an old maths olympiad question
an old maths olympiad question
hey guys! today I found a question paper that is almost four years old. Our teacher chose four of us for this Olympiad. there were a total of 10 questions with total 100 marks. I couldn't solve any of them. I still cant solve them. I want your insights for one of the question. if you want I can post others as well.
The question:
Prove that for no integer n, n6+3n5-5n4-15n3+4n2+12n+3 is a perfect square
What should i read to solve problems like this?
The question:
Prove that for no integer n, n6+3n5-5n4-15n3+4n2+12n+3 is a perfect square
What should i read to solve problems like this?
Last edited by S_r on Sat Jul 07, 2018 5:46 pm, edited 1 time in total.
- kenbrooker
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Re: an old maths olympiad question
Metaquestion: Is there perhaps an equal or inequality sign missing?
And yes, I am interested in any other questions...
And yes, I am interested in any other questions...
"Good Judgment comes from Experience;
Experience comes from Bad Judgment..."
Experience comes from Bad Judgment..."
Re: an old maths olympiad question
Whatever the missing question is, its answer probably relies on the fact that that polynomial can be written as
(n-2)(n-1)n(n+1)(n+2)(n+3)+3.
Re: an old maths olympiad question
I have updated ot now. Is it my inexperience that I cannot see how to factor the polynomial?
Re: an old maths olympiad question
Other one: suppose A1A2A3....An is an n sided regular polygon such that :
1/A1A2= 1/A1A3 + 1/A1A4. Determine number of sides of the polygon.
Re: an old maths olympiad question
Use the fact that all squares are 0 or 1 modulo 4 (and non squares 2 or 3 mod 4).S_r wrote: ↑Sat Jul 07, 2018 4:38 am hey guys! today I found a question paper that is almost four years old. Our teacher chose four of us for this Olympiad. there were a total of 10 questions with total 100 marks. I couldn't solve any of them. I still cant solve them. I want your insights for one of the question. if you want I can post others as well.
The question:
Prove that for no integer n, n6+3n5-5n4-15n3+4n2+12n+3 is a perfect square
What should i read to solve problems like this?
War ruins the life and health of untold numbers of innocent children.
Re: an old maths olympiad question
No. I cheated and typed it into Wolfram Alpha.
If I were in a competition, I would certainly have evaluated the polynomial (call it p(x) ) at a few small values of x, like 0, +-1, +-2, just to get a feel for what it is like, I would then have noticed the unusual fact that the result was 3 every time. That means that p(x)-3 has linear factors x, (x-1), (x+1), (x-2), (x+2) and one other.
- kenbrooker
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Re: an old maths olympiad question
Again, if I understand the question right (and someone's gotta bite : ),
without divulging my massive derivation,
I think the polygon has zero sides (of
any length : )...
(Unless 1 = 2)
"Good Judgment comes from Experience;
Experience comes from Bad Judgment..."
Experience comes from Bad Judgment..."
Re: an old maths olympiad question
I don't know if what you think is right or wrong. unless you can prove it it ain't right. Until then I can't believe if has zero sides.
- kenbrooker
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Re: an old maths olympiad question
Far as I know, a "regular" polygon has sides of equal length,
call it length S, so:
1/A1A2 = 1/A1A3 + 1/A1A4 is the same as
1/S^2 = 1/S^2 + 1/S^2, the same as
1 = 1 + 1 so
No polygon
satisfies...
I could be missing something but
did you say there were
10 questions?
call it length S, so:
1/A1A2 = 1/A1A3 + 1/A1A4 is the same as
1/S^2 = 1/S^2 + 1/S^2, the same as
1 = 1 + 1 so
No polygon
satisfies...
I could be missing something but
did you say there were
10 questions?
"Good Judgment comes from Experience;
Experience comes from Bad Judgment..."
Experience comes from Bad Judgment..."
Re: an old maths olympiad question
3-sided:
A1A2 = A1A3 = A2A3 = 1 = S
And if A4 didn't exist, A1A4 = infinity
-> 1/1 = 1/1 + 1/infinity ...
-> 1 = 1 + 0
The usage of capital letters is typical for Points, small letters are used for lengths.
A1A2 = A1A3 = A2A3 = 1 = S
And if A4 didn't exist, A1A4 = infinity
-> 1/1 = 1/1 + 1/infinity ...
-> 1 = 1 + 0
The usage of capital letters is typical for Points, small letters are used for lengths.
Re: an old maths olympiad question
A1, A2, A3, A4 are consecutive vertices of the polygon.kenbrooker wrote: ↑Mon Jul 09, 2018 9:21 am Far as I know, a "regular" polygon has sides of equal length,
call it length S, so:
1/A1A2 = 1/A1A3 + 1/A1A4 is the same as
1/S^2 = 1/S^2 + 1/S^2, the same as
1 = 1 + 1 so
No polygon
satisfies...
I could be missing something but
did you say there were
10 questions?
A1A2 is presumably supposed to mean the length of the side from A1 to A2, which you dubbed S. Note however that A1A3 and A1A4 represent the lengths of diagonals of the polygon, which will be larger than S.
Re: an old maths olympiad question
I take my thanks back
Last edited by S_r on Mon Jul 09, 2018 12:51 pm, edited 1 time in total.
Re: an old maths olympiad question
If n>3, there is no integer solution:
a = (n-2)/n * pi
1/1 = 1/(2-2cos(a)) + 1/(3-2cos(2a-pi))
-> n = 13.757
a = (n-2)/n * pi
1/1 = 1/(2-2cos(a)) + 1/(3-2cos(2a-pi))
-> n = 13.757
Re: an old maths olympiad question
I can derive a cubic equation that cos(x) must satisfy where x is half the central angle, i.e. x=180/n.
It is straightforward to see that:
A1A2 = 2sin(x)
A1A3 = 2sin(2x)
A1A4 = 2sin(3x)
Substituting this in the equation, and using the angle sum formulae sin(a+b)=sin(a)cos(b)+sin(b)cos(a) and cos(2a)=2cos(a)^2-1 you eventually get 8z^3-4z^2-z+1=0, where z=cos(x).
You can easily verify with a calculator that n=7 works.
Actually proving it is harder. There is probably some clever way to do this with complex numbers.
It is straightforward to see that:
A1A2 = 2sin(x)
A1A3 = 2sin(2x)
A1A4 = 2sin(3x)
Substituting this in the equation, and using the angle sum formulae sin(a+b)=sin(a)cos(b)+sin(b)cos(a) and cos(2a)=2cos(a)^2-1 you eventually get 8z^3-4z^2-z+1=0, where z=cos(x).
You can easily verify with a calculator that n=7 works.
Actually proving it is harder. There is probably some clever way to do this with complex numbers.
Re: an old maths olympiad question
I probably could but I can't be bothered. Spending a couple of minutes here and there during breaks at work is rather different from actually solving it in Olympiad. Olympiad questions are often hard until some specific insight hits you, and it can take many failed attempts and hard work before you get that insight. For now, the only way I can see of proving it takes too much time. In a real Olympiad you have lots of questions, and you can choose which ones to tackle as no one gets them all.
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Re: an old maths olympiad question
Thanks! jaap, I totally missed that and I totally agree...jaap wrote: ↑Mon Jul 09, 2018 10:02 am A1, A2, A3, A4 are consecutive vertices of the polygon.
A1A2 is presumably supposed to mean the length of the side from A1 to A2, which you dubbed S. Note however that A1A3 and A1A4 represent the lengths of diagonals of the polygon, which will be larger than S.
"Good Judgment comes from Experience;
Experience comes from Bad Judgment..."
Experience comes from Bad Judgment..."