649def2942f.png)
Checking to see if it works…
![\[(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right) = 64 \cdot \sqrt[3]{x^4 \cdot \left(\frac{1+x}{1-x}\right)^2 \cdot \left(\frac{1-x}{1+x}\right)^2 \cdot \frac{1}{x}} = 64x\]](http://latex.artofproblemsolving.com/9/2/9/9295dd47da421d6010235f52cd81b94601b8b37f.png)
~Archieguan
