an old maths olympiad question

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S_r
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an old maths olympiad question

Post by S_r » Sat Jul 07, 2018 3:38 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?
Last edited by S_r on Sat Jul 07, 2018 4:46 pm, edited 1 time in total.

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kenbrooker
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Re: an old maths olympiad question

Post by kenbrooker » Sat Jul 07, 2018 10:22 am

Metaquestion: Is there perhaps an equal or inequality sign missing?

And yes, I am interested in any other questions...
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Re: an old maths olympiad question

Post by jaap » Sat Jul 07, 2018 2:58 pm

S_r wrote:
Sat Jul 07, 2018 3:38 am
The question:
Prove that for no integer n, n6+3n5-5n4-15n3+4n2+12n+3
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.

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Re: an old maths olympiad question

Post by S_r » Sat Jul 07, 2018 4:47 pm

jaap wrote:
Sat Jul 07, 2018 2:58 pm
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.
I have updated ot now. Is it my inexperience that I cannot see how to factor the polynomial?

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Re: an old maths olympiad question

Post by S_r » Sat Jul 07, 2018 4:49 pm

kenbrooker wrote:
Sat Jul 07, 2018 10:22 am


And yes, I am interested in any other questions...
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.

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Re: an old maths olympiad question

Post by hk » Sat Jul 07, 2018 8:15 pm

S_r wrote:
Sat Jul 07, 2018 3:38 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?
Use the fact that all squares are 0 or 1 modulo 4 (and non squares 2 or 3 mod 4).
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Re: an old maths olympiad question

Post by jaap » Sun Jul 08, 2018 6:27 am

S_r wrote:
Sat Jul 07, 2018 4:47 pm
jaap wrote:
Sat Jul 07, 2018 2:58 pm
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.
I have updated ot now. Is it my inexperience that I cannot see how to factor the polynomial?
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.

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Re: an old maths olympiad question

Post by kenbrooker » Mon Jul 09, 2018 12:38 am

S_r wrote:
Sat Jul 07, 2018 4:49 pm
kenbrooker wrote:
Sat Jul 07, 2018 10:22 am
And yes, I am interested in any other questions...
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.
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)
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Re: an old maths olympiad question

Post by S_r » Mon Jul 09, 2018 4:26 am

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.

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Re: an old maths olympiad question

Post by kenbrooker » Mon Jul 09, 2018 8:21 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?
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Re: an old maths olympiad question

Post by v6ph1 » Mon Jul 09, 2018 8:30 am

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.
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Re: an old maths olympiad question

Post by jaap » Mon Jul 09, 2018 9:02 am

kenbrooker wrote:
Mon Jul 09, 2018 8:21 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?
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.

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Re: an old maths olympiad question

Post by S_r » Mon Jul 09, 2018 9:59 am

I take my thanks back
Last edited by S_r on Mon Jul 09, 2018 11:51 am, edited 1 time in total.

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Re: an old maths olympiad question

Post by v6ph1 » Mon Jul 09, 2018 10:21 am

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
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Re: an old maths olympiad question

Post by S_r » Mon Jul 09, 2018 11:27 am

I can't still see it can you make it clear?

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Re: an old maths olympiad question

Post by S_r » Mon Jul 09, 2018 11:47 am

The answer is 7 but I don't know how :(

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Re: an old maths olympiad question

Post by jaap » Mon Jul 09, 2018 12:28 pm

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.

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Re: an old maths olympiad question

Post by S_r » Mon Jul 09, 2018 2:23 pm

So people here can't solve an Olympiad question

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Re: an old maths olympiad question

Post by jaap » Mon Jul 09, 2018 3:21 pm

S_r wrote:
Mon Jul 09, 2018 2:23 pm
So people here can't solve an 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

Post by kenbrooker » Mon Jul 09, 2018 6:20 pm

jaap wrote:
Mon Jul 09, 2018 9:02 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.
Thanks! jaap, I totally missed that and I totally agree...
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Experience comes from Bad Judgment
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