HSC 2013 MX2 Marathon (archive) (9 Viewers)

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Sy123

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Re: HSC 2013 4U Marathon

Slightly offended that Fermat is worth more than Goldbach.
hahahahaha

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

I will post a new question soon though, I just need to verify the result for myself.
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

Here's a sexy beast of a geometry question, by yours truly.



 
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GoldyOrNugget

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Re: HSC 2013 4U Marathon

I originally thought of using a^x > log_a x somehow but this came to me first.

Do you mind posting your method? I'm curious
It's a neat trick that I've seen crop up in a number of places, so it's definitely worth keeping in the back of your mind:



:)
 

Sy123

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Re: HSC 2013 4U Marathon

Here's a sexy beast of a geometry question, by yours truly.



Awesome question, loved it.

So first of all, we must define the rotating circle equation:



h, k are variable, they are variable with respect to the locus of the centre, and we are actually rotating the centre of the circle around the origin.

The parametric equation of the centre (considering the initial conditions)




So subbing this back in:



Expansion, then simplification, we also establish the fact that if the point exists inside the circle, then LHS < RHS

After simplification we get:



Using auxilliary angle rule, then more simplification we arrive at:



If we make inequality equality, then solve for t, we arrive at:





Testing points out:





Hope its right lol






It's a neat trick that I've seen crop up in a number of places, so it's definitely worth keeping in the back of your mind:



:)
oo, nice.
 

Carrotsticks

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Re: HSC 2013 4U Marathon

hey guys I made a new question:





Hahahaha.

Though really, if the mark allocations were to conform to the mark allocation of a HSC paper, each of those would be worth several thousand marks.
 

seanieg89

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Re: HSC 2013 4U Marathon

Hahahaha.

Though really, if the mark allocations were to conform to the mark allocation of a HSC paper, each of those would be worth several thousand marks.
I would fancy most peoples chances of drawing several thousand 1 mark graphs over their chances of solving one of those three problems, and not by a small margin.
 

Carrotsticks

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Re: HSC 2013 4U Marathon

Here's one that doesn't require much working out, but a bit of thinking to do.

Prove that an n-sided convex polygon can have its diagonals intersecting in AT MOST nC4 interior points.
 

Sy123

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Re: HSC 2013 4U Marathon

Here's one that doesn't require much working out, but a bit of thinking to do.

Prove that an n-sided convex polygon can have its diagonals intersecting in AT MOST nC4 interior points.
Ok, I decided to undergo a proof by induction:



If it is quadrilateral, it is obvious that there is only at most 1 point the diagonals can all intersect, and:



Hence it is true for n=4



We assume true that a k sided convex polygon can have at most diagonal intersections.

k sided polygon diagram



Ok, so we need to construct a k+1 sided polygon from the k sided polygon. We can do this by 'breaking one side into 2 pieces and joining it on' as shown:

k+1 sided polygon diagram

So we have done so as shown above. Now keep in mind that we already have kC4 intersections from before (I didn't show it).
So lets construct the k-2 number of diagonals from the new point we have constructed.



Ok, now notice the first red line I have constructed. Indeed there are: k-2 intersections from diagonals FROM THE OTHER POINTS
Look at the next red line, it is you could say 'defending' 2 points behind the second red line, there are (k-3) + (k-3) intersections from other points (k-3) for each point.
The pattern continues, the new intersections created by the new point is:



So lets evaluate this:



I can evaluate sum of squares through telescoping sum but I will skip this (or I could just rote the formula and put it down)



I plug in the formula, simplify and I end up with:



(srs)



And remember we already have intersections from before.
So NOW we have:

(by Pascal or otherwise)




=======

So satisfying to finally get the answer out of all that
 

seanieg89

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Re: HSC 2013 4U Marathon

Ok, I decided to undergo a proof by induction:



If it is quadrilateral, it is obvious that there is only at most 1 point the diagonals can all intersect, and:



Hence it is true for n=4



We assume true that a k sided convex polygon can have at most diagonal intersections.

k sided polygon diagram



Ok, so we need to construct a k+1 sided polygon from the k sided polygon. We can do this by 'breaking one side into 2 pieces and joining it on' as shown:

k+1 sided polygon diagram

So we have done so as shown above. Now keep in mind that we already have kC4 intersections from before (I didn't show it).
So lets construct the k-2 number of diagonals from the new point we have constructed.



Ok, now notice the first red line I have constructed. Indeed there are: k-2 intersections from diagonals FROM THE OTHER POINTS
Look at the next red line, it is you could say 'defending' 2 points behind the second red line, there are (k-3) + (k-3) intersections from other points (k-3) for each point.
The pattern continues, the new intersections created by the new point is:



So lets evaluate this:



I can evaluate sum of squares through telescoping sum but I will skip this (or I could just rote the formula and put it down)



I plug in the formula, simplify and I end up with:



(srs)



And remember we already have intersections from before.
So NOW we have:

(by Pascal or otherwise)




=======

So satisfying to finally get the answer out of all that
Alternatively:

Each pair of intersecting diagonals gives rise to a convex quadrilateral with vertices among the vertices of your polygon P (the convex quadrilateral with these two diagonals).

Different pairs of intersecting diagonals give rise to different convex quadrilaterals through the above construction.

A convex quadrilateral is uniquely determined by its 4 vertices. (This statement is NOT true for non-convex quadrilaterals.)

So #{pts. of intersection} =< #{pairs of intersecting diagonals } =< #{of subsets of the vertices of P of size 4} = nC4.
 
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Carrotsticks

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Re: HSC 2013 4U Marathon

Alternatively:

Each pair of intersecting diagonals gives rise to a convex quadrilateral with vertices among the vertices of your polygon P (the convex quadrilateral with these two diagonals).

Different pairs of intersecting diagonals give rise to different convex quadrilaterals through the above construction.

A convex quadrilateral is uniquely determined by its 4 vertices. (This statement is NOT true for non-convex quadrilaterals.)

So #{pts. of intersection} =< #{pairs of intersecting diagonals } =< #{of subsets of the vertices of P of size 4} = nC4.
Yep, is the solution I had in mind, though very nice work Sy123 for managing to get that initially-messy solution to actually work out!
 

Sy123

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Re: HSC 2013 4U Marathon

Alternatively:

Each pair of intersecting diagonals gives rise to a convex quadrilateral with vertices among the vertices of your polygon P (the convex quadrilateral with these two diagonals).

Different pairs of intersecting diagonals give rise to different convex quadrilaterals through the above construction.

A convex quadrilateral is uniquely determined by its 4 vertices. (This statement is NOT true for non-convex quadrilaterals.)

So #{pts. of intersection} =< #{pairs of intersecting diagonals } =< #{of subsets of the vertices of P of size 4} = nC4.
Yep, is the solution I had in mind, though very nice work Sy123 for managing to get that initially-messy solution to actually work out!

Ahaha, yeah my solution isn't as elegant as your one but I like it because it actually worked out.

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

A question that is a little easier to other people don't get scared off or anything:





(it isn't that long if you can create shortcuts for yourself.)
 
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Sy123

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Re: HSC 2013 4U Marathon







 
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Sy123

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Re: HSC 2013 4U Marathon

First define that 1+x > 0 therefore x > -1

Step 1:

n=1 -> 1+x = 1+x

Hence true for n=1

Step 2:

Assumption

Step 3:



From step 2:








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

My questions are still unanswered.
 

HeroicPandas

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Re: HSC 2013 4U Marathon

First define that 1+x > 0 therefore x > -1

Step 1:

n=1 -> 1+x = 1+x

Hence true for n=1

Step 2:

Assumption

Step 3:



From step 2:








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

My questions are still unanswered.
Nice. I'm trying to do ur insane questions lol
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

at work, so ill be answering these intermittently

i) use sin(2a) = 2sin(a)cos(a) repeatedly, expanding the sin each time.

ii) i don't get this question.
 
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Sy123

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Re: HSC 2013 4U Marathon

at work, so ill be answering these intermittently



i) use sin(2a) = 2sin(a)cos(a) repeatedly, expanding the sin each time.

ii) i don't get this question.
Find the other expression for it using part i (just divide by theta then take n to infinity)
 
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