HSC 2013 MX2 Marathon (archive) (7 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.

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