HSC 2012 MX2 Marathon (archive) (5 Viewers)

[Nooblet94]

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)

Hmmm okay. I'll try it tomorrow after my goddamn legal assessment (which I possibly might fail).

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>

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

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

1, 1, 2, 3, 5. What do you mean? I just used this sequence and that is how I knew?

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Hmmm okay. I'll try it tomorrow after my goddamn legal assessment (which I possibly might fail).

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>

Get studying. Every assessment counts!

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

1, 1, 2, 3, 5. What do you mean? I just used this sequence and that is how I knew?

How come when you substitute n=0 into your final formula you get 0 when F(0)=1?

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

so confused...

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

so confused...

Sorry that's a good point haha! I don't know what I did wrong, I will revise the solution soon. But its now off to saving the world! Goodbye Bleak.

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

k thanks.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

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]

Re: 2012 HSC MX2 Marathon

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.

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?

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]

Re: 2012 HSC MX2 Marathon

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]

Re: 2012 HSC MX2 Marathon

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]

Re: 2012 HSC MX2 Marathon

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]

Re: 2012 HSC MX2 Marathon

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]

Re: 2012 HSC MX2 Marathon

nice, what year are you in?

 

[AAEldar]

Re: 2012 HSC MX2 Marathon

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

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Prove that all curves in the form:



are tangential to each other at at LEAST one point.

Hint: Implicit differentiation.

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

Prove that all curves in the form:



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]

Re: 2012 HSC MX2 Marathon

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]

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]

Re: 2012 HSC MX2 Marathon