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HSC 2013 MX2 Marathon (archive) (7 Viewers)

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seanieg89

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Re: HSC 2013 4U Marathon

Oh I see, my answer was way off.
Pretty amazing how that strategy increases probability by a lot though....

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Yeah, definitely one of my favourite puzzle type questions. It's been proven relatively recently that that strategy is actually optimal, but that is MUCH harder I think.
 

Sy123

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Re: HSC 2013 4U Marathon

This is a good inequality:



 

spyris

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Re: HSC 2013 4U Marathon

Hi, I have a general question, I heard somewhere that 1/z = z(bar) but im not sure if its true because i am unable to prove it...so is it true and if so how do i show it.
 

HeroicPandas

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Re: HSC 2013 4U Marathon

Hi, I have a general question, I heard somewhere that 1/z = z(bar) but im not sure if its true because i am unable to prove it...so is it true and if so how do i show it.
That is true for roots of unity, where |z| = 1

So if |z| = 1, then 1/z = z(bar)
 
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seanieg89

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Re: HSC 2013 4U Marathon

Practising for the BOS trial? :p (btw that was HSC 2007 4U 13 marks compressed).







I don't want to be shown up by any current MX2 students :p.

Haha yeah, I recognised it, although I forgot the way the question guided you through it (I did my HSC in 07).

P.S. Obviously, I would supply some intermediate reasoning in an actual exam haha.
 
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seanieg89

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Re: HSC 2013 4U Marathon

My first thought would be just 2 since the first and last terms are 0!, and the rest all as there will be an 'n' in the denominator.
But there are an increasing number of terms, so that isn't quite rigorous.

Eg.



You can check this counterexample yourself by mx2 level methods, but heuristically it is obvious...we are adding n objects, each of which is roughly 1/root(n) (roughly means bounded between two constant multiples of) so we expect the sum to be roughly root(n), which tends to infinity.
 
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Sy123

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Now, we must prove that:



Yeah........so I'm not too sure yet on how to do this within the scope of the HSC.

But lets complete the solution, assuming that the limit converges:







 

seanieg89

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Re: HSC 2013 4U Marathon

Hmm yeah, cannot think of any immediate way to show that the limit is finite.


My method was to assume n was large, (which is all we care about anyway) and use that binomial coefficients get bigger towards the middle.

So:

 

Sy123

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---- (proof not yet complete)

One I think would then have to subsequently prove that s_(n+1) < s_n (which is indeed true), problem is this is only true for n =>4 (the partial sums increase until 3 and 4)
So I don't know of any way to do this....

EDIT: Or you can just do your way which is much easier :p
 

seanieg89

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Re: HSC 2013 4U Marathon





































---- (proof not yet complete)

One I think would then have to subsequently prove that s_(n+1) < s_n (which is indeed true), problem is this is only true for n =>4 (the partial sums increase until 3 and 4)
So I don't know of any way to do this....

EDIT: Or you can just do your way which is much easier :p
Well we only need EVENTUAL monotonicity to use the theorem you are thinking of, but that theorem isn't really MX2 assumed knowledge. But yeah, syllabus aside...if your working is all correct then the method is valid. Well done for bashing it out :).
 

Sy123

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Re: HSC 2013 4U Marathon

Well we only need EVENTUAL monotonicity to use the theorem you are thinking of, but that theorem isn't really MX2 assumed knowledge. But yeah, syllabus aside...if your working is all correct then the method is valid. Well done for bashing it out :).
Ahhh ok

=============

Something a little easier:

For a triangle ABC, with point P, Q, R on AB, BC, CA respectively
Such that, AQ bisects BC, CP bisects AB, and BR bisects CA. These 3 lines intersect at 1 point (a property which you may assume), at a point K, this point is called the centroid of a triangle.

For a general hyperbola, when the tangent at a point S intersects the asymptotes at points M and N.
Show that the locus of the centroid of the triangle OMN (O being origin) is a hyperbola.
 
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