HSC 2013 MX2 Marathon (archive) (6 Viewers)

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RealiseNothing

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Re: HSC 2013 4U Marathon

True (provided you can prove the product is not periodic) I didn't think about convergence anyway haha
That wouldn't be too hard.

Anyway I'll come back and give your questions a proper go Tuesday after English is finished. Will have a week of maths so should be fun.
 

seanieg89

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Re: HSC 2013 4U Marathon

With MX2 a week away, I will post one or two difficult questions each day, hopefully covering all topics (polynomials, integration, etc). As usual, this is more to stretch good students and hone their problemsolving skill for Q16s than to provide exercises at doing the routine stuff quickly and accurately...practice and mindset is all there is to the latter.

First, polynomials. A result similar in flavour to the conjugate root theorem:

 

Sy123

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Re: HSC 2013 4U Marathon

With MX2 a week away, I will post one or two difficult questions each day, hopefully covering all topics (polynomials, integration, etc). As usual, this is more to stretch good students and hone their problemsolving skill for Q16s than to provide exercises at doing the routine stuff quickly and accurately...practice and mindset is all there is to the latter.

First, polynomials. A result similar in flavour to the conjugate root theorem:

I'm not sure if this is what you're looking for:



By cubing, then subbing x back in



Considering the graph of this cubic, we find that the maximum turning point at x=-1 is



Since,



















Forming a quadratic equation



Subbing that in, in terms of alpha





Those are the other 2 roots that must be there in order to create a rational polynomial p(x) with those roots.
 

seanieg89

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Re: HSC 2013 4U Marathon

I'm not sure if this is what you're looking for:



By cubing, then subbing x back in



Considering the graph of this cubic, we find that the maximum turning point at x=-1 is



Since,



















Forming a quadratic equation



Subbing that in, in terms of alpha





Those are the other 2 roots that must be there in order to create a rational polynomial p(x) with those roots.
Cool, so you are mostly done. What you have done is constructed a particular rational polynomial (your cubic) with alpha as a root. You have then found the two other roots of this poly (which look like the two roots I am looking for, although I haven't double-checked).

But how do we show that ANY rational polynomial with alpha as a root has at least the same two other roots?
 
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Sy123

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Re: HSC 2013 4U Marathon

Cool, so you are mostly done. What you have done is constructed a particular rational polynomial (your cubic) with alpha as a root. You have then found the two other roots of this poly (which look like the two roots I am looking for, although I haven't double-checked).

But how do we show that ANY rational polynomial with alpha as a root has at least the same two other roots?
So for a general polynomial Q(x), with a root , in order to 'ensure rationality', then





R(x) being a rational polynomial, therefore any rational polynomial Q(x) that has as a root, must have P(x) as a factor in order to ensure rationality......
---
I have a feeling that I haven't actually proven anything here lol
 

seanieg89

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Re: HSC 2013 4U Marathon

So for a general polynomial Q(x), with a root , in order to 'ensure rationality', then





R(x) being a rational polynomial, therefore any rational polynomial Q(x) that has as a root, must have P(x) as a factor in order to ensure rationality......
---
I have a feeling that I haven't actually proven anything here lol
So close, but why "must" we have Q=pR? Why must any Q be divisible by p? This needs proof but otherwise good.
 

rural juror

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Re: HSC 2013 4U Marathon

im not quite sure if this is right?? i got the expression I = k^n n!/(m+1)(m+n+1)(m+2n+1)...(m+kn+1)

plus if you sub in m = 1, n= 1, k = 100, the inequality doesn' work. i dunno, i might be worng about this
 

Sy123

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Re: HSC 2013 4U Marathon

im not quite sure if this is right?? i got the expression I = k^n n!/(m+1)(m+n+1)(m+2n+1)...(m+kn+1)

plus if you sub in m = 1, n= 1, k = 100, the inequality doesn' work. i dunno, i might be worng about this
For I, I'm getting:



Which is a little different to yours

The inequality is still wrong though my bad, it should be:

 

RealiseNothing

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Re: HSC 2013 4U Marathon

Consider a circle with equally spaced points around it's circumference.

I choose of these points where

I then make amount of polygons using my chosen amount of points such that no point is used more than once and so that all polygons have an equal amount of sides, where and

Find the probability that NO polygons share a common area.
 
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Sy123

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Re: HSC 2013 4U Marathon

Consider a circle with equally spaced points around it's circumference.

I choose of these points where

I then make amount of polygons using my chosen amount of points such that no point is used more than once and so that all polygons have an equal amount of sides, where and

Find the probability that NO polygons share a common area.
To have pk points make k polygons, then there are p points per polygon, in order for there to be no polygons with no common area, then all points for each polygon must be 'grouped together', that is if I have A1 A2 A3 belonging to one polygon and B1 B2 B3 belonging to another, in cyclic oritentation:

A1 B1 A2 A3 B2 B3, will be such that they overlap

The only way that they dont overalp is if they are grouped together, i.e.

A1 A2 A3 B1 B2 B3

Therefore, for the k groups of p points, we are simply going to find the number of ways to arrange pk points into k groups of p points, and divide by total ways to pick pk points from pn points

Number of ways to arrange k groups of p points around a circle is:






Hopefully that is correct
 

RealiseNothing

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Re: HSC 2013 4U Marathon

To have pk points make k polygons, then there are p points per polygon, in order for there to be no polygons with no common area, then all points for each polygon must be 'grouped together', that is if I have A1 A2 A3 belonging to one polygon and B1 B2 B3 belonging to another, in cyclic oritentation:

A1 B1 A2 A3 B2 B3, will be such that they overlap

The only way that they dont overalp is if they are grouped together, i.e.

A1 A2 A3 B1 B2 B3

Therefore, for the k groups of p points, we are simply going to find the number of ways to arrange pk points into k groups of p points, and divide by total ways to pick pk points from pn points

Number of ways to arrange k groups of p points around a circle is:






Hopefully that is correct
I don't think that's correct. Consider if so I choose all the points on the circle. Then the denominator is 1 since pn=pk. So the probability becomes:



Which will give you a probability > 1

(correct me if I have misinterpreted you)
 

RealiseNothing

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Re: HSC 2013 4U Marathon

I came up with this question as a generalised version of the "find the probability that two chords intersect" question, so I don't know the answer for sure. But what I got was:



The way I'm seeing it too, is that since we've already chosen 'pk' points, we could just construct another circle with 'pk' points and so every single point on the circle will be used (the cyclic order stays the same). And so the answer should be independent of 'n' I think.
 

Sy123

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Re: HSC 2013 4U Marathon

I don't think that's correct. Consider if so I choose all the points on the circle. Then the denominator is 1 since pn=pk. So the probability becomes:



Which will give you a probability > 1

(correct me if I have misinterpreted you)
That's true, I think my calculation is wrong, I am still intent on the grouping idea.
I do agree that its also independent of n

-----

Take our selected points:

P1 P2 P3 ..... P(nk)

The number of ways to split these points into k groups without moving them is number of ways to put k number of '|'s in between them:

_P1 _ P2 _ P3 _ P4 ..... _P(pk) -> (there is nothing to the right of P(pk) since its a circle)

So put k 'barriers' in those free slots, there are

ways to do so

^No, Picking the position of one barrier automatically chooses positions for everything else, so there are simply only (pk) choices, with k duplicates, making it only p choices

The number of ways to then group the points without restriction is



^No, we pick p every time not k, my bad!

It should be:



Hence the probability is:



^That was using my faulty logic, using my amended thinking I yield your answer

Which is quite similar to yours but not quite it.....I wonder where I went wrong

EDIT: Wait yes the calculation for the numerator is wrong, the position of 1 barrier determines the positions of everything else...


Oh and also my grouping for the denominator is wrong, i picked k points every time instead of p


In the end I yield your answer lol, good question!
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

My way was slightly different. First suppose that all groups of 'p' points were on the circle in cyclic order. Now we can move every point 1 to the right and make a new arrangement. We can do this 'p' times before reaching our original arrangement. Hence there are a total of 'p' arrangements.

Now we find the total arrangements possible without restriction. Consider now that instead of points already being on the circle, there are 'pk' amount of "holes" and we place each point in a hole.

We place the first group of 'p' points in the 'pk' holes to get

Now we place the second group of 'p' points in the remaining 'p(k-1)' points to get

We do this 'k' times until no holes are left and thus all points are used.

But we must also divide by k! to get rid of repetitions since there are 'k' groups, and we get the total arrangements:



Hence the probability is:



 
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