HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

Status
Not open for further replies.

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
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?
 
Last edited:

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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

Member
Joined
Oct 1, 2014
Messages
90
Gender
Male
HSC
2015
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

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
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

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
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

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
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

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
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.
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level

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





 

simpleetal

Member
Joined
Apr 6, 2015
Messages
54
Gender
Male
HSC
2016
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

Member
Joined
Apr 6, 2015
Messages
54
Gender
Male
HSC
2016
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

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
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

Member
Joined
Apr 6, 2015
Messages
54
Gender
Male
HSC
2016
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

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
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

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
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.)
 
Last edited:

simpleetal

Member
Joined
Apr 6, 2015
Messages
54
Gender
Male
HSC
2016
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).
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top