HSC 2012 MX2 Marathon (archive) (1 Viewer)

[seanieg89]

Re: 2012 HSC MX2 Marathon

A pretty fast way to prove the "reflective" properties of conics.

Let ABC be a triangle. Let X be a point on BC.

i) Prove that AB/BX = AC/CX if and only if the line segment AX bisects the angle BAC.

ii) Using the above, state and prove the reflective property of the hyperbola: x^2/a^2 - y^2/b^2 = 1.

 

[math man]

Re: 2012 HSC MX2 Marathon

fastest way to prove reflection property, from hsc a few years ago:

 

[math man]

Re: 2012 HSC MX2 Marathon

also here are some nice integration questions:

 

[math man]

Re: 2012 HSC MX2 Marathon

Here's a nice one:

i really like (b) a lot, ill let the 4u kids have a go first, but if no one posts a solution ill post one tonight

 

[zeebobDD]

Re: 2012 HSC MX2 Marathon

can someone explain to me why a point say Q( a sec(90-x), b tan(90-x)) = a cosec x , b cot x?

 

[deswa1]

Re: 2012 HSC MX2 Marathon

sec(90-x)=1/cos(90-x)=1/sinx=cosecx. Similarly, tan(90-x)=cotx

 

[deswa1]

Re: 2012 HSC MX2 Marathon

This is a question from our 4U test this afternoon (I hope I remembered it correctly...):

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}@plus;a_{2}@plus;...@plus;a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" title="\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" /></a>

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

This is a question from our 4U test this afternoon (I hope I remembered it correctly...):

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}@plus;a_{2}@plus;...@plus;a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" title="\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" /></a>

haha.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Well really I suppose more justification is needed etc but meh.

 

[Internet_Police]

Re: 2012 HSC MX2 Marathon

Well really I suppose more justification is needed etc but meh.

not good enough son.

 

[kingkong123]

Re: 2012 HSC MX2 Marathon

haha.

m8.

Well really I suppose more justification is needed etc but meh.

please explain

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

haha.

It's not clear at all why the whole second part thingo after the sum ai is necessarily positive.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

It's not clear at all why the whole second part thingo after the sum ai is necessarily positive.

Haha I know, that's why I said it needs justification, because whether it's a plus or a minus is dependent on what n is.

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

Haha I know, that's why I said it needs justification, because whether it's a plus or a minus is dependent on what n is.

It doesnt just depend on whether its a plus or minus though. Like, you have I think n-2 or n-3 sums after that, and its not clear that given a certain sum it has say, a bigger absolute value, than the one that comes after it for instance. Sorry but I just dont buy the argument at all

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

It doesnt just depend on whether its a plus or minus though. Like, you have I think n-2 or n-3 sums after that, and its not clear that given a certain sum it has say, a bigger absolute value, than the one that comes after it for instance. Sorry but I just dont buy the argument at all

We are told that a_k E (0,1), so this raises the issue to prove that:



And it can the inequality follows.

Sorry if I'm being vague. The new *fabulous* Carslaw rooms closes in 5 minutes and I'm already pushing the security guard's patience =D

I'll put a more rigorous proof when I get home and have dinner. Sydney Uni Union sandwiches just don't cut it nowadays.

 

[seanieg89]

Re: 2012 HSC MX2 Marathon

The induction is straightforward.

 

[largarithmic]

Re: 2012 HSC MX2 Marathon

We are told that a_k E (0,1), so this raises the issue to prove that:



And it can the inequality follows.

Sorry if I'm being vague. The new *fabulous* Carslaw rooms closes in 5 minutes and I'm already pushing the security guard's patience =D

I'll put a more rigorous proof when I get home and have dinner. Sydney Uni Union sandwiches just don't cut it nowadays.

I assume the inequality signs are supposed to be the other way; but youve got a problem because there are by no means the same number of summands in sum aiaj and sum aiajak.

 

[math man]

Re: 2012 HSC MX2 Marathon

This is a question from our 4U test this afternoon (I hope I remembered it correctly...):

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}@plus;a_{2}@plus;...@plus;a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" title="\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}+a_{2}+...+a_{n})\\ \textup{for positive integers n}\geq 2 \textup{ where}\\ 0<a_{k}<1 \textup{ for }1<k<n" /></a>

this the tech test?

 

[deswa1]

Re: 2012 HSC MX2 Marathon

this the tech test?

Yep

 

[jenslekman]

Re: 2012 HSC MX2 Marathon

mate, how long was teh solution for that?