HSC 2013 MX2 Marathon (archive) (4 Viewers)

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Suppose three complex numbers $z_1~z_2~z_3$ make an equilateral triangle on an argand diagram. Show that:

$z_1^2 + z_2^2 + z_3^2 = |z_1||z_2| + |z_1||z_3| + |z_2||z_3|$

remove modulus signs.

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
remove modulus signs.

Thanks for that, I couldn't remember the exact condition/equation.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Also, more interestingly, the condition is equivalent to the three points forming an equilateral triangle. (If and only if).

cutemouse · Nov 14, 2012

Re: HSC 2013 4U Marathon

I think an 'elegant' approach to this is setting one of the points to 0, finding a condition and then using a change of coordinates.

Carrotsticks · Nov 14, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
I think an 'elegant' approach to this is setting one of the points to 0, finding a condition and then using a change of coordinates.

That is the fastest method that I know of at the moment, though there are most likely faster ones. A typical 'school test proof' would be to actually draw out arbitrary vectors z_1, z_2 and z_3 then work with the condition, though this causes problems because it only proves one direction of the statement.

w.l.o.g you can shift one of the points z_1, z_2 or z_3 to the origin, then solve the quadratic. Upon doing so, you should immediately acquire that the other two are plus/minus cis(pi/3) away whilst preserving modulus. Since the solving of the quadratic works both ways, you have consequently proved both directions of the 'iff' statement.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Carrotsticks said:
That is the fastest method that I know of at the moment, though there are most likely faster ones. A typical 'school test proof' would be to actually draw out arbitrary vectors z_1, z_2 and z_3 then work with the condition, though this causes problems because it only proves one direction of the statement.

w.l.o.g you can shift one of the points z_1, z_2 or z_3 to the origin, then solve the quadratic. Upon doing so, you should immediately acquire that the other two are plus/minus cis(pi/3) away whilst preserving modulus. Since the solving of the quadratic works both ways, you have consequently proved both directions of the 'iff' statement.

That used to be my fastest method. I just thought of something arguably faster though.

Let a=z1-z2, b=z2-z3, c=z3-z1. Observe that a+b+c=0 always.

The equation given is equivalent to a^2+b^2+c^2=0. But since a+b+c is also zero, this is equivalent to the fact that the monic polynomial with roots a,b,c must have zero coefficients for z and z^2. This is equivalent to a,b and c having the same modulus. This is equivalent to the triangle with vertices z1,z2,z3 being equilateral.

seanieg89 · Nov 14, 2012

Re: HSC 2013 4U Marathon

They are roughly equal in length probably.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Well that was an excellent question.

Solution

I feel like my solution is unnecessarily long however.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1)$

$Show that $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right = \frac{n}{6}(n+1)(n+2) \\ \\ $Hence prove that the product of any three consecutive integers is divisible by 6$

Note:

$\sum_{k=1}^r k = \frac{r}{2}(r+1)$

(Arithmetic series)

$\sum_{r=1}^n \frac{r}{2}(r+1) \\ \\ = 0.5 \sum_{r=1}^n r^2 +r \\ \\ = 0.5 [\sum_{r=1}^n r^2 + \sum_{r=1}^n r] \\ \\ = 0.5[\frac{n}{6}(n+1)(2n+1)+\frac{n}{2}(n+1)] \\ \\ = 0.5(n/6 (n+1) (2n+1+3) = \frac{n}{6}(n+1)(n+2)$

EDIT: If you want to read about this
http://en.wikipedia.org/wiki/Figurate_number

==============

$$Given the polynomial $ x^3-3x^2+5x+2012 $ has roots $ \alpha, \beta, \gamma$

$$i) Find $ (\alpha-1)(\beta-1)(\gamma -1)$

$$ii) Hence or otherwise find $ (\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)$

twinklegal19 · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Note:

$\sum_{k=1}^r k = \frac{r}{2}(r+1)$

(Arithmetic series)

$\sum_{r=1}^n \frac{r}{2}(r+1) \\ \\ = 0.5 \sum_{r=1}^n r^2 +r \\ \\ = 0.5 [\sum_{r=1}^n r^2 + \sum_{r=1}^n r] \\ \\ = 0.5[\frac{n}{6}(n+1)(2n+1)+\frac{n}{2}(n+1)] \\ \\ = 0.5(n/2 (n+1) (\frac{2n+4}{3}) = \frac{n}{6}(n+1)(n+2)$

==============

$$Given the polynomial $ x^3-3x^2+5x+2012 $ has roots $ \alpha, \beta, \gamma$

$$i) Find $ (\alpha-1)(\beta-1)(\gamma -1)$

$$ii) Hence or otherwise find $ (\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)$

$\\x^3-3x^2+5x+2012=0\Rightarrow x=\alpha,\beta,\gamma\\\sum \alpha=3\\\sum \alpha\beta=-5\\\sum\alpha\beta\gamma=-2012\\$i. $(\alpha-1)(\beta-1)(\gamma -1)\\=\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\alpha\gamma)+(\alpha+\beta+\gamma)-1\\=(-2012)-(-5)+(3)-1)\\=-2005\\$ii. $(\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)\\=(3-2\alpha)(3-2\beta)(3-2\gamma)\\=-8(\alpha\beta\gamma)+12(\alpha\beta+\beta\gamma+\alpha\gamma)-18(\alpha+\beta+\gamma)+27\\=-8(-2012)+12(-5)-18(3)+27\\=16009$

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

twinklegal19 said:
$\\x^3-3x^2+5x+2012=0\Rightarrow x=\alpha,\beta,\gamma\\\sum \alpha=3\\\sum \alpha\beta=-5\\\sum\alpha\beta\gamma=-2012\\$i. $(\alpha-1)(\beta-1)(\gamma -1)\\=\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\alpha\gamma)+(\alpha+\beta+\gamma)-1\\=(-2012)-(-5)+(3)-1)\\=-2005\\$ii. $(\beta+ \gamma - \alpha)(\alpha+\beta-\gamma)(\alpha+\gamma -\beta)\\=(3-2\alpha)(3-2\beta)(3-2\gamma)\\=-8(\alpha\beta\gamma)+12(\alpha\beta+\beta\gamma+\alpha\gamma)-18(\alpha+\beta+\gamma)+27\\=-8(-2012)+12(-5)-18(3)+27\\=16009$

Ah my second question was not built nicely. Sorry about that. (Was supposed to lead on from part i)
It should of been.
$x^3-2x^2+5x+2012$

Will think of another question soon.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

You are given:

$0<a_0, a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_0, a_1, a_2, .... , a_n \in \mathbb{N}$

$a_0+a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+1000a_3+100a_2+10a_1+a_0 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
You are given:

$0<a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_1, a_2, .... , a_n \in \mathbb{N}$

$a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+100a_3+10a_2+a_1 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

You sure that's right?

The general term is $10^ka_k$ so shouldn't it be $1000a_3$ and $100a_2$ and $10a_1$ ?

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
You sure that's right?

The general term is $10^ka_k$ so shouldn't it be $1000a_3$ and $100a_2$ and $10a_1$ ?

Apologies, will fix it now

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
You are given:

$0<a_0, a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_0, a_1, a_2, .... , a_n \in \mathbb{N}$

$a_0+a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N}$

$Prove by process of mathematical induction or otherwise$

$10^n a_n+10^{n-1}a_{n-1}+10^{n-2}a_{n-2}+...+1000a_3+100a_2+10a_1+a_0 = 3k \ \ \ \ k \in \mathbb{N}$

(So basically prove that expression is divisible by 3)

I'm going to skip the "prove true for n=0" bits as they are just trivial really.

Well we can assume:

$10^na_n + ... + a_0=3k$

Now we want to prove that:

$10^{n+1}a_{n+1} + 10^na_n + ... + a_0 = 3m$

By substituting in our assumption:

$10^{n+1}a_{n+1} + 3k$

Now we are given that:

$a_0 + a_1 +...+ a_n = 3p$

Hence:

$a_{n+1} = 3q - a_n - a_{n-1} - ... - a_1 - a_0$

Substituting this in gives us:

$10^{n+1}(3q - a_n - a_{n-1} - ... - a_1 - a_0) + 3k$

$10^{n+1}(3q - (a_n + a_{n-1} + ... + a_1 + a_0)) + 3k$

But we were given:

$a_n + a_{n-1} + ... + a_1 + a_0 = 3p$

By substituting this in:

$10^{n+1}(3q - 3p) + 3k$

$3(q - p)10^{n+1} + 3k$

$3((q - p)10^{n+1} + k)$

$=3m$ where $m=((q - p)10^{n+1} + k)$

Therefore it is divisible by 3 for all blah blah, etc.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I'm going to skip the "prove true for n=0" bits as they are just trivial really.

Well we can assume:

$10^na_n + ... + a_0=3k$

Now we want to prove that:

$10^{n+1}a_{n+1} + 10^na_n + ... + a_0 = 3m$

By substituting in our assumption:

$10^{n+1}a_{n+1} + 3k$

Now we are given that:

$a_0 + a_1 +...+ a_n = 3p$

Hence:

$a_{n+1} = 3q - a_n - a_{n-1} - ... - a_1 - a_0$

Substituting this in gives us:

$10^{n+1}(3q - a_n - a_{n-1} - ... - a_1 - a_0) + 3k$

$10^{n+1}(3q - (a_n + a_{n-1} + ... + a_1 + a_0)) + 3k$

But we were given:

$a_n + a_{n-1} + ... + a_1 + a_0 = 3p$

By substituting this in:

$10^{n+1}(3q - 3p) + 3k$

$3(q - p)10^{n+1} + 3k$

$3((q - p)10^{n+1} + k)$

$=3m$ where $m=((q - p)10^{n+1} + k)$

Therefore it is divisible by 3 for all blah blah, etc.

Nice work. I will post another question if I can think of one (unless someone else can post one?)

RealiseNothing · Nov 14, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work. I will post another question if I can think of one (unless someone else can post one?)

I would post one, but I'm saving them for the little complex numbers quiz I'm writing lol.

Sy123 · Nov 14, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
I would post one, but I'm saving them for the little complex numbers quiz I'm writing lol.

Alright haha. Ill post another one tomorrow (I will try to make it a Complex Numbers/Polynomials one)

Sy123 · Nov 15, 2012

Re: HSC 2013 4U Marathon

$$Given the polynomial $ S(x)=ax^3+bx^2 + bx+a$

$The polynomial has 3 real roots all of which have a multiplicity of 2 or less.$

$i) Factorise the polynomial as one linear and one quadratic factor$

$$ii) Show that if the polynomial has 3 real roots then $ b \geq 3a$

$$iii) What happens when $ b=3a$

twinklegal19 · Nov 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Given the polynomial $ S(x)=ax^3+bx^2 + bx+a$

$The polynomial has 3 real roots all of which have a multiplicity of 2 or less.$

$i) Factorise the polynomial as one linear and one quadratic factor$

$$ii) Show that if the polynomial has 3 real roots then $ b \geq 3a$

$$iii) What happens when $ b=3a$

~~still editing~~

$\\$i) by inspection, $(x+1)$ is a factor$\\S(x)=(x+1)(ax^2+cx+a)\\$equating coefficients of $x\\b=a+c\\c=b-a\\S(x)=(x+1)[ax^2+(b-a)x+a]$
$\\ax^2+(b-a)x+a=0\\\bigtriangleup =(b-a)^2-4a^2\\$For real roots, \bigtriangleup \geq 0$\\(b-a)^2-4a^2\geq 0\\(b-a-2a)(b-a+2a)\geq 0\\(b-3a)(b+a)\geq 0$

Edit: sorry got to study for an English exam omorrow, someone feel free to take over lol.

HSC 2013 MX2 Marathon (archive) (4 Viewers)

Well-Known Member

what is that?It is Cowpea

Well-Known Member

Account Closed

Retired

Well-Known Member

Well-Known Member

This too shall pass

This too shall pass

.

This too shall pass

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

This too shall pass

This too shall pass

.

Users Who Are Viewing This Thread (Users: 0, Guests: 4)