Difference between revisions of "1963 IMO Problems/Problem 4"
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Summarizing, if <math>y=2</math>, then the answer is of the form <math>x_1=x_2=x_3=x_4=x_5</math>. Otherwise, <math>x_1=x_2=x_3=x_4=x_5=0</math>. | Summarizing, if <math>y=2</math>, then the answer is of the form <math>x_1=x_2=x_3=x_4=x_5</math>. Otherwise, <math>x_1=x_2=x_3=x_4=x_5=0</math>. | ||
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+ | ==Mistake== | ||
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+ | While doing this question, I found out that the answer is actually wrong, <math>y</math> can equal <math>\frac{-1-\sqrt{5}}{2}</math> and <math>\frac{-1+\sqrt{5}}{2}</math> and still produce an infinite number of solutions in the form <math>(n,n,-\frac{ny}{y+1},-2ny,-\frac{ny}{y+1})</math> where <math>n</math> is a real number and the set is cyclic (Ex: The set can correspond to <math>(x_{1},x_{2},x_{3},x_{4},x_{5})</math> or <math>(x_{2},x_{3},x_{4},x_{5},x_{1})</math>, either works. Order matters, but not starting position.). For example, if <math>n=1</math> and <math>y=\frac{-1+\sqrt{5}}{2}</math> the set will be <math>(1,1,\frac{\sqrt{5}-3}{2},1-\sqrt{5},\frac{\sqrt{5}-3}{2})</math>, which you can test and find out that it still works even though the set isn't symmetric. | ||
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+ | Can someone change this answer so it's correct? | ||
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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}} |
Latest revision as of 16:54, 19 November 2020
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?
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 |