Difference between revisions of "1963 IMO Problems/Problem 4"
Sblnuclear17 (talk | contribs) m (→Mistake) |
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Edit: 亲爱的中国盆友，我找到错误了。 | Edit: 亲爱的中国盆友，我找到错误了。 | ||
| − | If <math>x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0</math>, y can be anything. | + | If <math>x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0</math>, y can be anything(不一定要轮换对称). |
==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 02:07, 2 January 2023
Contents
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, .
Mistake
While doing this question, I found out that the answer is actually wrong, can equal 
and 
and still produce an infinite number of solutions in the form 
where 
is a real number and the set is cyclic (Ex: The set can correspond to 
or 
, either works. Order matters, but not starting position.). For example, if 
and 
the set will be 
, which you can test and find out that it still works even though the set isn't symmetric.
Can someone change this answer so it's correct?
Edit: 亲爱的中国盆友，我找到错误了。
If 
, y can be anything(不一定要轮换对称).
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 | ||
