HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

RealiseNothing · Jun 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
I haven't gotten around to grounding myself well yet, I still don['t know what eigenvectors/values are

I did them today, if you go beyond rote learning them it's very interesting.

Especially in the textbook when it briefly brings up isomorphisms.

Sy123 · Jun 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Let$ \ a,b,c,d \ $be positive reals that sum up to 1$ \\ \\ $Show that$ \ \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}$

aDimitri · Jun 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Using Prime numbers gives all the board:

Black blocks are blocks where the knight has not ended
Red blocks are blocks where the knight has ended.
The number of iterations shows how many times it has been done, and shows the various positions it has ended.

The number of moves is set at 19 (prime)

10 iterations:

20 iterations:

100 iterations:

1000 iterations (ignore black dot in corner, I think its just my program, I think, not sure why its there)

EDIT: This works with most odd numbers that I'm trying, I don't think its something special with primes

By alternating colours though how can this be true? If A1 is black, surely any large odd number could only be white squares (or less if you don't allow moves back) than the whole board?

Sy123 · Jun 10, 2014

Re: HSC 2014 4U Marathon - Advanced Level

This is a good inequality

$\\ $Let$\ n \geq 3 \ $be an integer, and$ \ a_2,\dots,a_n \ $be positive reals so that$ \ a_2a_3\dots a_n = 1 \\ \\ $Prove that$ \ (1+a_2)^2(1+a_3)^3(1+a_4)^4 \dots (1+a_n)^n > n^n$

Sy123 · Jun 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
This is a good inequality

$\\ $Let$\ n \geq 3 \ $be an integer, and$ \ a_2,\dots,a_n \ $be positive reals so that$ \ a_2a_3\dots a_n = 1 \\ \\ $Prove that$ \ (1+a_2)^2(1+a_3)^3(1+a_4)^4 \dots (1+a_n)^n > n^n$

$a_k + 1 = a_k + \underbrace{\left(\frac{1}{k-1} + \frac{1}{k-1} + \dots + \frac{1}{k-1} \right)}_{(k-1)}$

$\\ \therefore \ (a_k + 1)^k \geq \frac{k^k}{(k-1)^{(k-1)}} a_k$

$\prod_{k=2}^{n} (1+a_k)^k > \frac{2^2}{1^1} \frac{3^3}{2^2} \dots \frac{n^n}{(n-1)^{(n-1)}} a_2 a_3 \dots a_n \\ \\ \therefore \ \prod_{k=2}^n (1+a_k)^k > n^n$

Sy123 · Jun 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find$ \ a_{n} \ $if$ \\ \\ a_1 = 2 \\ a_{n} = \frac{n+1}{n-1} (a_1 + a_2 + \dots + a_{n-1})$

Davo_01 · Jun 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find$ \ a_{n} \ $if$ \\ \\ a_1 = 2 \\ a_{n} = \frac{n+1}{n-1} (a_1 + a_2 + \dots + a_{n-1})$

$$let \; \sum_{k=1}^n a_k=S_n$

$S_n-S_{n-1}=\dfrac{n+1}{n-1}S_{n-1}$

$S_n=\dfrac{2n}{n-1}S_{n-1}=\dfrac{(2n)(2n-2)(2n-4)...(4)}{(n-1)(n-2)(n-3)...1}S_1=2^n\,n$

$\therefore a_n=S_{n}-S_{n-1}=2^{n-1}(n+1)$

TL1998 · Jun 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

The lengths of the sides of a triangle form an arithmetic progression and the largest angle of the triangle exceeds the smallest by 90 degrees. show that the lengths of the sides of the triangle are in the ratio 7^(1/2) -1 : 7^(1/2):7^(1/2)+1

Sy123 · Jun 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
$$let \; \sum_{k=1}^n a_k=S_n$

$S_n-S_{n-1}=\dfrac{n+1}{n-1}S{n-1}$ (*)

$S_n=\dfrac{2n}{n-1}S_{n-1}=\dfrac{(2n)(2n-2)(2n-4)...(4)}{(n-1)(n-2)(n-3)...1}S_1=2^n\,n$

$\therefore a_n=S_{n}-S_{n-1}=2^{n-1}(n+1)$

(*): Did you make a mistake here?

Davo_01 · Jun 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
(*): Did you make a mistake here?

The lowercase thing? I fixed it but is the answer right?

Sy123 · Jun 19, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
The lowercase thing? I fixed it but is the answer right?

Ah yes I see
Yes your answer was correct, I got a_n = (n+1) but I think I made a mistake, my proof was very similar to yours though

RealiseNothing · Jun 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Two people are playing a game of cards. The rules of the game are as follows:

1) The deck is fully randomised and shuffled (52 card deck).

2) A dealer takes the top card of the deck and turns it over.

3) This is done continuously until the game is won by either of the competitors.

A game is won by the following:

1) Each card has a certain value. These values are:

Ace = 1
Number cards = The number is the value
Jack/Queen/King = 10

2) If the card is red, that value is positive. If the card is black, that value is negative.

3) As each card is turned over, the total sum is calculated (so a red 9 adds 9 whilst a black queen subtracts 10).

4) Any of the players can call "stop" after a card has been turned over.

5) You win if the card you called "stop" after gives a higher total sum than the card your opponent called stop after.

Question:

Does there exist an optimal strategy to this game, and if so what is that strategy, if:

i) Both competitors are aware of when their opponent calls stop.

ii) Neither competitor knows when their opponent has called stop.

seanieg89 · Jun 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Two people are playing a game of cards. The rules of the game are as follows:

1) The deck is fully randomised and shuffled (52 card deck).

2) A dealer takes the top card of the deck and turns it over.

3) This is done continuously until the game is won by either of the competitors.

A game is won by the following:

1) Each card has a certain value. These values are:

Ace = 1
Number cards = The number is the value
Jack/Queen/King = 10

2) If the card is red, that value is positive. If the card is black, that value is negative.

3) As each card is turned over, the total sum is calculated (so a red 9 adds 9 whilst a black queen subtracts 10).

4) Any of the players can call "stop" after a card has been turned over.

5) You win if the card you called "stop" after gives a higher total sum than the card your opponent called stop after.

Question:

Does there exist an optimal strategy to this game, and if so what is that strategy, if:

i) Both competitors are aware of when their opponent calls stop.

ii) Neither competitor knows when their opponent has called stop.

Presumably a "strategy" is just a set of rules determining whether or not to say stop after a given sequence of cards (and for (i) also with the knowledge of when an opponent says stop.)

Care to define "optimal"?

In the game theoretic sense of having the best guaranteed outcome, then saying stop as soon as your opponent does (but before the next card is dealt) guarantees you a tie every time, and by symmetry is an optimal solution for i).

For (ii), it seems like any good strategy would be based on rather ugly probabilistic expressions (Eg the probability of an arbitrary ordering of the remaining cards always having non-positive sum). Such a strategy would also be hard to specify, given the large number of possible situations. It sounds like the sort of thing you would use a computer for.

Did you make this question yourself / are you sure it has a nice solution?

RealiseNothing · Jun 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
Did you make this question yourself / are you sure it has a nice solution?

Yes/no.

Just pulled it from the top of my head to get this thread alive again. Thought even if the answer wasn't nice it would still produce decent discussion.

RealiseNothing · Jun 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Also optimal as in, would give you the best chance of winning.

I added the two parts to see if there was any difference in the optimal strategy with the knowledge of when your opponent says stop (maybe a bit of game theory? idk). My intuition says there should be.

seanieg89 · Jun 24, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Also optimal as in, would give you the best chance of winning.

I added the two parts to see if there was any difference in the optimal strategy with the knowledge of when your opponent says stop (maybe a bit of game theory? idk). My intuition says there should be.

Well by symmetry, no strategy can assure you better than a 50% result. (Here a draw counts as half of a win). I stated a trivial optimal solution for (i), but if we have no information about our opponents action then the theory behind this game is entirely probabilistic...and looks quite messy.

Perhaps someone else will be able to say something more useful about it.

abecina · Jun 29, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ a,b,c,d \ $be positive reals that sum up to 1$ \\ \\ $Show that$ \ \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq \frac{1}{2}$

$\begin{align*}\frac{a^2}{a+b}-\frac{a}{2}&= \frac{2a^2-a(a+b)}{2(a+b)}\\&= \frac{a^2-ab}{2(a+b)}\\&\geq \frac{a^2-(\frac{a^2}{2}+\frac{b^2}{2})}{2(a+b)}\\&= \frac{a^2-b^2}{4(a+b)}\\&=\frac{a-b}{4}, a+b>0\\\therefore \Big(\frac{a^2}{a+b}-\frac{a}{2}\Big) + \Big(\frac{b^2}{b+c} -\frac{b}{2}\Big)+ \Big(\frac{c^2}{c+d}-\frac{c}{2}\Big) + \Big(\frac{d^2}{d+a}-\frac{d}{2}\Big)&\geq \frac{a-b}{4}+\frac{b-c}{4}+\frac{c-d}{4}+\frac{d-a}{4}\\&=0\\\therefore \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} &\geq \frac{a+b+c+d}{2}\\&=\frac{1}{2} \end{align*}$

Sy123 · Jun 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

abecina said:
$\begin{align*}\frac{a^2}{a+b}-\frac{a}{2}&= \frac{2a^2-a(a+b)}{2(a+b)}\\&= \frac{a^2-ab}{2(a+b)}\\&\geq \frac{a^2-(\frac{a^2}{2}+\frac{b^2}{2})}{2(a+b)}\\&= \frac{a^2-b^2}{4(a+b)}\\&=\frac{a-b}{4}, a+b>0\\\therefore \Big(\frac{a^2}{a+b}-\frac{a}{2}\Big) + \Big(\frac{b^2}{b+c} -\frac{b}{2}\Big)+ \Big(\frac{c^2}{c+d}-\frac{c}{2}\Big) + \Big(\frac{d^2}{d+a}-\frac{d}{2}\Big)&\geq \frac{a-b}{4}+\frac{b-c}{4}+\frac{c-d}{4}+\frac{d-a}{4}\\&=0\\\therefore \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} &\geq \frac{a+b+c+d}{2}\\&=\frac{1}{2} \end{align*}$

Elegant solution, well done!

Here is mine:

$\frac{a^2}{a+b} = \frac{a^2+ab-ab}{a+b} = a - \frac{ab}{a+b}$

$\\ (a+b)^2 \geq 4ab \Rightarrow \ \frac{a+b}{4} \geq \frac{ab}{a+b} \\ \\ \Rightarrow \ a - \frac{ab}{a+b} \geq a - \frac{a+b}{4}$

$\therefore \ \frac{a^2}{a+b} + \frac{b^2}{b+c} + \frac{c^2}{c+d} + \frac{d^2}{d+a} \geq (a+b+c+d) - \frac{2a + 2b + 2c + 2d}{4} = \frac{1}{2} \ \ \square$

Sy123 · Jun 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find a polynomial with integer co-efficients, with one of its roots being$ \ \left( \frac{5}{3} \right)^{1/6} + \left(\frac{3}{5} \right)^{1/6}$

seanieg89 · Jun 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$$Let $a,b,c\geq 0.\\ \\$Prove that$: \\ \\ abc\geq (a+b-c)(a+c-b)(b+c-a).$

HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

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