HSC 2013 MX2 Marathon (archive) (2 Viewers)

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rumbleroar

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Re: HSC 2014 4U Marathon

R is the centroid of triangle OPQ. By symmetry, triangle PQS is a reflection of triangle OPQ. Consider R' is the centroid of triangle PQS. A property of all triangles is the centroid divides the median (i.e. OM, QN, MS, etc) into the ratio 2:1. Thus OR:RM is 2:1 and similarly SR':MR' is 2:1.

Thus since OM=MS, then it follows that SR:OR is 2:1, hence
Nice!!
However, I think one person tried doing that and got marked down, because the question specifically stated hence and that meant they needed to use the previous parts.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Nice!!
However, I think one person tried doing that and got marked down, because the question specifically stated hence and that meant they needed to use the previous parts.
It stated deduce not hence, not sure if the same still applies.
 

rumbleroar

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Re: HSC 2014 4U Marathon

It stated deduce not hence, not sure if the same still applies.
Oh crap, sorry, my bad. But one of the teachers went through it and told us it was marked down because it didn't relate to the previous parts, so idk if they would be marks back or not :/ but apparently the q was quite hard, as some teachers couldn't do it on the first go and stuff
 

Carrotsticks

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Re: HSC 2014 4U Marathon

View attachment 29383

This was from our term 1 complex numbers assessment

Much tear
It got ripped off Sydney Grammar though so some of you may have seen it before
No one got iii

Have fun :))))


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I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?
 

Sy123

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Re: HSC 2014 4U Marathon

I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?
Take the 2 equations from part (i), and re-arrange both so that is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that
 

Carrotsticks

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Re: HSC 2014 4U Marathon

Take the 2 equations from part (i), and re-arrange both so that is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that
I understand how to get the result, but what I meant was that I do not understand how the result is supposed to work geometrically.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Take the 2 equations from part (i), and re-arrange both so that is the subject, equate the equations, to yield an equation only in z,v which re-arranged gives you the required result, I did it and the algebra matches with the Show that
I just did equation 1 minus equation 2, though I guess they are essentially the same thing.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

I could be derping atm, but I don't see how (3-2l)v = 2(k-l)z.

Because l and k are real numbers so if we re-arrange it, we have v=(3-2l)/(2(k-l))z, so v = A*z for some real number A, meaning that v and z are collinear, which is not the case as per the construction?
Have you re-arranged it right?
 

Sy123

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Re: HSC 2014 4U Marathon

I think I've found the problem, when we consider the final result,

Substituting this into the first 2 equations, we get:



Meaning the division or multiplication by cannot be done, since its zero.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

The other problem is that from (iii), if:



and

then it implies

Also from part (ii) we get:

when we let

Thus and so

But we chose arbitrary positive real numbers for (k,l) and so that would imply which is definitely not true.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

I think I've found the problem, when we consider the final result,

Substituting this into the first 2 equations, we get:



Meaning the division or multiplication by cannot be done, since its zero.
ok b10
 

Carrotsticks

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Re: HSC 2014 4U Marathon

Yep I get it now, I was derping, it's part of the proof.

Since z cannot be v, the only alternative is that (3-2l) and (k-l) = 0, so k = l = 3/2, so using (i) that instantly yields (iii).
 

Carrotsticks

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Re: HSC 2014 4U Marathon

The other problem is that from (iii), if:



and

then it implies

Also from part (ii) we get:

when we let

Thus and so

But we chose arbitrary positive real numbers for (k,l) and so that would imply which is definitely not true.
We aren't choosing arbitrary positive reals.

The construction is enough to uniquely define one value each of K and L, and it just so happens that the values are the same (3/2).
 

RealiseNothing

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Re: HSC 2014 4U Marathon

We aren't choosing arbitrary positive reals.

The construction is enough to uniquely define one value each of K and L, and it just so happens that the values are the same (3/2).
Yep I mistook it as arbitrary values for k and l.
 

Carrotsticks

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Re: HSC 2014 4U Marathon

Haha I just realised, k=3/2 was on the photo of the question.

Rumbleroar, surely you would have seen what to do from here? Or did you write that in afterwards?
 
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