1963 IMO Problems/Problem 4

Revision as of 16:54, 19 November 2020 by Circlegeometrygang (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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, $x_1=x_2=x_3=x_4=x_5=0$.


While doing this question, I found out that the answer is actually wrong, $y$ can equal $\frac{-1-\sqrt{5}}{2}$ and $\frac{-1+\sqrt{5}}{2}$ and still produce an infinite number of solutions in the form $(n,n,-\frac{ny}{y+1},-2ny,-\frac{ny}{y+1})$ where $n$ is a real number and the set is cyclic (Ex: The set can correspond to $(x_{1},x_{2},x_{3},x_{4},x_{5})$ or $(x_{2},x_{3},x_{4},x_{5},x_{1})$, either works. Order matters, but not starting position.). For example, if $n=1$ and $y=\frac{-1+\sqrt{5}}{2}$ the set will be $(1,1,\frac{\sqrt{5}-3}{2},1-\sqrt{5},\frac{\sqrt{5}-3}{2})$, which you can test and find out that it still works even though the set isn't symmetric.

Can someone change this answer so it's correct?

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
Invalid username
Login to AoPS