HSC 2012 MX2 Marathon (archive) (2 Viewers)

[crazy_paki123]

Re: 2012 HSC MX2 Marathon

hey, hey, hey im new. can sum1 tell me how to start new threads nd rite my courses nd atar aim. thanx. much appreciated

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Yep, entirely. The only remotely challenging part is knowing to apply the triangle inequality and it's extension 2 maths, so everyone should be able to make basic logical connections like that.

EDIT:
For nightweavers question I got , but I honestly wouldn't be surprised if that's wrong considering how I've been going this evening.

Actual Edit:
FUCK. Meant to edit my post, not reply to it. I really need to go to bed.

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

^ I got 0.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

EDIT:
For nightweavers question I got , but I honestly wouldn't be surprised if that's wrong considering how I've been going this evening.

Actual Edit:
FUCK. Meant to edit my post, not reply to it. I really need to go to bed.

Haha.. not correct. Try again tomorrow when you're feeling fresh.

 

[bleakarcher]

Re: 2012 HSC MX2 Marathon

Dont worry Nooblet, get some rest. I have bad days as well.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

^ I got 0.

I got 0.

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Correct. I hope no-one tried modulus argument lol.

 

[Aesytic]

Re: 2012 HSC MX2 Marathon

(i) angle NPC = angle CBP (alternate segment theorem)
angle PNC = 90 (given)
.'. angle NCP = 180 - 90 - angle NPC (angle sum of a triangle is 180)
= 90 - angle NPC
angle ACB = 90 (angle subtended at the circumference from a diameter is a right angle)
since angle ACB = angle NCP + angle PCB,
.'. 90 = 90 - angle NPC + angle PCB
.'. angle NPC = angle PCB
.'. angle CBP = angle PCB
.'. PC = PB (sides opposite equal angles in a triangle are equal)

(ii) angle PNC = 90 (given)
.'. angle NPC + angle NCP = 90 (angle sum of a triangle is 180)
since angle NPC = angle APC + angle NPA,
.'. angle APC + angle NPA + angle ACP = 90
angle NPA = angle ACP (alternate segment theorem)
.'. angle APC + 2*angle ACP = 90

(iii) let angle PAB = x
angle APB = 90 (angle subtended at circumference by a diameter is a right angle)
.'. angle PBA = 90 - x (angle sum of a triangle is 180)
angle PBA = angle PCA (angles subtended by the same arc are equal)
.'. angle PCA = 90 - x
angle PNC = 90 (given)
.'. angle NPC = 90 - angle PCA (angle sum of a triangle is 180)
.'. angle NPC = 90 - (90 - x)
= x
.'. angle PAB = angle NPC

 

[nightweaver066]

Re: 2012 HSC MX2 Marathon

Qn:

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

Haha.. not correct. Try again tomorrow when you're feeling fresh.

Oh god. That's so embarrasing. The legal notes I was writing last night are complete shit as well. Guess I really need to sleep more

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Always sleep.

I got pwned in the 2U test lolololol as a result.

 

[deswa1]

Re: 2012 HSC MX2 Marathon

Always sleep.

I got pwned in the 2U test lolololol as a result.

Not sure if serious. Regardless it doesn't matter for you. I got 42/55 in my 2U test because I did it without a calculator to see how I'd go (trig became my new enemy...)

 

[SpiralFlex]

Re: 2012 HSC MX2 Marathon

Not sure if serious. Regardless it doesn't matter for you. I got 42/55 in my 2U test because I did it without a calculator to see how I'd go (trig became my new enemy...)

I am serious. >_>

I came near bottom out of 117 hahaha. I thought I got everything correct...I skipped questions (Oops).

 

[deswa1]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

 

[IamBread]

Re: 2012 HSC MX2 Marathon

I am serious. >_>

I came near bottom out of 117 hahaha. I thought I got everything correct...I skipped questions (Oops).

That's what happens when you go penguin

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

Oh I love recursive relations!

 

[IamBread]

Re: 2012 HSC MX2 Marathon

Oh I love recursive relations!

Me too! <3

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" title="U_{1}=2, U_{2}=16 \textup{ and }U_{n}=8U_{n-1}-15U_{n-2} \textup{ for } n\geq 3\\ \textup{Show that }U_{n}=5^n-3^n \textup{ for n}\geq 1" /></a>

Never seen anything like this before. Can someone post a worked solution? I'm curious.

 

[Carrotsticks]

Re: 2012 HSC MX2 Marathon

Never seen anything like this before. Can someone post a worked solution? I'm curious.

There are a couple of ways to do the question.

The most 'mindless' of which being induction.

 

[Nooblet94]

Re: 2012 HSC MX2 Marathon

There are a couple of ways to do the question.

The most 'mindless' of which being induction.

That's what I immediately thought of, but I figured there must be a nicer way.