HSC 2015 MX2 Marathon (archive) (1 Viewer)

Status
Not open for further replies.

braintic

Well-Known Member
Joined
Jan 20, 2011
Messages
2,137
Gender
Undisclosed
HSC
N/A
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

be.
Joined
Jul 7, 2014
Messages
798
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

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
 

glittergal96

Active Member
Joined
Jul 25, 2014
Messages
418
Gender
Female
HSC
2014
Re: HSC 2015 4U Marathon

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.
 

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
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.
 
Last edited:

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

I think the initial conditions eliminate that "two complex numbers"
But we can see that if at least one of is the complex number , then the given condition is satisfied. We could just tweak the wording to 'non-zero complex numbers' though.
 
Last edited:

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

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

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon

Any circle geometry?
 

Speed6

Retired '16
Joined
Jul 31, 2014
Messages
2,949
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

Let and be complex numbers. You are told that

Prove that or
 

Speed6

Retired '16
Joined
Jul 31, 2014
Messages
2,949
Gender
Male
HSC
N/A
Re: HSC 2015 4U Marathon

There are a multiple of ways doing it right?
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top