HSC 2013 MX2 Marathon (archive) (2 Viewers)

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Also for the 957 question, I am currently trying to find pythagorean triads that have multiples as 957 lol.

Like for the triple 3, 4, 5

957 is divisible by 3 into 319 times

Multiply 4 by 319 =/= square number

And it didnt work for the next couple of triples that would of worked, so back to square one I guess

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
Lol, is the answer a = 479, b=478; c=31, d=2; a=45, b = 42; c=161, d=158

That is one answer, good work.

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Also for the 957 question, I am currently trying to find pythagorean triads that have multiples as 957 lol.

Like for the triple 3, 4, 5

957 is divisible by 3 into 319 times

Multiply 4 by 319 =/= square number

And it didnt work for the next couple of triples that would of worked, so back to square one I guess

I would recommend swapping 957 for much smaller numbers, then trying to see a pattern (which is a really cool one at that imo).

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Actually I just remembered an awesome question (aye aye spiral )

If $a ~ b ~ c ~ d$ are all distinct positive integers, find what they are so that:

$a^2 - b^2 = c^2 - d^2 = 957$

Expressing 957 in its prime factors: 3*11*29

Now, a^2 - b^2 = (a-b)(a+b)=957

So now, take turns making a+b and a-b, from the factors of 957.

1.
let a+b=957
a-b=1
2a=958
therefore, a = 479, b=478

2.
let a+b=3*11
a-b=29
2a=62
therefore, a=31, b=2

3.
let a+b=11*29
a-b=3
2a=322
therefore, a=161, b=158

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
Expressing 957 in its prime factors: 3*11*29

Now, a^2 - b^2 = (a-b)(a+b)=957

So now, take turns making a+b and a-b, from the factors of 957.

1.
let a+b=957
a-b=1
2a=958
therefore, a = 479, b=478

2.
let a+b=3*11
a-b=29
2a=62
therefore, a=31, b=2

3.
let a+b=11*29
a-b=3
2a=322
therefore, a=161, b=158

Very nice solution.

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

The pattern I also wanted you to see was that any odd number can be expressed as a difference of two squares. To find these squares, you just add 1 and divide by 2, minus 1 and divide by 2. For example:

Say we want $a^2 - b^2 = 129$

$a = \frac{129+1}{2} = 65$

$b = \frac{129-1}{2} = 64$

$65^2 - 64^2 = 129$

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

a) Find, in modulus-argument form, the roots of the equation z^(2n+1) =1.
b) Hence factorise z^2n + z^(2n-1) + ... + z^2 + z +1 into quadratic factors with real coefficients.
c) Deduce that 2^n * sin(pi/(2n+1)) * sin(2pi/(2n+1)) * sin(3pi/(2n+1)) ... sin(n*pi/(2n+1)) = sqrt(2n+1)

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
The pattern I also wanted you to see was that any odd number can be expressed as a difference of two squares. To find these squares, you just add 1 and divide by 2, minus 1 and divide by 2. For example:

Say we want $a^2 - b^2 = 129$

$a = \frac{129+1}{2} = 65$

$b = \frac{129-1}{2} = 64$

$65^2 - 64^2 = 129$

aaah, I see =/

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
aaah, I see =/

Your way is good since it gets all the possible combinations for 'a' and 'b'.

RealiseNothing · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
That is really cool, and the algebraic way to show it is:

$a^2-b^2=k$

$(\frac{k+1}{2})^2-(\frac{k-1}{2})^2=\frac{k^2+2k+1-k^2+2k-1}{4}=k$

$\therefore a=\frac{k+1}{2}$
$b=\frac{k-1}{2}$

The only way a and b can be integral is if k is odd.

The way I found it was I used the fact that squares go up by odd numbers. Like to get from $1^2$ to $2^2$ we add 3, then from $2^2$ to $3^2$ we add 5. Then we add 7, then 9, then 11, etc. So we are always adding $2x+1$ if we let the number we are squaring be $x$ . Which is obvious by observing that:
$(x+1)^2 - x^2 = x^2 + 2x + 1 - x^2 = 2x + 1$

Also this can be observed by knowing the series:

$1 + 3 + 5 + ... + (2n-1) = n^2$

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
The way I found it was I used the fact that squares go up by odd numbers. Like to get from $1^2$ to $2^2$ we add 3, then from $2^2$ to $3^2$ we add 5. Then we add 7, then 9, then 11, etc. So we are always adding $2x+1$ if we let the number we are squaring be $x$ . Which is obvious by observing that:
$(x+1)^2 - x^2 = x^2 + 2x + 1 - x^2 = 2x + 1$

Also this can be observed by knowing the series:

$1 + 3 + 5 + ... + (2n-1) = n^2$

Nice.

Also bob for part b do we need to factorise the whole thing into quadratic factors or only 1 quadratic factor?

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

oh, factorise into a bunch of quadratic factors

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Is it:

$1+z+z^2+...+z^{2n}=(z^2+\frac{z}{n}+1)^n$

Making sure before I continue

bobmcbob365 · Nov 15, 2012

Re: HSC 2013 4U Marathon

not quite; but i'm curious as to how you came to that

Sy123 · Nov 16, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
not quite; but i'm curious as to how you came to that

Assume that the factorisation is: (something that probably was wrong in the first place)
$(z^2+bz+1)^n=1+z+z^2+...+z^{2n}$

Equate co-efficient of z

on LHS we can get z an n number of times

$(z^2+bz+1)(z^2+bz+1)...(z^2+bz+1)=1+z+...$

$nbz=z \\ \therefore b=\frac{1}{n}$

I did something completely wrong somewhere

bobmcbob365 · Nov 16, 2012

Re: HSC 2013 4U Marathon

Oh right, not sure if 1st line is right. But I think the question is leading on into expressing z^(2n+1) -1 in (z-z1)(z-z2)(z-z3)... form, where z1, z2, z3... are roots of z^(2n+1)=1

SpiralFlex · Nov 16, 2012

Re: HSC 2013 4U Marathon

I'm back from gaming. 14 hrs of L4D, let me think of something kwl.

Sy123 · Nov 16, 2012

Re: HSC 2013 4U Marathon

bobmcbob365 said:
Oh right, not sure if 1st line is right. But I think the question is leading on into expressing z^(2n+1) -1 in (z-z1)(z-z2)(z-z3)... form, where z1, z2, z3... are roots of z^(2n+1)=1

Oh wow, I cant believe I could not see that :/

$1+z+z^2+...+z^{2n}=\frac{z^{2n+1}-1}{z-1}=\frac{(z-1)(z-z_1)(z-\bar{z_1})...}{z-1}=(z^2-2\cos \frac{2\pi}{2n+1}+1)(z^2-2\cos \frac{4\pi}{2n+1}+1)...(z^2-2\cos \frac{2\pi n}{2n+1}+1)$

SpiralFlex · Nov 16, 2012

Re: HSC 2013 4U Marathon

$$(a) Prove$\;\frac{1+\cos \theta+i\sin \theta}{1-\cos \theta-i\sin \theta}=i\cot \frac{\theta}{2}$

$(b) Let$\;n \;$be a positive integer. Show that the roots of the equation$

$(z-1)^n+(z+1)^n=0......(*)$

$$can be written as$\;ia_{k}, \;$where$\; a_{k}\in \mathbb{R} , k=0,1....,n-1$

$(c) If$\;ia_{k} \;$are the roots of (*) in (b), show that$

$\sum_{k=0}^{n-1}(a_{k})^2=n(n-1)$

$$(d) Let$\;P_{0}, P_{1},....,P_{n-1}\;$be the n points of an Argand plane representing the roots in (*) in (b), and O be the origin. Q is the point representing$\;r(\cos \beta+i\sin \beta)\; $where$\; r \geq 0\;$and$\;\beta \in \mathbb{R}. \;$If$\; d_{k} \;$is the distance between$\;P_{k} \;$and$\;Q, \;$show that$\;\sum_{k=0}^{n-1}(d_{k})^2\;$is independent of$\; \beta.$

bobmcbob365 · Nov 16, 2012

Re: HSC 2013 4U Marathon

haha yeah, dw its 12:30. Heads get all screwy.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

Member

This too shall pass

what is that?It is Cowpea

Member

Member

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

Member

This too shall pass

Member

This too shall pass

Member

Well-Known Member

This too shall pass

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)