Difference between revisions of "1963 IMO Problems/Problem 4"
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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 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,若,最终答案为,否则答案为
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 |