HSC 2012 MX2 Marathon (archive) (2 Viewers)

SpiralFlex · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Wut?^

Think twas a typo.

bleakarcher said:
Spiral, where did you get the expression after you wrote system of linear equations from?

Substituted the F0, F1

bleakarcher · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
$F_{n+2}=F_{n+1}+F_{n}$

$$Let F_{n}=a^n$

Forming a polynomial

$a^{n+2}=a^{n+1}+a^n$

$a^2-a-1=0$

$a=\psi, \varphi$

$\therefore F_{n}=\varphi^n$

System of linear equations,

$F_{n}=A \varphi^n +B\psi^n=1$

Since we know,

$F_{0}=A+B=1$

$F_{1}=A \varphi +B\psi=1$

Solve and arrange.

$F_{n}=\frac{\varphi^n - \psi^n}{\sqrt{5}}$

Now you can deduce Lucas my neighbour.

Why is it that when I sub n=0 into F(n) I get 1/sqrt(5) when it is meant to be one?

kingkong123 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
How can a modulus result in an imaginary number? lol..

trueeee. there was a typo in the exercise. i think it should be Re(z) + 2 not 2i. In that case it will be a parabola, focal length 1, vertex (-1,-2) , concave up. right?

cutemouse · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

A bit of several variables calculus...

(i) Use the method of Lagrange multipliers, or otherwise, to find the maximum value of f(x,y)=4xy on the ellipse x^2 + 4y^2=4.

(ii) Give a geometrical interpretation of the result in (i).

Aesytic · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
trueeee. there was a typo in the exercise. i think it should be Re(z) + 2 not 2i. In that case it will be a parabola, focal length 1, vertex (-1,-2) , concave up. right?

mine ended up being a sideways parabola, so i had one that was "concave right"

D94 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Aesytic said:
mine ended up being a sideways parabola, so i had one that was "concave right"

This is right.

Trebla · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Another reminder that this thread is intended for HSC students. If you want to post some uni maths questions, go to the extracurricular topics forum.

kingkong123 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Aesytic said:
mine ended up being a sideways parabola, so i had one that was "concave right"

yea sorry, concave right. vertex and focal length were correct?

Carrotsticks · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
A bit of several variables calculus...

(i) Use the method of Lagrange multipliers, or otherwise, to find the maximum value of f(x,y)=4xy on the ellipse x^2 + 4y^2=4.

(ii) Give a geometrical interpretation of the result in (i).

To be fair I'll answer the question (because I hate it when I post a question and nobody does it).

$\\ f(x,y) = 4xy $ and $ g(x,y) = x^{2} + 4y^{2} = 4 \\\\ F(x,y,\lambda) = f(x,y) - \lambda \left ( g(x,y) - 4 \right ) \\\\ F(x,y,\lambda) = 4xy - \lambda \left (x^{2} + 4y^{2} - 4 \right ) \\\\ \frac{\partial}{\partial x} F_{x,y,\lambda} = 4y - 2x \lambda = 0 \Rightarrow 2y - x \lambda = 0 \\\\ \frac{\partial}{\partial y} F_{x,y,\lambda} = 4x - 8y \lambda = 0 \Rightarrow x - 2y \lambda = 0 \\\\ \frac{\partial}{\partial \lambda} F_{x,y,\lambda} = -(x^{2} + 4y^{2} - 4) = 0 \Rightarrow x^{2} + 4y^{2} - 4 \\\\ $Solving simultaneously yields: $ \\\\ x = \pm \sqrt{2} \\\\ y = \pm \frac{1}{\sqrt{2}} \\\\ $And so we see that the maximum value occurs at $ \left ( \pm \sqrt{2} , \pm \frac{1}{\sqrt{2}} \right )$

$\\ $Geometrically, this means that the highest possible z value of the curve $ f_{x,y} = 4xy $ within the boundries of the ellipse $ x^{2} + 4y^{2} = 4 $ occurs at $ \left ( \pm \sqrt{2} , \pm \frac{1}{\sqrt{2}} \right ), $ which consequently yields a z value of 4$$

Now for Larg's question. I did it a different way to what I presume you expected given that you defined the modulus and gave a basic property (which I didn't use lol).

This method was inspired by an idea that I got a few years ago when I was in the library reading up on various sequences including the Fibbonaci series, where they asserted that proving divisibility of homogenous linear recurrence relations of order n requires grouping of the modulo into groups of n (in this case in pairs since the Fibbonaci sequence is of degree 2, resulting in p^2 pairs)

$\\ $Let the nth Fibbonaci number be $ F_n $ and define some arbitrary constant k such that $ F_n \mod(k) = F_n ' \\\\ $Consider the sequence $ F_1 ',F_2 ',F_3 ',...,F_n ',F_{n+1} ',... $We re-arrange the sequence into pairs such that we have $ (F_1 ',F_2 '),(F_2 ',F_3 '),...,(F_n ',F_{n+1} '),... $ and we assume that $(F_n ',F_{n+1} ') $ is the first pair of remainders that appears more than once.$ \\\\ $Suppose that $(F_n ',F_{n+1} ') $ were the most trivial and firstly unique pair for some positive integer(s), naturally being the pair (1,1).$ \\\\ $We assume the opposite is true ie: there exists some $ m \in \mathbb{N}$ such that $ (F_n ',F_{n+1} ') = (F_m ',F_{m+1} ') $ where $ m > n \\\\ $It can easily be shown using the Recursive definition of the Fibbonaci Sequence $ F_n = F_{n-1} + F_{n-2} $ that $ F_{m-1} '=F_{n-1}' $ and as a consequence of this, we find that (F_{m-1} ',F_{m} ') = (F_n ',F_{n-1} ') \\\\$

$\\ $HOWEVER, this contradicts our our original assumption that $ (F_n ',F_{n+1} ') $ is the foremost pair in the sequence. As a result, the first repeated (and equal) pair of remainders is indeed our trivial case $ (1,1). $ So we can safely assume $ (F_n ',F_{n+1} ') = (1,1) $ so hence $ F_n $ and $ F_{n+1} \mod (k) = 1 \\\\ $But this directly implies that $ \left [F_{n+1} - F_{n} \right ] \mod (k) = 0 $ using the recursive definition $ F_{n+1} - F_{n} = F_{n-1} \\\\ $ Thus there exists some $ F_j $ where $ j \in \mathbb{N} $ such that $ F_j \mod (k) = 0 $ where $ k \in \mathbb{N}$$

Carrotsticks · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Oh and here's a new 4U question:

Prove that the ellipse and hyperbola in canonical form intersect orthogonally IFF they share the same focus.

Carrotsticks · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Or alternatively (much easier): Prove that there does not exist a right angled triangle such that the length of all 3 sides are Fibbonaci numbers.

Nooblet94 · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

kingkong123 said:
yea sorry, concave right. vertex and focal length were correct?

Yep

Carrotsticks · Mar 3, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Oh and here's a new 4U question:

Prove that the ellipse and hyperbola in canonical form intersect orthogonally IFF they share the same focus.

Many many thanks to Math Man for taking the liberty of making this question more do-able for everybody!

cutemouse · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
$\\ $Geometrically, this means that the highest possible z value of the curve $ f_{x,y} = 4xy $ within the boundries of the ellipse $ x^{2} + 4y^{2} = 4 $ occurs at $ \left ( \pm \sqrt{2} , \pm \frac{1}{\sqrt{2}} \right ), $ which consequently yields a z value of 4$$

Close.

Geometrically, f(x,y)=4xy is the area of a rectangle inscribed inside the given ellipse. Thus the result found geometrically is area of the largest possible rectangle that can be inscribed in the ellipse.

Nooblet94 · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Many many thanks to Math Man for taking the liberty of making this question more do-able for everybody!

Can you give me a hint? I'm meant to be studying for legal but I keep coming back to this question and I'm not getting anywhere

EDIT: Also, for the fibonacci numbers question, I might be wrong, but isn't that true for all triangles, not only right angled ones?

math man · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

A hint is the reason I wrote Part one was
To show you that if the hyperbola and ellipse
have the Same a and b value they can't share
the same Focus.

Nooblet94 · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

math man said:
A hint is the reason I wrote Part one was
To show you that if the hyperbola and ellipse
have the Same a and b value they can't share
the same Focus.

Yep, I'd realised that. I guess I'll just wait till someone else posts the answer

math man · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

For the hyperbola and ellipse use the formula
To work out eccentricity and use it to deduce
A nice relation. Also don't actually solve them
simultaneously, let the point of intersection be
P(x1,y1)

SpiralFlex · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Maff Man, get ready for that prize that you are going to give me. I solved the probability question on da board.

bleakarcher · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
Why is it that when I sub n=0 into F(n) I get 1/sqrt(5) when it is meant to be one?

Spiral how did u know that F(sub 0)=1? There is one such thing as the 0th term so isnt it F1?

HSC 2012 MX2 Marathon (archive) (2 Viewers)

Well-Known Member

Active Member

Member

Account Closed

Member

New Member

Administrator

Member

Retired

Retired

Retired

Premium Member

Retired

Account Closed

Premium Member

Member

Premium Member

Member

Well-Known Member

Active Member

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