Difference between revisions of "1963 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
Notice: The following words are Chinese. | Notice: The following words are Chinese. | ||
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首先,我们可以将以上5个方程相加,得到: | 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) | + | |
| − | 当<math>x_1+x_2+x_3+x_4+x_5 | + | <math>2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)</math> |
| − | <math>x_1=x_2=x_3=x_4=x_5=0</math> | + | |
| + | 当<math>x_1+x_2+x_3+x_4+x_5=0</math>时,因为<math>x_1,x_2,x_3,x_4,x_5</math>关于原方程组轮换对称,所以 | ||
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| + | <math>x_1=x_2=x_3=x_4=x_5=0</math> | ||
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若反之,则方程两边同除以<math>(x_1+x_2+x_3+x_4+x_5)</math>,得到<math>y=2</math>,显然解为 | 若反之,则方程两边同除以<math>(x_1+x_2+x_3+x_4+x_5)</math>,得到<math>y=2</math>,显然解为 | ||
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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> | ||
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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,最终答案为<math>x_1=x_2=x_3=x_4=x_5</math> | ||
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==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 03:16, 21 February 2010
Problem
Find all solutions 
of the system
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 
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:
当
时,因为关于原方程组轮换对称,所以
若反之,则方程两边同除以
,得到
,显然解为
综上所述,最终答案为
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 | ||
