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HSC 2013 MX2 Marathon (archive) (9 Viewers)

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HeroicPandas

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Re: HSC 2014 4U Marathon

Yep

Alternatively just use

and just a bit more.
Yep

Alternatively just use

and just a bit more.
Another method (using triangle inequality):
Use triangle inequality on the triangle with sides a, b, c and re-arranging them and subbing it into A results in:



Now to prove that







then let x = a, b, c and y = b, c, a, add up all 3 inequalities to get



which is greater or equal to ab (because since a, b, c are positive then bc and ca are also positive)
 

RealiseNothing

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Re: HSC 2014 4U Marathon

These aren't that hard, but there are really intuitive solutions to them which I think are nice.

1) For integers show that has atleast one solution pair for all

Where

2) Prove that there are no solutions to the above problem such that
 

seanieg89

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Re: HSC 2014 4U Marathon

These aren't that hard, but there are really intuitive solutions to them which I think are nice.

1) For integers show that has atleast one solution pair for all

Where

2) Prove that there are no solutions to the above problem such that
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Yep your second solution was the same as mine, I'll post up my solution for the first though (imo it's a cool method).

These questions were just meant to take up some time before I thought of anything better, thanks for ruining that basically straight away lol.
 

seanieg89

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Re: HSC 2014 4U Marathon

Yep your second solution was the same as mine, I'll post up my solution for the first though (imo it's a cool method).

These questions were just meant to take up some time before I thought of anything better, thanks for ruining that basically straight away lol.
Haha I know, I normally don't answer these for a while so other people have a chance. Felt like spoiling the fun today though.

Am interested to see your first solution, did you do it nonconstructively or something?
 

RealiseNothing

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Re: HSC 2014 4U Marathon

A square number can be represented visually by a square, whilst a triangular number can be represented visually by a triangle. Any square can be broken into two triangles, and thus there will always be a solution to the equation:

 

seanieg89

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Re: HSC 2014 4U Marathon

A square number can be represented visually by a square, whilst a triangular number can be represented visually by a triangle. Any square can be broken into two triangles, and thus there will always be a solution to the equation:

Oh right cool, yeah those triangles are exactly the ones in my construction. Just numbers vs a picture.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

Oh right cool, yeah those triangles are exactly the ones in my construction. Just numbers vs a picture.
Yep I think it's nice to see how an equation like that can easily be done visually. I've been trying to make a harder question which uses a similar method (constructing a visual of a certain type of number - square/triangular numbers for example) which turns out to be a lot easier than any algebraic method. No luck yet though.
 

seanieg89

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Re: HSC 2014 4U Marathon

I can't really imagine such a thing existing, but I might well be wrong of course. I guess "easier" is pretty subjective.
 

dunjaaa

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Re: HSC 2014 4U Marathon

Major arc on top half of the real axis, with open circles at -1+0i and 1+0i. Cartesian equation of locus: x^2 + (y-1)^2 = 2 for y>0
 

seanieg89

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Re: HSC 2014 4U Marathon

Alternatively:







You would of course have to check the b=0 case separately, but it is obvious this won't give us any issues thanks to symmetry.

Regardless, here is a third way :p.

By Cauchy-Schwartz twice:

 
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RealiseNothing

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Re: HSC 2014 4U Marathon

Consider a regular 8x8 chessboard. 8 knights are placed on the chessboard, 4 at each end as such:

_ _ K K K K _ _

So in the middle of the very top and very bottom rows is 4 knights.

If all knights move simultaneously, prove whether or not it is possible for either of the diagonals of the chessboard to be completely filled by all 8 knights.
 
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