HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

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RealiseNothing

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Re: HSC 2015 4U Marathon - Advanced Level

How can we have the example of ? Didn't you say needed to be integer?
Forget the condition that needed to be integer. It was late and I'm not too sure why I said that.
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

No, I'm saying isn't necessarily integer.
What do you mean by the highest power of p that divides k? Why wouldn't we just have that be , for ?

And what do we mean by 'divides k' if k isn't an integer?
 
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RealiseNothing

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Re: HSC 2015 4U Marathon - Advanced Level

What do you mean by the highest power of p that divides k? Why wouldn't we just have that be , for ?

And what do we mean by 'divides k' if k isn't an integer?
I didn't worry too much about the fine details since it's a 4U thread, but I'll clarify:

The domain of the function is the positive rationals.

The function can only be calculated directly if is a positive integer, otherwise we just use the property given.

Also is the highest integer power of that divides , so it's a bit different to the usual log function.
 

Chlee1998

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Re: HSC 2015 4U Marathon - Advanced Level

I didn't worry too much about the fine details since it's a 4U thread, but I'll clarify:

The domain of the function is the positive rationals.

The function can only be calculated directly if is a positive integer, otherwise we just use the property given.

Also is the highest integer power of that divides , so it's a bit different to the usual log function.
wait, now youre saying that m is an integer?
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

Also, what was the meaning of etc., since k may not be a prime number, and we defined the function for a prime number in the subscript?
 

RealiseNothing

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Re: HSC 2015 4U Marathon - Advanced Level

Also, what was the meaning of etc., since k may not be a prime number, and we defined the function for a prime number in the subscript?
as the highest power of 2 that goes into 12 is 2 (2^2=4|12)
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

as the highest power of 2 that goes into 12 is 2 (2^2=4|12)
No, I mean in that property of the function, you gave like , what is 's meaning here, if is non-prime (or non-integer even)?
 

RealiseNothing

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Re: HSC 2015 4U Marathon - Advanced Level

No, I mean in that property of the function, you gave like , what is 's meaning here, if is non-prime (or non-integer even)?
You only need prime values of for the question so we don't even have to define them.

If you want to define them, non-prime integer can still be defined in the same way without any issues from what I can see.

If we take rational , then we'd only have to extend to being negative for this to work.

We can probably extend it to real as well by adjusting the values can take.
 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Suppose a is a non-square positive integer with a rational root.

Let p be a prime divisor of a with p^k the largest power of p that divides a. Since a is nonsquare, we can choose p such that k is odd. (Otherwise FTA implies a is a product of square primary factors and hence a square.)

Then a=p^kq with p coprime to q.

So p^kq=m^2/n^2 for some pair (m,n) of positive integers.

So m^2=n^2p^kq

Applying f:=f_p to both sides and using the f(ab)=f(a)+f(b) property, we get:

2f(m)=2f(n)+k+0

The LHS is even but the RHS is odd, contradiction.

---

Alternative proof of the same fact:

Suppose p/q is a rational root of x^2-a=0 for some positive integer a.

The rational root theorem tells us q|1, ie p/q is an integer. So any a for which this is possible must be the square of an integer.
 
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Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

Not too difficult (but better suited for this thread than the other):





 

simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

Not too difficult (but better suited for this thread than the other):





given that i have interpreted the question correctly, d1= {1,3,6,8} results in dn = 2 if n even or 6 if n odd. d1{2 and 7} results in dn=6 for even n and 2 for odd n. d1{4,5,9} have dn =0 for all n>1.
 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

given that i have interpreted the question correctly, d1= {1,3,6,8} results in dn = 2 if n even or 6 if n odd. d1{2 and 7} results in dn=6 for even n and 2 for odd n. d1{4,5,9} have dn =0 for all n>1.
Yep well done!
 

simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

Here's the first question from a paper called the UNSW mathematics competition for this year, (not the ICAS but a 3 hour exam).

Around a spherical planet, there are 37 satellites. Prove that, for any point on the planet,
there are at most 17 satellites visible.
 

InteGrand

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Re: HSC 2015 4U Marathon - Advanced Level

What if the satellites were all close to each other and you were on a point on the planet near the satellites; wouldn't we be able to see all of them?
 

simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

What if the satellites were all close to each other and you were on a point on the planet near the satellites; wouldn't we be able to see all of them?
Ah sorry, mistyped the question, it should be, Around a spherical planet, there are 37 satellites. Prove that, there exists a point on the surface such that at most 17 satellites are visible
 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

Ah sorry, mistyped the question, it should be, Around a spherical planet, there are 37 satellites. Prove that, there exists a point on the surface such that at most 17 satellites are visible
Wouldn't it matter how close the satellites are orbiting this planet? Or am I just misunderstanding the question
 

glittergal96

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Re: HSC 2015 4U Marathon - Advanced Level

Cool question, it's a shame there isn't more solid geometry or combinatorial geometry in the HS syllabus :(.

Take the origin 0 to be the centre of the sphere.

If all of the satellites are co-linear, then the problem is trivial (any point on the sphere that also lies on the plane through 0 which is orthogonal to this line does not have vision of any of the satellites).

Suppose then, that x,y are two linearly independent satellites (ie 0,x,y form a non-degenerate triangle, and hence define a plane E.)

This plane intersects the sphere in a great circle that after rotation we can view as the equator of the sphere (just for ease of nomenclature).

Then we draw tangent planes N and S at the "north" and "south" poles, relative to to this equator.

An observer at the north pole can only see everything lying above N, and an observer at S can only see everything lying below S.

So if the claim were not true there would have to be at least 18 points above N and at least 18 points below S.

As these sets are disjoint, there can be at most 1 point between the planes N and S.

But x and y both lie between N and S...contradiction! (note that the planes E, S and N are all parallel.)
 
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simpleetal

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Re: HSC 2015 4U Marathon - Advanced Level

Cool question, it's a shame there isn't more solid geometry or combinatorial geometry in the HS syllabus :(.

Take the origin 0 to be the centre of the sphere.

If all of the satellites are co-linear, then the problem is trivial (any point on the sphere that also lies on the plane through 0 which is orthogonal to this line does not have vision of any of the satellites).

Suppose then, that x,y are two linearly independent satellites (ie 0,x,y form a non-degenerate triangle, and hence define a plane E.)

This plane intersects the sphere in a great circle that after rotation we can view as the equator of the sphere (just for ease of nomenclature).

Then we draw tangent planes N and S at the "north" and "south" poles, relative to to this equator.

An observer at the north pole can only see everything lying above N, and an observer at S can only see everything lying below S.

So if the claim were not true there would have to be at least 18 points above N and at least 18 points below S.

As these sets are disjoint, there can be at most 1 point between the planes N and S.

But x and y both lie between N and S...contradiction! (note that the planes E, S and N are all parallel.)
yeah that was the basic idea behind my method during the exam (proving that there had to be at least 38 points if the claim was false).
 
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