Difference between revisions of "1963 IMO Problems/Problem 4"
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==Problem== | ==Problem== | ||
Find all solutions <math>x_1,x_2,x_3,x_4,x_5</math> of the system | Find all solutions <math>x_1,x_2,x_3,x_4,x_5</math> of the system | ||
− | < | + | <cmath>\begin{eqnarray*} |
x_5+x_2&=&yx_1\\ | x_5+x_2&=&yx_1\\ | ||
x_1+x_3&=&yx_2\\ | x_1+x_3&=&yx_2\\ | ||
x_2+x_4&=&yx_3\\ | x_2+x_4&=&yx_3\\ | ||
x_3+x_5&=&yx_4\\ | x_3+x_5&=&yx_4\\ | ||
− | x_4+x_1&=&yx_5,\end{eqnarray}</ | + | x_4+x_1&=&yx_5,\end{eqnarray*}</cmath> |
where <math>y</math> is a parameter. | where <math>y</math> is a parameter. | ||
Revision as of 18:57, 10 March 2015
Problem
Find all solutions of the system 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
The solution in English (translated by Google Translate):
First of all, we can add the five equations to get:
When , Because is symmetric in the original equations,
Otherwise, dividing both sides by , we get , and clearly
Summarizing, if , then the answer is of the form . Otherwise, .
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 |