HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

[Examine]

Re: 2012 HSC MX1 Marathon

1. Using Latex. You use this to write up equations:
www.codecogs.com/latex/eqneditor.php

Then you input them in to this forum by inserting whatever you typed in to these brackets

 

[Examine]

Re: 2012 HSC MX1 Marathon

All this maths seems rather interesting. Can't wait for Extension 1 Prelims and HSC.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

All this maths seems rather interesting. Can't wait for Extension 1 Prelims and HSC.

*is

 

[hayabusaboston]

Re: 2012 HSC MX1 Marathon

*is

lol right on man

 

[deswa1]

Re: 2012 HSC MX1 Marathon

lol nobody has yet tried my question ='(

Don't worry Mr. Carrotsticks. I just have to finish my belonging essay (I know ...) and then I'll have a go.

 

[SpiralFlex]

Re: 2012 HSC MX1 Marathon

I leave you with more things to ponder.

















Have to dash and do some Physics, I will attempt the questions afterwards. (Be back at 9 PM) Keep them coming!

 

[tywebb]

Re: 2012 HSC MX1 Marathon

2 and e²

Next question:

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Here you go

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Next question:

Wait a second... this looks like the Ruse Extension 2 2011 paper lol.

 

[tywebb]

Re: 2012 HSC MX1 Marathon

Correct! But Ext. 1 students can do it.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Correct! But Ext. 1 students can do it.

Use the tan(A+B) thing where tanA=5 and tanB = 3, but twice. After some simplification, we have tan(A+B+C) = 0.

Thus A + B + C = arctan(0) = pi

I'm only saying this because my tablet just broke down lol.

 

[jdnRof]

Re: 2012 HSC MX1 Marathon

I'd be very careful about tanning and then arc tanning it since you need to consider the restrictions between -pi/2 and pi/2. Probably should include an argument about it.

Just to clarify carrotsticks, is one of your steps:

arctan (-4/7) + arctan (4/7)?

 

[RishBonjour]

Re: 2012 HSC MX1 Marathon

Did you guys finish the whole course already? FUCK I'm SO BEHIND

ohh yeah, and lazy

 

[tywebb]

Re: 2012 HSC MX1 Marathon

Here's James Ruse's solution:

But it can be generalised as follows:



And of course, James Ruse's result follows by letting x=4, y=1.

 

[zeebobDD]

Re: 2012 HSC MX1 Marathon

Last Q of one of our mx1 papers ;

Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.

Prove that PBCR is a cyclic quadrilateral

 

[Sanical]

Re: 2012 HSC MX1 Marathon

Last Q of one of our mx1 papers ;

Two circles intersect at A and B. P is a point on the first circle and Q is a point on the second circle such that PAQ is a straight line. C is a point on the second circle. The tangent to the first circle at P meets QC produced to R.

Prove that PBCR is a cyclic quadrilateral

Construct BA and point D (I forgot to write point Q which is produced from PA and lies on second circle)
let /_DPB = x
/_PAB = x (angle between tangent and chord equal to angle in alternate segment)
/_BAQ = 180 - x (adjacent supplementary angles)
/_BCQ = 180 - x (angles equal on the same arc)
/_RCB = x (adjacent supplementary angles)

Therefore PBCR is cyclic since angle is equal to opposite external angle /_RCB=/_DPB

 

[Sanical]

Re: 2012 HSC MX1 Marathon

New question:

 

[cutemouse]

Re: 2012 HSC MX1 Marathon

2 and e²

Should really justify the 3rd last line being because the exponential function is continuous.

 

[DL9559]

Re: 2012 HSC MX1 Marathon

Yahoo hoo! This one looks fun.

Hmmm...

1/2(ln(x^2+4)) + tan*-1 (x/2) + c (BTW that tan thing is meant to be inverse tan.)

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Yahoo hoo! This one looks fun.

Hmmm...

1/2(ln(x^2+4)) + tan*-1 (x/2) + c (BTW that tan thing is meant to be inverse tan.)

For the record, to indicate inverse tan, we use arctan (x)

ie: arctan (1) = pi/4

EDIT: I'm a sped and typed it wrong way around.