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

Revision as of 02:18, 21 February 2010

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.

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

@@ Line 24: / Line 24: @@
 <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==
 {{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
All IMO Problems and Solutions