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

(Solution)
(Solution)
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<math>x_1=x_2=x_3=x_4=x_5</math>
 
<math>x_1=x_2=x_3=x_4=x_5</math>
  
综上所述,最终答案为<math>x_1=x_2=x_3=x_4=x_5</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>
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1963|num-b=3|num-a=5}}
 
{{IMO box|year=1963|num-b=3|num-a=5}}

Revision as of 03:18, 21 February 2010

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}$ (Error compiling LaTeX. Unknown error_msg)

where $y$ is a parameter.

Solution

Notice: The following words are Chinese.

首先,我们可以将以上5个方程相加,得到:

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

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

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