HSC 2013 MX2 Marathon (archive) (5 Viewers)

braintic · Dec 24, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
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.

After how many moves? As many as you need?

RealiseNothing · Dec 24, 2013

Re: HSC 2014 4U Marathon

braintic said:
After how many moves? As many as you need?

Yep as many as you need.

seanieg89 · Dec 24, 2013

Re: HSC 2014 4U Marathon

Nope, there will always be four knights on dark squares and four knights on light squares (as this is the case to start with, and each move changes the colour of each knights square).

Sy123 · Dec 24, 2013

Re: HSC 2014 4U Marathon

$\\ $Suppose for positive real$ \ a,b,c,d \ \ $the polynomial$ \\ p(x) = x^4 - 4ax^3 + 6b^2x^2 - 4c^3x + d^4 \ \ \ \ $has 4 distinct positive real roots$$

$\\ $Show that$ \\ \\ $i) $ \ c>d \\ \\ $ii)$ \ b >c \\ \\ $iii)$ \ a>b$

RealiseNothing · Dec 24, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Suppose for positive real$ \ a,b,c,d \ \ $the polynomial$ \\ p(x) = x^4 - 4ax^3 + 6b^2x^2 - 4c^3x + d^4 \ \ \ \ $has 4 distinct positive real roots$$

$\\ $Show that$ \\ \\ $i) $ \ c>d \\ \\ $ii)$ \ b >c \\ \\ $iii)$ \ a>b$

Just wanted to point out from first thoughts, a>d comes straight from AM-GM.

Sy123 · Dec 24, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
Just wanted to point out from first thoughts, a>d comes straight from AM-GM.

That is the general idea for all 3 haha (and a bit more)

RealiseNothing · Dec 24, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
That is the general idea for all 3 haha (and a bit more)

$b=\sqrt{\frac{AB+AC+AD+BC+BD+CD}{6}}$

$\geq \sqrt[12]{A^3B^3C^3D^3 } = \sqrt[4]{ABCD} = d$

I see what you mean.

RealiseNothing · Dec 24, 2013

Re: HSC 2014 4U Marathon

Ok yer you do the same thing and you get c>d too.

Then you will probably do something similar and get all the results.

Ikki · Dec 25, 2013

Re: HSC 2014 4U Marathon

Sy123 said:
Yep

Alternatively just use $\frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

How do you derive this?

asianese · Dec 25, 2013

Re: HSC 2014 4U Marathon

$$Using $ a+b\ge 2\sqrt{ab} $ twice on $ a, b, c, d, $ we get $ \\ \dfrac{a+b+c+d}{4} \ge \dfrac{2\sqrt{ab} + 2\sqrt{cd}}{4} = \dfrac{ \sqrt{ab}+\sqrt{cd}}{2} \ge \sqrt{\sqrt{ab}\sqrt{cd}} =\sqrt[4]{abcd}.$

Ikki · Dec 25, 2013

Re: HSC 2014 4U Marathon

asianese said:
$$Using $ a+b\ge 2\sqrt{ab} $ twice on $ a, b, c, d, $ we get $ \\ \dfrac{a+b+c+d}{4} \ge \dfrac{2\sqrt{ab} + 2\sqrt{cd}}{4} = \dfrac{ \sqrt{ab}+\sqrt{cd}}{2} \ge \sqrt{\sqrt{ab}\sqrt{cd}} =\sqrt[4]{abcd}.$

Wow that's pretty awesome.
Thanks

dunjaaa · Dec 26, 2013

Re: HSC 2014 4U Marathon

Do we learn these inequality results later when we cover harder 3U?

Carrotsticks · Dec 26, 2013

Re: HSC 2014 4U Marathon

dunjaaa said:
Do we learn these inequality results later when we cover harder 3U?

Yep =)

seanieg89 · Dec 26, 2013

Re: HSC 2014 4U Marathon

$Let $x_1,x_2,\ldots,x_n$ be positive real numbers. Let \\ \\$M_p:=\left(\frac{1}{n}\sum_{j=1}^n x_j^p\right)^{1/p}$\\ \\ for positive real $p$.\\ \\ Prove that $M_p\leq M_q$ whenever $p\leq q$.\\ \\ Prove that $\lim_{p\rightarrow 0^+} M_p = \left(\prod_{j=1}^n x_j\right)^{1/n}.$$

Note: This question should probably be split into several parts in order to make the difficulty more reasonable, but I would like to see what you guys can come up with off the bat.

RealiseNothing · Dec 26, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
$Let $x_1,x_2,\ldots,x_n$ be positive real numbers. Let \\ \\$M_p:=\left(\sum_{j=1}^n x_j^p\right)^{1/p}$\\ \\ for positive real $p$.\\ \\ Prove that $M_p\leq M_q$ whenever $p\leq q$.\\ \\ Prove that $\lim_{p\rightarrow 0^+} M_p = \left(\prod_{j=1}^n x_j\right)^{1/n}.$$

Note: This question should probably be split into several parts in order to make the difficulty more reasonable, but I would like to see what you guys can come up with off the bat.

$(M_p)^p = \sum_{j=1}^{n} (x_j)^p$

$\geq n\sqrt[n]{\prod_{j=1}^{n} (x_j)^p}$

$=n(\prod_{j=1}^{n} x_j)^{\frac{p}{n}}$

$M_p \geq n^{\frac{1}{p}}(\prod_{j=1}^{n} x_j)^{\frac{1}{n}}$

As $p \to 0^+$ then the RHS $\to \infty$ as $n\geq1$

What am I doing wrong?

If this is actually correct then I'm going to try find an upper bound for $M_p$ then call on my best friend Mr.Squeeze Theorem.

asianese · Dec 26, 2013

Re: HSC 2014 4U Marathon

It should be

$=n\left( \prod_{j=1}^n x_j^{p/n}\right)$

on the second last line

RealiseNothing · Dec 26, 2013

Re: HSC 2014 4U Marathon

asianese said:
It should be

$=n\left( \prod_{j=1}^n x_j^{p/n}\right)$

on the second last line

Isn't that what I have?

asianese · Dec 26, 2013

Re: HSC 2014 4U Marathon

Hm I'm confused myself.

seanieg89 · Dec 26, 2013

Re: HSC 2014 4U Marathon

RealiseNothing said:
$(M_p)^p = \sum_{j=1}^{n} (x_j)^p$

$\geq n\sqrt[n]{\prod_{j=1}^{n} (x_j)^p}$

$=n(\prod_{j=1}^{n} x_j)^{\frac{p}{n}}$

$M_p \geq n^{\frac{1}{p}}(\prod_{j=1}^{n} x_j)^{\frac{1}{n}}$

As $p \to 0^+$ then the RHS $\to \infty$ as $n\geq1$

What am I doing wrong?

If this is actually correct then I'm going to try find an upper bound for $M_p$ then call on my best friend Mr.Squeeze Theorem.

Whoops, my bad, I forgot to include a factor inside the brackets. You need to average the power sum before taking the p-th root. Edited appropriately.

RealiseNothing · Dec 26, 2013

Re: HSC 2014 4U Marathon

seanieg89 said:
Whoops, my bad, I forgot to include a factor inside the brackets. You need to average the power sum before taking the p-th root. Edited appropriately.

Ok so that gives me my lower bound:

$M_p \geq (\prod_{j=1}^{n} x_j)^\frac{1}{n}$

Now I need to find an upper bound to use Squeeze Theorem on (well at least that is what I'm thinking will happen).

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