1963 IMO Problems/Problem 4

Revision as of 18:57, 10 March 2015 by Mathgeek2006 (talk | contribs) (Problem)


Find all solutions $x_1,x_2,x_3,x_4,x_5$ of the system \begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*} where $y$ is a parameter.


Notice: The following words are Chinese.








The solution in English (translated by Google Translate):

First of all, we can add the five equations to get:


When $x_1+x_2+x_3+x_4+x_5=0$, Because $x_1,x_2,x_3,x_4,x_5$ is symmetric in the original equations,


Otherwise, dividing both sides by $(x_1+x_2+x_3+x_4+x_5$, we get $y=2$, and clearly


Summarizing, if $y=2$, then the answer is of the form $x_1=x_2=x_3=x_4=x_5$. Otherwise, $x_1=x_2=x_3=x_4=x_5=0$.

See Also

1963 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions
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