HSC 2015 MX2 Marathon (archive) (2 Viewers)

braintic · Jun 20, 2015

Re: HSC 2015 4U Marathon

You can easily rotate about oblique lines. But creating a solid by rotating about a curve is pretty meaningless. Assuming you want to take slices perpendicular to the 'axis' of revolution, these slices won't be parallel, so they'll crash into each other. It's like drawing a wavy line on an aeroplane, then telling the pilot 'try to rotate around that'. What would it even mean?

Kaido · Jun 21, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I just want to go further than rotating around straight x= and y= lines...

To the best of 4U knowledge, you can rotate around oblique lines (e.g. y=x)
Use y=1/x and set yourself up with a couple of boundaries and try rotating around y=x

Sy123 · Jun 22, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ n_1,n_2, \dots , n_m \ $be non-negative integers such that$ \ n_1 + n_2 + \dots + n_m = n \\ \\ $i) How many ways can one distribute$ \ n \ $chickens into$ \ m \ $pens, with pen$ \ i \ $having$ \ n_i \ $chickens for$ \ i = 1,2,\dots,m \\ \\ $ii) Hence show that$ \ (n_1 + n_2 + \dots + n_m)! \ $is divisible by$ \ n_1!n_2!n_3!\dots n_m!$

Sy123 · Jun 27, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ S(n,k) \ $be the number of ways to divide the collection of numbers$ \ \{1,2,3,\dots,n \} \ $into$ \ k \ $non-empty groups$$

$\\ $e.g.$ \ S(3,2) = 3 \ $since one can divide$ \ (A,B,C) \ $into groups$ \ (AB, C) , (AC, B) \ $and$ \ (BC, A)$

$\\ $By considering whether the number$ \ n \ $is in its own group or not, show that$ \\ S(n,k) = S(n-1,k-1) + k\cdot S(n-1,k)$

glittergal96 · Jun 28, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ S(n,k) \ $be the number of ways to divide the collection of numbers$ \ \{1,2,3,\dots,n \} \ $into$ \ k \ $non-empty groups$$

$\\ $e.g.$ \ S(3,2) = 3 \ $since one can divide$ \ (A,B,C) \ $into groups$ \ (AB, C) , (AC, B) \ $and$ \ (BC, A)$

$\\ $By considering whether the number$ \ n \ $is in its own group or not, show that$ \\ S(n,k) = S(n-1,k-1) + k\cdot S(n-1,k)$

Well if n is in it's own group, we obtain a partitioning of the set {1,...,n} by partitioning the remaining n-1 numbers into k-1 nonempty subsets, hence the first term.

If n isn't, that means we need to partition the remaining n-1 numbers into k sets, and then choose one of these k sets to slot in the number "n", hence the second term.

glittergal96 · Jun 28, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ n_1,n_2, \dots , n_m \ $be non-negative integers such that$ \ n_1 + n_2 + \dots + n_m = n \\ \\ $i) How many ways can one distribute$ \ n \ $chickens into$ \ m \ $pens, with pen$ \ i \ $having$ \ n_i \ $chickens for$ \ i = 1,2,\dots,m \\ \\ $ii) Hence show that$ \ (n_1 + n_2 + \dots + n_m)! \ $is divisible by$ \ n_1!n_2!n_3!\dots n_m!$

i) $\binom{n}{n_1}\binom{n-n_1}{n_2}\ldots\binom{n_m}{n_m}=\frac{n!}{\prod n_i!}$

by expanding out the binomial coefficients in terms of factorials.

ii) This quotient is precisely the RHS and hence integral.

Drsoccerball · Jun 28, 2015

Re: HSC 2015 4U Marathon

$$ If $ z_1 $ and $ z_2 $ are complex numbers such that $ |z_1+z_2| = |z_1 -z_2|. $ With the aid of a diagram, show that $ arg(z_1) $ and $ arg(z_2) $ differ by either $ \frac{\pi}{2} $ or $ \frac{3\pi}{2}$

leehuan · Jun 28, 2015

Re: HSC 2015 4U Marathon

Too lazy to LaTeX

Let A represent z1 and B represent z2

Construct the paralleogram OACB, where C represents (z1+z2). The diagonal OC has length mod(z1+z2) and the diagonal AB has length mod(z1-z2)

But this means that OC = AB. Hence, since the diagonals are equal in parallelogram OACB, OACB is actually a rectangle.

As all angles are right in a rectangle, arg(z1) and arg(z2) differ by pi/2 or 3pi/2, as such differences in arguments are essentially 90 degree rotations on the Argand diagram (pi/2 being counter-clockwise), noting that all angles in a rectangle are right.

braintic · Jun 29, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ If $ z_1 $ and $ z_2 $ are complex numbers such that $ |z_1+z_2| = |z_1 -z_2|. $ With the aid of a diagram, show that $ arg(z_1) $ and $ arg(z_2) $ differ by either $ \frac{\pi}{2} $ or $ \frac{3\pi}{2}$

OR ... at least one of z1 and z2 is zero.

Drsoccerball · Jun 29, 2015

Re: HSC 2015 4U Marathon

braintic said:
OR ... at least one of z1 and z2 is zero.

I think the initial conditions eliminate that "two complex numbers"

InteGrand · Jun 29, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I think the initial conditions eliminate that "two complex numbers"

But we can see that if at least one of $z_1,z_2$ is the complex number $0$ , then the given condition is satisfied. We could just tweak the wording to 'non-zero complex numbers' though.

leehuan · Jun 29, 2015

Re: HSC 2015 4U Marathon

EDIT: Don't worry, I realised that my confusion was actually somewhat related to the previous comments

braintic · Jun 29, 2015

Re: HSC 2015 4U Marathon

leehuan said:
Does 0 actually have an argument? I thought that was undefined.

That's right, it doesn't. But I'm not sure what your point is.

RealiseNothing · Jun 29, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I think the initial conditions eliminate that "two complex numbers"

0 is a complex number.

leehuan · Jul 2, 2015

Re: HSC 2015 4U Marathon

Any circle geometry?

Drsoccerball · Jul 2, 2015

Re: HSC 2015 4U Marathon

leehuan said:
Any circle geometry?

Its easier to post algbraic questions

Speed6 · Jul 3, 2015

Re: HSC 2015 4U Marathon

Let $p$ and $x$ be complex numbers. You are told that
$px=0$
Prove that $p=0$ or $x=0$

leehuan · Jul 3, 2015

Re: HSC 2015 4U Marathon

$\\ $Let $ p=a+ib $ and $ x=c+id \\ $where $ a,b,c,d\in \mathbb R \quad a,b,c,d\neq 0\\ \therefore px=ac-bd+i\left( bc+ad \right) \\ $If $ px=0+0i,$ then$\\ ac=bd\quad \quad ---(1)\\ bc=-ad\quad ---(2)\\ (1) $ rearranged $ b=\frac { ac }{ d } \quad ---(3)\\ $Sub $ (3) $ into $ (2):\\ c\left( \frac { ac }{ d } \right) =-ad\\ { c }^{ 2 }=-{ d }^{ 2 }\\ \therefore c=\pm id \\ $But this is a contradiction, as $c,d\in \mathbb R , \quad c,d\neq 0\\ $Hence $px\neq 0+0i\quad if\quad x\neq 0,\quad p\neq 0$

$\\ $Whereas if $x=0$, $c=d=0 $ satisfy ${ c }^{ 2 }=-{ d }^{ 2 }$ and by a similar means ${ a }^{ 2 }=-{ b }^{ 2 } \\ $Hence, $p=0$ or $x=0$

InteGrand · Jul 3, 2015

Re: HSC 2015 4U Marathon

Speed6 said:
Let $p$ and $x$ be complex numbers. You are told that
$px=0$
Prove that $p=0$ or $x=0$

$If $x\neq 0$, then $p=\frac{0}{x}=0$. Similarly if $p\neq 0$, then $x=\frac{0}{p}=0$.$

Speed6 · Jul 3, 2015

Re: HSC 2015 4U Marathon

There are a multiple of ways doing it right?

HSC 2015 MX2 Marathon (archive) (2 Viewers)

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