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

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).

@@ Line 8: / Line 8: @@
 x_4+x_1&=&yx_5,\end{eqnarray*}</cmath>
 where <math>y</math> is a parameter.
+==Video Solution==
+https://youtu.be/n02OTRpHO20?si=p7WSt-NZNU-LsAPe [Video Solution by little-fermat]
 ==Solution==
@@ Line 42: / Line 45: @@
 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==
 {{IMO box|year=1963|num-b=3|num-a=5}}

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