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Difference between revisions of "1963 IMO Problems/Problem 4"

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==Problem==
 
==Problem==
 
Find all solutions <math>x_1,x_2,x_3,x_4,x_5</math> of the system
 
Find all solutions <math>x_1,x_2,x_3,x_4,x_5</math> of the system
−
<center><math>\begin{eqnarray}
+
<cmath>\begin{eqnarray*}
 
x_5+x_2&=&yx_1\\
 
x_5+x_2&=&yx_1\\
 
x_1+x_3&=&yx_2\\
 
x_1+x_3&=&yx_2\\
 
x_2+x_4&=&yx_3\\
 
x_2+x_4&=&yx_3\\
 
x_3+x_5&=&yx_4\\
 
x_3+x_5&=&yx_4\\
−
x_4+x_1&=&yx_5,\end{eqnarray}</math></center>
+
x_4+x_1&=&yx_5,\end{eqnarray*}</cmath>
 
where <math>y</math> is a parameter.
 
where <math>y</math> is a parameter.
 +
 +
==Video Solution==
 +
https://youtu.be/n02OTRpHO20?si=p7WSt-NZNU-LsAPe [Video Solution by little-fermat]
  
 
==Solution==
 
==Solution==
Line 25: Line 28:
  
 
综上所述,若<math>y=2</math>,最终答案为<math>x_1=x_2=x_3=x_4=x_5</math>,否则答案为<math>x_1=x_2=x_3=x_4=x_5=0</math>
 
综上所述,若<math>y=2</math>,最终答案为<math>x_1=x_2=x_3=x_4=x_5</math>,否则答案为<math>x_1=x_2=x_3=x_4=x_5=0</math>
 +
 +
 +
The solution in English (translated by Google Translate):
 +
 +
First of all, we can add the five equations to get:
 +
 +
<math>2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)</math>
 +
 +
When <math>x_1+x_2+x_3+x_4+x_5=0</math>, Because <math>x_1,x_2,x_3,x_4,x_5</math> is symmetric in the original equations,
 +
 +
<math>x_1=x_2=x_3=x_4=x_5=0</math>
 +
 +
Otherwise, dividing both sides by <math>(x_1+x_2+x_3+x_4+x_5</math>, we get <math>y=2</math>, and clearly
 +
 +
<math>x_1=x_2=x_3=x_4=x_5</math>
 +
 +
Summarizing, if <math>y=2</math>, then the answer is of the form <math>x_1=x_2=x_3=x_4=x_5</math>. Otherwise, <math>x_1=x_2=x_3=x_4=x_5=0</math>.
 +
 +
==Mistake==
 +
 +
While doing this question, I found out that the answer is actually wrong, <math>y</math> can equal <math>\frac{-1-\sqrt{5}}{2}</math> and <math>\frac{-1+\sqrt{5}}{2}</math> and still produce an infinite number of solutions in the form <math>(n,n,-\frac{ny}{y+1},-2ny,-\frac{ny}{y+1})</math> where <math>n</math> is a real number and the set is cyclic (Ex: The set can correspond to <math>(x_{1},x_{2},x_{3},x_{4},x_{5})</math> or <math>(x_{2},x_{3},x_{4},x_{5},x_{1})</math>, either works. Order matters, but not starting position.). For example, if <math>n=1</math> and <math>y=\frac{-1+\sqrt{5}}{2}</math> the set will be <math>(1,1,\frac{\sqrt{5}-3}{2},1-\sqrt{5},\frac{\sqrt{5}-3}{2})</math>, 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?
 +
 +
Edit: 亲爱的中国盆友，我找到错误了。
 +
If <math>x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0</math>, y can be anything(不一定要轮换对称).
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1963|num-b=3|num-a=5}}
 
{{IMO box|year=1963|num-b=3|num-a=5}}

Latest revision as of 23:51, 14 September 2023

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.

Video Solution

https://youtu.be/n02OTRpHO20?si=p7WSt-NZNU-LsAPe [Video Solution by little-fermat]

Solution

Notice: The following words are Chinese.

首先,我们可以将以上5个方程相加,得到:

$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$

当$x_1+x_2+x_3+x_4+x_5=0$时,因为$x_1,x_2,x_3,x_4,x_5$关于原方程组轮换对称,所以

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

若反之,则方程两边同除以$(x_1+x_2+x_3+x_4+x_5)$,得到$y=2$,显然解为

$x_1=x_2=x_3=x_4=x_5$

综上所述,若$y=2$,最终答案为$x_1=x_2=x_3=x_4=x_5$,否则答案为$x_1=x_2=x_3=x_4=x_5=0$


The solution in English (translated by Google Translate):

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

$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$

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,

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

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

$x_1=x_2=x_3=x_4=x_5$

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

Mistake

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?

Edit: 亲爱的中国盆友，我找到错误了。 If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0$, y can be anything(不一定要轮换对称).

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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