Difference between revisions of "1963 IMO Problems/Problem 4"
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Can someone change this answer so it's correct? | Can someone change this answer so it's correct? | ||
− | + | 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. | ||
==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 01:04, 2 January 2023
Contents
[hide]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.
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 |