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HSC 2015 MX2 Marathon (archive) (5 Viewers)

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braintic

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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

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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

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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

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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.
 
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InteGrand

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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.
 
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leehuan

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Re: HSC 2015 4U Marathon

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

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Re: HSC 2015 4U Marathon

Any circle geometry?
 

Speed6

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Re: HSC 2015 4U Marathon

Let and be complex numbers. You are told that

Prove that or
 

Speed6

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Re: HSC 2015 4U Marathon

There are a multiple of ways doing it right?
 
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