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HSC 2012 MX2 Marathon (archive) (4 Viewers)

Trebla

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Re: 2012 HSC MX2 Marathon

is the riemann hypothesis thing actually syllabus?
lol, this is a Riemann sum for an integral. The Riemann hypothesis is a completely different thing altogether.
 

deswa1

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Re: 2012 HSC MX2 Marathon

Since we are on important mathematical problems, prove P is not NP.
P=NP if P=0 or N=1. So as long as neither of those cases apply, P does not equal NP. Do I get $1,000,000 now? :)
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Here's a nice simple one.

Find the locus of:



Note: Try to use a method that minimises algebra.
 

zeebobDD

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Re: 2012 HSC MX2 Marathon

Assume two people are playing a game with a Dice, X and Y, to determine the winner, the player who rolls a 6 first will win the game

If X starts rolling first, Find the probablility that X will win the game
 

nightweaver066

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Re: 2012 HSC MX2 Marathon

How did you come by this answer?
|z - i| starts from 1 on the imaginary axis.

|z - 1| starts from 1 on the real axis.

Perpendicular bisector of the chord joining those two points is y = x

For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore
 

nightweaver066

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Re: 2012 HSC MX2 Marathon

Assume two people are playing a game with a Dice, X and Y, to determine the winner, the player who rolls a 6 first will win the game

If X starts rolling first, Find the probablility that X will win the game
P(winning on first go) = 1/6

P(winning on second go) = 5/6 x 5/6 x 1/6

P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6

By continuing the trend and adding all the possible cases..

P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ...

Common ratio of (5/6)^2 therefore it is a geometric progression.

=

 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

|z - i| starts from 1 on the imaginary axis.

|z - 1| starts from 1 on the real axis.

Perpendicular bisector of the chord joining those two points is y = x

For |z - i| >= |z - 1|, has to be the region below and including y = x, therefore
Nice.

The reason why I asked is because as I was waiting for a MATH1001 (Differential Calculus) tutorial, I observed the tutor letting z=x+iy etc and had about half a board of working out before arriving at the same answer, which could have been acquired within a few seconds *after thinking for a bit*.
 

nightweaver066

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Re: 2012 HSC MX2 Marathon

Nice.

The reason why I asked is because as I was waiting for a MATH1001 (Differential Calculus) tutorial, I observed the tutor letting z=x+iy etc and had about half a board of working out before arriving at the same answer, which could have been acquired within a few seconds *after thinking for a bit*.
lol.. I hope whatever locus you were looking at in your tutorial was more complex than the one you gave.
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Oh sorry, I wasn't clear. I was waiting outside for the first year tutorial to finish.

This subject MATH1001 is mostly done by people who did not do Extension 2 Maths, so the idea of finding locus with complex numbers etc is alien to the vast majority.

Anyway new question, a little bit different from what you guys/girls are used to seeing =)



Hint: Use the equivalent of a telescoping sum, but for products.
 
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zeebobDD

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Re: 2012 HSC MX2 Marathon

P(winning on first go) = 1/6

P(winning on second go) = 5/6 x 5/6 x 1/6

P(winning on third go) = 5/6 x 5/6 x 5/6 x 5/6 x 1/6

By continuing the trend and adding all the possible cases..

P(X winning) = 1/6 + 1/6 x (5/6)^2 + 1/6 x (5/6)^4 + ...

Common ratio of (5/6)^2 therefore it is a geometric progression.

=

uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc
 

nightweaver066

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Re: 2012 HSC MX2 Marathon

uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc
I thought that they rolled the dice one at a time.

For X to win on the second go, he has to miss the first time, Y then has to miss, and then he gets 6 on his next roll.

Apply the same thing for the next go.
 

Nooblet94

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Re: 2012 HSC MX2 Marathon

uhmm why do you multiply it by 5/6 * 5/6 on the second go.. etc
Because for X to win the second time, he must lose on his roll (probability 5/6), and Y must also lose (probability 5/6).

EDIT: beat me to it :(
 

Nooblet94

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Re: 2012 HSC MX2 Marathon

Oh sorry, I wasn't clear. I was waiting outside for the first year tutorial to finish.

This subject MATH1001 is mostly done by people who did not do Extension 2 Maths, so the idea of finding locus with complex numbers etc is alien to the vast majority.

Anyway new question, a little bit different from what you guys/girls are used to seeing =)



Hint: Use the equivalent of a telescoping sum, but for products.
<a href="http://www.codecogs.com/eqnedit.php?latex=\prod_{k=2}^n\left(1@plus;\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k@plus;1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n@plus;1}{n} =\frac{n@plus;1}{2}\\ ~\\ $For the product to be an integer $n@plus;1$ must be divisible by 2.\\ $\therefore n$ must be odd" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" title="\prod_{k=2}^n\left(1+\frac{1}{k}\right) = \prod_{k=2}^n\left(\frac{k+1}{k}\right) = \frac{3}{2}\cdot\frac{4}{3}\cdot \frac{5}{4}\cdot ...\cdot \frac{n}{n-1}\cdot\frac{n+1}{n} =\frac{n+1}{2}\\ ~\\ $For the product to be an integer $n+1$ must be divisible by 2.\\ $\therefore n$ must be odd" /></a>

New question:

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}@plus;\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." title="\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." /></a>
 
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Carrotsticks

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Re: 2012 HSC MX2 Marathon

Haha that question has been asked to death.

But nonetheless it is indeed a good one to practise!
 

jet

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Re: 2012 HSC MX2 Marathon

Here's a nice one:

 

kingkong123

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Re: 2012 HSC MX2 Marathon

New question:

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}@plus;\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." title="\\ P(x_1,y_1)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ \\~\\A line drawn from the centre $O(0,0)$ parallel to the tangent at $P$ meets the ellipse at $Q$. \\~\\Prove that the area of $\triangle OPQ$ is independent of the position of $P." /></a>
 
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