HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

Zen2613 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
.

I like that one. Nice introduction to 4 unit problems. But since you made the problem, how did you actually find the roots for that quartic equation in x ? The 1/2sqrt(2-sqrt(3)) ones.

Drsoccerball · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Zen2613 said:
I like that one. Nice introduction to 4 unit problems. But since you made the problem, how did you actually find the roots for that quartic equation in x ? The 1/2sqrt(2-sqrt(3)) ones.

$$ So if you want to make an equation with a root $ \sqrt{2} $ you say, let $ x=\sqrt{2}$

$0=x-\sqrt{2}$

$$ This is the linear equation with root $ x=\sqrt{2}$

$$ Similarly $ x=\sqrt{2}$

$x^2=2$

$x^2 -2=0 $ is the quadratic equation with a root $ = \sqrt{2}$

and so on...

dan964 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
Here's a crappy one I made

$i) $ Prove that the equation $ 16x^4 -16x^2 +1 $ has a root $ x=\frac{1}{2}\sqrt{2-\sqrt{3}} $ by letting $ x= \frac{1}{2}\sqrt{2-\sqrt{3}}$

$ii) $ Using De Moivre's theorem find the quartic equation with a root $ cos(\frac{5 \pi}{12})$

$iii) $ Explain the significance of the results found above. $$

Wrong thread, part (i) is 2-unit.
Part (ii) doesn't even require De Moivre's, even if do use it, it is not Q16 material.
Part (iii) is good though.
This should be in the normal marathon.

lita1000 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

dan964 said:
Wrong thread, part (i) is 2-unit.
Part (ii) doesn't even require De Moivre's, even if do use it, it is not Q16 material.
This should be in the normal marathon.

Maybe he accidentally posted in the advanced thread because there's normal marathon for 2016ers yet.....

Drsoccerball · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

dan964 said:
Wrong thread, part (i) is 2-unit.
Part (ii) doesn't even require De Moivre's, even if do use it, it is not Q16 material.
Part (iii) is good though.
This should be in the normal marathon.

i) Usually in extension 2 they ask you to prove a result whether it is 2 unit, extension 1 or 2 it doesn't matter.
ii) I never said this was a question 16.
ii) Thank you.
There's no normal marathon yet.

dan964 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Zen2613 said:
I think he wants us to use demoivre to find an expression for cos(4theta) in terms of cos(theta) because that will be a quartic which will probably resemble his 16x^4 - 16x^2 +1 = 0 thingo.

It is +1; looks like you forgot something.

dan964 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
i) Usually in extension 2 they ask you to prove a result whether it is 2 unit, extension 1 or 2 it doesn't matter.
ii) I never said this was a question 16.
ii) Thank you.
There's no normal marathon yet.

Never mind; part (iii) is not as good as first stated, but still reasonable; part (ii) is alright. Sorry a mistake in my working in forgetting in missing a +1 meant that part (iii) was impossible. I'll post the sol in a sec.
The only unique thing about it was the Demoivre's theorem, I reckon an Ext 1 student could do it easily.

dan964 · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
i) Usually in extension 2 they ask you to prove a result whether it is 2 unit, extension 1 or 2 it doesn't matter.
ii) I never said this was a question 16.
ii) Thank you.
There's no normal marathon yet.

I kind of have to sharpen my maths skills before the course next year starts.
Solution:

Drsoccerball · Oct 25, 2015

Re: HSC 2016 4U Marathon - Advanced Level

dan964 said:
I kind of have to sharpen my maths skills before the course next year starts.
Solution: View attachment 32602

Pretty much. Except for the first question I was looking for this :

$2x=\sqrt{2-\sqrt{3}}$

$4x^2 =2-\sqrt{3}$

$(4x^2-2)^2=3$

$16x^4 -16x^2 +1=0$

[]

dan964 · Oct 26, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
Pretty much. Except for the first question I was looking for this :

$2x=\sqrt{2-\sqrt{3}}$

$4x^2 =2-\sqrt{3}$

$(4x^2-2)^2=3$

$16x^4 -16x^2 +1=0$

[]

same thing, I just used substitution instead of rearrangement.

Sy123 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$\\ a,b,c \ $are integers so that$ \ a^2 + b^2 = c^2 \ $and$ \ a < b < c$
$\\ $Prove that$ \ a + b + c \ $is always even$$

Drongoski · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc) = c^2 + c^2 + 2(ab+ac+bc) = 2(c^2 + (ab+ac+bc)) = 2 \times (integer)$

$Since $ (odd)^2 = $ odd and $ (even)^2 = $ even, a+b+c = even$

Drsoccerball · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drongoski said:
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ac+bc) = c^2 + c^2 + 2(ab+ac+bc) = 2(c^2 + (ab+ac+bc)) = 2 \times (integer)$

But theres a square... Just because the square is divisible by 2 doesn't mean the root is also divisible by 2... I think we should try proving that
$c^2 +ab+ac+bc$ Is a perfect square divisible by 2.

calamebe · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

I just assumed:
Odd + odd= even
Odd - odd = even
Even + even = even
Even - even = even
Even + odd = odd
Even - odd = odd
Odd - even = odd

And here it is: http://m.imgur.com/zL8wwRU

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Can someone post a difficult problem?

Drsoccerball · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

calamebe said:
I just assumed:
Odd + odd= even
Odd - odd = even
Even + even = even
Even - even = even
Even + odd = odd
Even - odd = odd
Odd - even = odd

And here it is: http://m.imgur.com/zL8wwRU

You dont have to assume most of those results as they are given to be positive.

calamebe · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
You dont have to assume most of those results as they are given to be positive.

Ok thanks, I thought so, just wanted to make sure.

Drsoccerball · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

calamebe said:
Ok thanks, I thought so, just wanted to make sure.

But that thinking would be right if you continue from what the other guy did. But I still think we need to prove the even number is a perfect square.

lita1000 · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

Drsoccerball said:
But theres a square... Just because the square is divisible by 2 doesn't mean the root is also divisible by 2... I think we should try proving that
$c^2 +ab+ac+bc$ Is a perfect square divisible by 2.

Uhhhhhhh.........................

Show me an even square integer whose square root isn't even then.

Drsoccerball · Oct 28, 2015

Re: HSC 2016 4U Marathon - Advanced Level

lita1000 said:
Uhhhhhhh.........................

Show me an even square integer whose square root isn't even then.

If you noticed I commented on the unedited version of his post. The unedited one had $A^2 =2 (integer)$ Which wouldnt allow A to be a multiple of 2 for all real values of the RHS.

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