HSC 2013 MX2 Marathon (archive) (1 Viewer)

GoldyOrNugget · Jan 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
hey guys I made a new question:

$$i) Prove Goldbach's Conjecture$ \ \ \ \ \fbox{56}$

$$ii) Prove the Hodge Conjecture$ \ \ \ \ \fbox{23}$

$$iii) Prove Fermat's Last Theorem$ \ \ \ \ \fbox{442}$

Slightly offended that Fermat is worth more than Goldbach.

Sy123 · Jan 3, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
hey.

dont worry man, I scanned my solution up here you can check it if you want:

http://www.ams.org/notices/199507/faltings.pdf

Sy123 · Jan 3, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
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 · Jan 3, 2013

Re: HSC 2013 4U Marathon

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

$A circle on the Cartesian plane with diameter 5 and center $(0; \frac{5}{2})$ completes a $360^\circ$ clockwise revolution around the $z$-axis (as if you pinned a point of its circumference to $(0; 0)$ and rotated the rest of the circle around in 2 dimensions) in 1 second.$

$If the circle rotates at constant angular velocity, beginning its rotation at time $t=0$ seconds and ending at $t=1$ second, for what time interval is the point $(2; -3)$ strictly inside the circle?$

GoldyOrNugget · Jan 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
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:

$$\log(x)$ is strictly increasing, so $\log(a) > \log(b) \iff a > b, \quad a,b > 0$ \\ \\ $\log(9999^{10000}) = 10000 \log(9999) \approx 92102 > \log(10000^{9999}) = 9999 \log(10000) \approx 92094$ so $9999^{10000} > 10000^{9999}$

Sy123 · Jan 3, 2013

Re: HSC 2013 4U Marathon

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

$A circle on the Cartesian plane with diameter 5 and center $(0; \frac{5}{2})$ completes a $360^\circ$ clockwise revolution around the $z$-axis (as if you pinned a point of its circumference to $(0; 0)$ and rotated the rest of the circle around in 2 dimensions) in 1 second.$

$If the circle rotates at constant angular velocity, beginning its rotation at time $t=0$ seconds and ending at $t=1$ second, for what time interval is the point $(2; -3)$ strictly inside the circle?$

Awesome question, loved it.

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

$(x-h)^2+(y-k)^2 = 2.5^2$

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)

$x=2.5\sin 2\pi t$
$y=2.5 \cos 2\pi t$

So subbing this back in:

$(x-2.5\sin 2\pi t)^2 + (y-2.5\cos 2\pi t )^2 = 2.5^2$

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

After simplification we get:

$10 \sin 2\pi t + 15 \cos 2\pi t > 13$

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

$\cos (2\pi t - \tan^{-1} 2/3) > \sqrt{13}/5$

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

$t \doteq 0.21....$

$t \doteq 0.98...$

Testing points out:

$0<t<0.22...$

$0.98...<t<1$

Hope its right lol

GoldyOrNugget said:
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:

$$\log(x)$ is strictly increasing, so $\log(a) > \log(b) \iff a > b, \quad a,b > 0$ \\ \\ $\log(9999^{10000}) = 10000 \log(9999) \approx 92102 > \log(10000^{9999}) = 9999 \log(10000) \approx 92094$ so $9999^{10000} > 10000^{9999}$

oo, nice.

Carrotsticks · Jan 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
hey guys I made a new question:

$$i) Prove Goldbach's Conjecture$ \ \ \ \ \fbox{56}$

$$ii) Prove the Hodge Conjecture$ \ \ \ \ \fbox{23}$

$$iii) Prove Fermat's Last Theorem$ \ \ \ \ \fbox{442}$

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 · Jan 3, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
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 · Jan 3, 2013

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 · Jan 3, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
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:

$$Step 1: $ \ \ n=4$

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

$1=\binom{4}{4}$

Hence it is true for n=4

$$Step 2:$ \ \ n=k$

We assume true that a k sided convex polygon can have at most $\binom{k}{4}$ diagonal intersections.

k sided polygon diagram

$$Step 3:$ \ \ n=k+1$

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:

$\sum_{m=1}^{k-2} m(k-(m+1))$

So lets evaluate this:

$=k\sum_{m=1}^{k-2} m - \sum_{m=1}^{k-2} m^2 - \sum_{m=1}^{k-2} m$

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)

$= k/2 (k-1)(k-2) -1/2(k-1)(k-2) - \sum_{m=1}^{k-2}m^2$

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

$1/2(k-1)^2(k-2) - 1/6 (k-2)(k-1)(2k-3)$

$=\frac{1}{6} k (k-1)(k-2)$ (srs)

$=\binom{k}{3}$

And remember we already have $\binom{k}{4}$ intersections from before.
So NOW we have:

$\binom{k}{3} + \binom{k}{4} = \binom{k+1}{4}$ (by Pascal or otherwise)

$$Proof complete by mathematical Induction$ \ \ \square$

=======

So satisfying to finally get the answer out of all that

seanieg89 · Jan 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ok, I decided to undergo a proof by induction:

$$Step 1: $ \ \ n=4$

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

$1=\binom{4}{4}$

Hence it is true for n=4

$$Step 2:$ \ \ n=k$

We assume true that a k sided convex polygon can have at most $\binom{k}{4}$ diagonal intersections.

k sided polygon diagram

$$Step 3:$ \ \ n=k+1$

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:

$\sum_{m=1}^{k-2} m(k-(m+1))$

So lets evaluate this:

$=k\sum_{m=1}^{k-2} m - \sum_{m=1}^{k-2} m^2 - \sum_{m=1}^{k-2} m$

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)

$= k/2 (k-1)(k-2) -1/2(k-1)(k-2) - \sum_{m=1}^{k-2}m^2$

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

$1/2(k-1)^2(k-2) - 1/6 (k-2)(k-1)(2k-3)$

$=\frac{1}{6} k (k-1)(k-2)$ (srs)

$=\binom{k}{3}$

And remember we already have $\binom{k}{4}$ intersections from before.
So NOW we have:

$\binom{k}{3} + \binom{k}{4} = \binom{k+1}{4}$ (by Pascal or otherwise)

$$Proof complete by mathematical Induction$ \ \ \square$

=======

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.

Carrotsticks · Jan 3, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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!

HeroicPandas · Jan 3, 2013

Re: HSC 2013 4U Marathon

carrotsticks said:
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.

hahaha!

Sy123 · Jan 3, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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.

Carrotsticks said:
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:

$x^3+3x+2=0 \ \ \ \ \rightarrow $roots$ \ \ \ \ \alpha , \beta, \gamma$

$$Find the polynomial equation with roots$ \ \ \alpha + \frac{1}{\alpha} , \beta + \frac{1}{\beta} , \gamma + \frac{1}{\gamma}$

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

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

$$i) Show that$ \ \ \sin \theta = 2^{n} \cos (\theta /2) \cos (\theta /4) \cos (\theta /8) ... \cos (\theta / 2^n) \sin (\theta/ 2^n) \ \ \ \ \ \fbox{2}$

$$ii) Hence show that$ \ \ \lim_{n \to \infty} \frac{\sin \theta}{\theta}=\cos (\theta/2)\cos(\theta/4)\cos(\theta/8)...\cos(\theta/2^n)... \ \ \ \ \fbox{1}$

$iii) By utilising clever substitution and formula usage, show that$

$\pi = 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot ... \ \ \ \ \ \ \fbox{3}$

HeroicPandas · Jan 4, 2013

Re: HSC 2013 4U Marathon

$$Prove by mathematical induction for $ x > -1$ and $ n > 0 \\ \\$ $(1+x)^n \geq 1+nx$

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$$Prove by mathematical induction for $ x > -1$ and $ n > 0 \\ \\$ $(1+x)^n \geq 1+nx$

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:

$(1+x)^{k+1} \geq 1+(k+1)x \ \ \ ?$

From step 2:

$(1+x)^{k} (1+x) \geq (1+kx)(1+x)$

$(1+x)^{k+1} \geq 1+ (k+1)x + kx^2 > 1+(k+1)x$

$\therefore \ \ $Proof complete by Mathematical Induction$ \ \ \ \square$

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

My questions are still unanswered.

HeroicPandas · Jan 4, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
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:

$(1+x)^{k+1} \geq 1+(k+1)x \ \ \ ?$

From step 2:

$(1+x)^{k} (1+x) \geq (1+kx)(1+x)$

$(1+x)^{k+1} \geq 1+ (k+1)x + kx^2 > 1+(k+1)x$

$\therefore \ \ $Proof complete by Mathematical Induction$ \ \ \ \square$

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

My questions are still unanswered.

Nice. I'm trying to do ur insane questions lol

GoldyOrNugget · Jan 4, 2013

Re: HSC 2013 4U Marathon

at work, so ill be answering these intermittently

Sy123 said:
$$i) Show that$ \ \ \sin \theta = 2^{n} \cos (\theta /2) \cos (\theta /4) \cos (\theta /8) ... \cos (\theta / 2^n) \sin (\theta/ 2^n) \ \ \ \ \ \fbox{2}$

$$ii) Hence find$ \ \ \lim_{n \to \infty} \frac{\sin \theta}{\theta} \ \ \ \ \fbox{1}$

$iii) By utilising clever substitution and formula usage, show that$

$\pi = 2 \cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot ... \ \ \ \ \ \ \fbox{3}$

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

ii) i don't get this question. $\ \lim_{n \to \infty} \frac{\sin \theta}{\theta} = \frac{\sin \theta}{\theta}$

Sy123 · Jan 4, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
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. $\ \lim_{n \to \infty} \frac{\sin \theta}{\theta} = \frac{\sin \theta}{\theta}$

Find the other expression for it using part i (just divide by theta then take n to infinity)

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