HSC 2012 MX2 Marathon (archive) (2 Viewers)

Nooblet94 · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Get studying. Every assessment counts!

I know everything I need to know, the problem is that it's a "formal interview" - the teacher pretends to be someone who needs legal advice and we're a lawyer answering their questions.

I hate public speaking (it's in front of the entire class) so I'll be really nervous and I'm not incredibly quick on my feet, so if I get asked tricky questions (which is bound to happen, last year they asked ridiculous stuff like "if the marriage celebrant dies during the ceremony, is the marriage still valid?") I'll probably panic and get fucked over. Also, I'm not on the nicest terms with the teacher who's marking it, so that might be a bit interesting

.

Carrotsticks · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

cutemouse said:
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.

Ohh alright then, I only learnt Lagrange Multipliers a few days ago whilst reading up ahead. Thanks for this! I'll keep this in mind next time.

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

Is this the Letters one?

Nooblet94 said:
For the Fibonnacci/triangle question, is this right? From what I can see it is, but I've never actually done a proof by contradiction before so I might've fucked up somewhere.

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $Fibonacci numbers are defined by $F_n=F_{n-1}@plus;F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B@plus;C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B@plus;C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." title="\\ $Fibonacci numbers are defined by $F_n=F_{n-1}+F_{n-2}$ so it is clear that if $A,B,C$ are all Fibonacci numbers and A is the largest then $A\geq B+C\\ ~\\ $Assume there exists a triangle such that all three sides are Fibonacci numbers. Let $A$ be the length of the longest side and $B$ and $C$ the other two sides. We know from the triangle inequality that $A< B+C$ which is a clear contradiction of our assumption. Hence, there exists no triangle such that all three sides are Fibonacci numbers." /></a>

That is correct, but you said "Clear contradiction of assumption". You should have said that it contradicted the recursive definition of the Fibbonaci Numbers.

Good proof.

SpiralFlex · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
I know everything I need to know, the problem is that it's a "formal interview" - the teacher pretends to be someone who needs legal advice and we're a lawyer answering their questions.

I hate public speaking (it's in front of the entire class) so I'll be really nervous and I'm not incredibly quick on my feet, so if I get asked tricky questions (which is bound to happen, last year they asked ridiculous stuff like "if the marriage celebrant dies during the ceremony, is the marriage still valid?") I'll probably panic and get fucked over. Also, I'm not on the nicest terms with the teacher who's marking it, so that might be a bit interesting .

Weird assessment haha. Gl.

Carrotsticks · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:
Hey guys I need some help here. When a question asks prove... How do you go about it? When I see a proof at university my initial thoughts tend to be "how the fuck am I supposed to see that in an exam?".

Eventually you learn many different methods of proving things including Induction, contradiction etc.

When I do my proofs, I think about it step by step

ie: I need to prove X, but to prove X I need to prove Y, but to prove Y I need to prove Z etc etc and I keep working backwards until I need to prove something fairly trivial.

Then a 'chain reaction' occurs, leading to the final answer.

mnmaa · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Eventually you learn many different methods of proving things including Induction, contradiction etc.

When I do my proofs, I think about it step by step

ie: I need to prove X, but to prove X I need to prove Y, but to prove Y I need to prove Z etc etc and I keep working backwards until I need to prove something fairly trivial.

Then a 'chain reaction' occurs, leading to the final answer.

Thats a great way of thinking about it, thanks carrotsticks. Also I have a useless professor for intro to real analysis(calculus really), are there any books you would recommend that explain stuff simply and well?Im currently using Introduction to real analysis by william trench, its not that great tbh.

Carrotsticks · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:
Thats a great way of thinking about it, thanks carrotsticks. Also I have a useless professor for intro to real analysis(calculus really), are there any books you would recommend that explain stuff simply and well?Im currently using Introduction to real analysis by william trench, its not that great tbh.

Can't really help you out there as I am studying Analysis this semester too (first lecture of the year this week in 12 hours from now ^^)

But here are some online textbooks:

http://www.jirka.org/ra/realanal.pdf

http://www.math.harvard.edu/~ctm/home/text/class/harvard/114/07/html/home/course/course.pdf

http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF

http://classicalrealanalysis.com/Documents/BBT-hypertexted-SAMPLE-June2008.pdf

mnmaa · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

nice, what year are you in?

AAEldar · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

More 4U Maths, less Uni talk. Feel free to visitor message/PM.

Carrotsticks · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Prove that all curves in the form:

$x^{n} + y^{n} = 1, $ where $ n\geq 2$

are tangential to each other at at LEAST one point.

Hint: Implicit differentiation.

seanieg89 · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Prove that all curves in the form:

$x^{n} + y^{n} = k, $ where $ n,k \in \mathbb{R^+}$

Are mutually tangential.

Hint: Implicit differentiation.

Thats not true as stated. Let (n,k)=(1,1),(2,1). The circle is not tangential to the line...

Carrotsticks · Mar 4, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Thats not true as stated. Let (n,k)=(1,1),(2,1). The circle is not tangential to the line...

Thank you, corrected by adding the restriction for n and changed the question slightly after observing an interesting result.

Nooblet94 · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x^n@plus;y^n=1\\ nx^{n-1}@plus;ny^{n-1}\cdot \frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}~($for $n\geq 2)\\ ~\\ $Clearly, $(1,0)$ and $(0,1)$ both satisfy the equation $x^n@plus;y^n=1$ for all values of n and $\frac{dy}{dx}$ has the same values at those points for all $n \geq 2$. Hence, all curves of the form $x^n@plus;y^n=1~(n\geq 2)$ are tangential at at least two points, namely (1,0) and (0,1)\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x^n+y^n=1\\ nx^{n-1}+ny^{n-1}\cdot \frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}~($for $n\geq 2)\\ ~\\ $Clearly, $(1,0)$ and $(0,1)$ both satisfy the equation $x^n+y^n=1$ for all values of n and $\frac{dy}{dx}$ has the same values at those points for all $n \geq 2$. Hence, all curves of the form $x^n+y^n=1~(n\geq 2)$ are tangential at at least two points, namely (1,0) and (0,1)\\" title="\\ x^n+y^n=1\\ nx^{n-1}+ny^{n-1}\cdot \frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}~($for $n\geq 2)\\ ~\\ $Clearly, $(1,0)$ and $(0,1)$ both satisfy the equation $x^n+y^n=1$ for all values of n and $\frac{dy}{dx}$ has the same values at those points for all $n \geq 2$. Hence, all curves of the form $x^n+y^n=1~(n\geq 2)$ are tangential at at least two points, namely (1,0) and (0,1)\\" /></a>

zeebobDD · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
View attachment 24537

$C: x^2 + y^2 = R^2$

$L: ax + by + c = 0$

$i) For L to be tangent to C, the perpendicular distance from the centre of the circle (0,0) to L must equal to R$

$\therefore |\frac{c}{\sqrt{a^2 + b^2}}| = R$

$Squaring both sides$

$\frac{c^2}{a^2 + b^2} = R^2$

$\therefore R^2(a^2 + b^2) = c^2$

$$ii) $ |\frac{2a + c}{\sqrt{a^2 + b^2}}| + |\frac{-2a + c}{\sqrt{a^2 + b^2}}| = 6$

$Squaring both sides$

$\frac{4a^2 + 4ac + c^2}{a^2 + b^2} + \frac{2}{a^2 + b^2}|c^2 - 4a^2| + \frac{4a^2 - 4ac + c^2}{a^2 + b^2} = 36$

$\frac{8a^2 + 2c^2}{a^2 + b^2} + \frac{2}{a^2 + b^2}|c^2 - 4a^2| = 36$

$$As $ c > 2a => c^2 > 4a^2, \; c^2 - 4a^2 > 0$

$\frac{8a^2 + 2c^2}{a^2 + b^2} + \frac{2c^2 - 8a^2}{a^2 + b^2} = 36$

$\frac{4c^2}{a^2 + b^2} = 36$

$c^2 = 9(a^2 + b^2)$

$$Similar to part (i), L must be tangent to the circle $ x^2 + y^2 = 9$

SpiralFlex · Mar 6, 2012

Re: 2012 HSC MX2 Marathon

zeebobDD said:
View attachment 24537

Ewww maffs.

By perpendicular distance formula,

$\frac{|c|}{\sqrt{a^2+b^2}}=R$

$R^2(a^2+b^2)=c^2$

$\frac{|c-2a|}{\sqrt{a^2+b^2}}+\frac{|c+2a|}{\sqrt{a^2+b^2}}=6$

Squaring all sides,

$\frac{(c-2a)^2+(c+2a)^2+2(c-2a)(c+2a)}{\sqrt{a^2+b^2}}=36$

$=\frac{4c^2}{a^2+b^2}=36$

$9(a^2+b^2)=c^2$

As proven before, the line is a tangent if the condition holds true. Clearly the form is of equivalence to, what we have shown before.

Your equation of your circle is,

$x^2+y^2=9$

IamBread · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

Looks like we need more questions!

$z=1+i\sqrt{3}, w =1+i \\\\ $Show $ cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}}$

Aesytic · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

the complex number z and w in mod-arg form are 2cis(pi/3) and root2 cis(pi/4) respectively
.'. zw = 2root2 cis(pi/3 + pi/4)
=2root2cis(7pi/12)
= 2root2*cos(7pi/12) + i2root2sin(7pi/12)

also, zw = (1+iroot3)(1+i)
= 1 - root3 + i(root3 + 1)

equating real parts,
2root2*cos(7pi/12) = 1 - root3
.'.cos(7pi/12) = [1-root3]/[2root2]

zeebobDD · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Looks like we need more questions!

$z=1+i\sqrt{3}, w =1+i \\\\ $Show $ cos\frac{7\pi}{12} = \frac{1-\sqrt{3}}{2\sqrt{2}}$

LOL this came in our maths exam today, except it was show sin5pi/12:L

rolpsy · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:
$\\\textup{Let } z = r\, \mathrm{cis}(\theta).\\\textup{Show that } |z+1|\times|z-1| = 1 \textup{ implies } r^2 = 2\cos(2\theta).$

Nooblet94 · Mar 8, 2012

Re: 2012 HSC MX2 Marathon

rolpsy said:

You sure that question's right?

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