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HSC 2013 MX2 Marathon (archive) (7 Viewers)

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seanieg89

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Re: HSC 2014 4U Marathon

Yep that was my method, I thought it was a really nice use of AM-GM. I came across a nice inequalities document too on google, might link.
I must admit, the AM-GM solution would be a bit harder to find if I didn't know Lagrange multipliers. (As the appropriate definition of y_k could potentially take a little time to spot.)

LM told me the equality conditions (each y_k as defined above must be equal.) From this fact as well as the fact that the problem reeked of AM-GM I figured an MX2 method would just be defining y_k as I did and using AM-GM.

If you don't already know LM you should learn about it, it's a very powerful and general method.
 
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Re: HSC 2014 4U Marathon

I'm talking about assuming what was to be proved.
 

seanieg89

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Re: HSC 2014 4U Marathon

I'm talking about assuming what was to be proved.
It's fine, all steps are obviously biconditional so it doesn't matter what end you start at. And you need this biconditionality to get the "iff" statement.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

It's fine, all steps are obviously biconditional so it doesn't matter what end you start at. And you need this biconditionality to get the "iff" statement.
This, which is why I put as obviously the squaring will still preserve the direction of the inequality.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

This isn't too hard, will have to do though until I think of something better.

Consider a polynomial with roots for

Find the sum of the roots two at a time.
 

Carrotsticks

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Re: HSC 2014 4U Marathon

It's fine, all steps are obviously biconditional so it doesn't matter what end you start at. And you need this biconditionality to get the "iff" statement.
To add on to this, think of it as building a two-laned road all one go rather than building one lane and then once you reach one end, go backwards to build the other lane.

This is often a good trick to kill of 'IFF' problems, but obviously you have to be very careful that everything you do is 'two laned', otherwise the other direction is immediately 'compromised'.
 

Carrotsticks

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Re: HSC 2014 4U Marathon

I haven't tried the Q yet, but it looks pretty!
 

Sy123

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Re: HSC 2014 4U Marathon

This isn't too hard, will have to do though until I think of something better.

Consider a polynomial with roots for

Find the sum of the roots two at a time.












 
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RealiseNothing

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Re: HSC 2014 4U Marathon

Yep that's correct, but you double count I think, just divide by 2.
 

RealiseNothing

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Re: HSC 2014 4U Marathon

I like this sum:



So far I have this, not sure if it'll go anywhere. Will edit in my solution as I progress.

Consider the general k'th term:



Now we just sum this:



Alternative expression:



The second expression seems like it would be easier to evaluate atm.
 
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