MedVision ad

HSC 2012 MX2 Marathon (archive) (4 Viewers)

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Re: 2012 HSC MX2 Marathon

Correct, but it's debatable whether there should be equality for the -1/3 side of the inequality.

Reason being d=-1/3 gives the degenerate 'triangle', which is just a straight line.

And the 'axiom of triangle' you mentioned is more famously known as the "Triangle Inequality" haha.
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
Re: 2012 HSC MX2 Marathon

but lets all agree, that for a q8 from a previous trial, this question is rather pathetic in that sense, more suited for q4,5,6
 

Nooblet94

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
Re: 2012 HSC MX2 Marathon

but lets all agree, that for a q8 from a previous trial, this question is rather pathetic in that sense, more suited for q4,5,6
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.
 

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
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.
A question similar to this was in my 2U preliminary half yearly test last year. Only a handful of students got it lol.
 

crazy_paki123

Well-Known Member
Joined
Nov 3, 2011
Messages
1,129
Gender
Male
HSC
2012
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

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
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.
 

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
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

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
Re: 2012 HSC MX2 Marathon

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

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
Re: 2012 HSC MX2 Marathon

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

 

Aesytic

Member
Joined
Jun 19, 2011
Messages
141
Gender
Male
HSC
2012
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
 
Last edited:

Nooblet94

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
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

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Re: 2012 HSC MX2 Marathon

Always sleep.

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

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
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

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
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).
 

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top