HSC 2015 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Mar 8, 2015

Re: HSC 2015 4U Marathon

braintic said:
Connected to what?

To each other. They all meet at the destination but the roads that are connected you have to go through all pf them to make it to the destination

braintic · Mar 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
To each other. They all meet at the destination but the roads that are connected you have to go through all pf them to make it to the destination

I'm afraid I can't picture what you are describing. Surely if three roads start from here, then each 'connect' again at the destination, then you only need one road to get there. And these roads that aren't connected ... do they just end in a cul-de-sac? I don't know what you mean by 'not connected'.

Drsoccerball · Mar 8, 2015

Re: HSC 2015 4U Marathon

braintic said:
I'm afraid I can't picture what you are describing. Surely if three roads start from here, then each 'connect' again at the destination, then you only need one road to get there. And these roads that aren't connected ... do they just end in a cul-de-sac? I don't know what you mean by 'not connected'.

Lets say to get somewhere you have to use one road, that road uses another road and that uses another road so theyre connected. The other roads go directly to the destination

braintic · Mar 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Lets say to get somewhere you have to use one road, that road uses another road and that uses another road so theyre connected. The other roads go directly to the destination

So what's the difference between three connected roads and one long road made up of three segments? I'm still missing something here.

Ekman · Mar 8, 2015

Re: HSC 2015 4U Marathon

braintic said:
So what's the difference between three connected roads and one long road made up of three segments? I'm still missing something here.

What he is trying to say is that from your starting point, there are 2 direct roads to your destination and there is one path consisting of 3 roads. Thus in total there are 3 paths that you can take from your starting point till your destination.

braintic · Mar 9, 2015

Re: HSC 2015 4U Marathon

Ekman said:
What he is trying to say is that from your starting point, there are 2 direct roads to your destination and there is one path consisting of 3 roads. Thus in total there are 3 paths that you can take from your starting point till your destination.

So kind of like series and parallel electric circuits?

I'm getting (1-p)² - (1-p)⁵.

Ekman · Mar 10, 2015

Re: HSC 2015 4U Marathon

braintic said:
So kind of like series and parallel electric circuits?

I'm getting (1-p)² - (1-p)⁵.

Actually what Drsoccerball wanted was p^3 + p^3(1-p) + p^3(1-p)^2

glittergal96 · Mar 10, 2015

Re: HSC 2015 4U Marathon

Lol, most ambiguous question ever.

braintic · Mar 10, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Actually what Drsoccerball wanted was p^3 + p^3(1-p) + p^3(1-p)^2

Yes, my answer was wrong.
I'm getting your answer in a different form:
p^2 - p^2 (1-p)^3

It simplifies to the same thing, but could you explain your logic for that form of the answer.

FrankXie · Mar 10, 2015

Re: HSC 2015 4U Marathon

new question

$$ Find the limit $ \lim_{n\rightarrow\infty}\frac{2^{\frac{1}{n}}+2^{\frac{2}{n}}+2^{\frac{3}{n}}+\cdots+2^{\frac{n}{n}}}{n}.$

integral95 · Mar 10, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
new question

$$ Find the limit $ \lim_{n\rightarrow\infty}\frac{2^{\frac{1}{n}}+2^{\frac{2}{n}}+2^{\frac{3}{n}}+\cdots+2^{\frac{n}{n}}}{n}.$

By riemann sum we get

$\int_{0}^{1} 2^x dx = \frac{1}{ln(2)}$

Chlee1998 · Mar 10, 2015

Re: HSC 2015 4U Marathon

Lol glitter is back. Glitter y u no state rank.....

Chlee1998 · Mar 10, 2015

Re: HSC 2015 4U Marathon

Also Glitter, were there any maths questions which you struggled with/not able to do during your HSC year? If so, can u provide an example? Just curious

glittergal96 · Mar 10, 2015

Re: HSC 2015 4U Marathon

Silly mistakes. Better at solving hard problems than doing lots of easy ones without stuffing up.

Can't remember any HSC/trials qs I had major problems with, but I spent most of my time trying olympiad problems and teaching myself some uni maths. Some of those were too hard for me.

Ekman · Mar 12, 2015

Re: HSC 2015 4U Marathon

$$ Prove $ arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2} $ Given that $ |z|^2 = |w|^2$

Drsoccerball · Mar 12, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$ Prove $ arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2} $ Given that $ |z| = |w|$

omg i never knew this was that easy D : 3 marks gonw

Ekman · Mar 13, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$ Prove $ arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2} $ Given that $ |z|^2 = |w|^2$

Bump. Can someone still solve this question pls?

VBN2470 · Mar 13, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$$ Prove $ arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2} $ Given that $ |z|^2 = |w|^2$

$Since the moduli of $z$ and $w$ are equal, the parallelogram formed using these vectors actually becomes a rhombus. So the diagonals (given by $z+w$ and $z-w$) are perpendicular to each other, and hence the expression holds true.$

Ekman · Mar 13, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Since the moduli of $z$ and $w$ are equal, the parallelogram formed using these vectors actually becomes a rhombus. So the diagonals (given by $z+w$ and $z-w$) are perpendicular to each other, and hence the expression holds true.$

Is there a way to do it algebraically?

VBN2470 · Mar 13, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Is there a way to do it algebraically?

$Since the diagonals of a rhombus are perpendicular then $z+w=ki(z-w)$ \Rightarrow \, $\frac{z+w}{z-w}=ki$ \Rightarrow \, $\arg(\frac{z+w}{z-w}) = \pm\frac{\pi}{2}$ for all real $k$.$

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