HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

[barbernator]

Re: HSC 2012 Marathon

guys, this is a marathon... once ewe answer a question ewe post one... otherwise we get nowhere!

 

[SpiralFlex]

Re: HSC 2012 Marathon

Confused. This is MX1 or MX2 marathon?

 

[barbernator]

Re: HSC 2012 Marathon

Confused. This is MX1 or MX2 marathon?

X1, in that vein, i call upon carrotsticks to clarify the thread name

 

[SpiralFlex]

Re: HSC 2012 Marathon

Well, you guys were doing CS method so I thought it was MX2. In that case we need to tone is down a bit so MX1 people can participate.

 

[barbernator]

Re: HSC 2012 Marathon

Well, you guys were doing CS method so I thought it was MX2. In that case we need to tone is down a bit so MX1 people can participate.

read back a page, there was multiple ways to solve the question. It was a difficult off the cuff question, that was easiest solved through that method

 

[SpiralFlex]

Re: HSC 2012 Marathon

read back a page, there was multiple ways to solve the question. It was a difficult off the cuff question, that was easiest solved through that method

Okais, I should probably post new question. Give me 1 hour.

 

[Sy123]

Re: HSC 2012 Marathon

Sorry for not posting a question, I was in the process of latexing it.

Also, is my way an X2 way? I thought it was 2 unit. (I dont even know anything from X2 math either...yet)

EDIT: Are we allowed to post "Harder 2 Unit" questions here, even if they involve no 3 unit knowledge at all?

 

[SpiralFlex]

Re: HSC 2012 Marathon

Make sure you post stuff reasonable that can be attempted by 3U students alone. There aren't really any rules besides have fun and collaborate ideas/results.

 

[Sy123]

Re: HSC 2012 Marathon

Ok since this thread is awesome, and its dead, I will revive it with a more reasonable question that is actually fun to do.

I dont actually know the answer, but this is a question someone gave to me. In the exact wording and everything

 

[bleakarcher]

Re: HSC 2012 Marathon

Ok since this thread is awesome, and its dead, I will revive it with a more reasonable question that is actually fun to do.

I dont actually know the answer, but this is a question someone gave to me. In the exact wording and everything

By division transformation, P(x)=(x^2-k^2)Q(x)+R(x) where Q is the quotient
Since the divisor is quadratic and the deg[R(x)]=deg[Q(x)]-1 it follows that Q(x) is linear
Let Q(x)=ax+b.
=>P(x)=(x^2-k^2)Q(x)+ax+b
When x=k,
P(k)=ak+b (1)
When x=-k,
P(-k)=-ak+b (2)
(1)+(2):=> P(k)+P(-k)=2b :. b=[P(k)+P(-k)]/2
(1)-(2): => P(k)-P(-k)=2ak :. a=[P(k)-P(-k)]/2k
:. R(x)=(1/2k)[P(k)-P(-k)]x+(1/2)[P(k)+P(-k)] as requested

 

[barbernator]

Re: HSC 2012 Marathon

By division transformation, P(x)=(x^2-k^2)Q(x)+R(x) where Q is the quotient
Since the divisor is quadratic and the deg[R(x)]=deg[Q(x)]-1 it follows that Q(x) is linear
Let Q(x)=ax+b.
=>P(x)=(x^2-k^2)Q(x)+ax+b
When x=k,
P(k)=ak+b (1)
When x=-k,
P(-k)=-ak+b (2)
(1)+(2):=> P(k)+P(-k)=2b :. b=[P(k)+P(-k)]/2
(1)-(2): => P(k)-P(-k)=2ak :. a=[P(k)-P(-k)]/2k
:. R(x)=(1/2k)[P(k)-P(-k)]x+(1/2)[P(k)+P(-k)] as requested

learn latex!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! dude it takes like 10 minutes to learn

 

[bleakarcher]

Re: HSC 2012 Marathon

^ lol i cbf bro

 

[Sy123]

Re: HSC 2012 Marathon

Here is another question because no one decided to post one.

http://www.codecogs.com/latex/eqneditor.php

Use that link man, you dont even need to learn anything about latex to use it.

 

[bleakarcher]

Re: HSC 2012 Marathon

A+B+C=180
(A/2)+(B/2)+(C/2)=90
(A/2)+(B/2)=90-(C/2)
sin[(A+B)/2]=sin[90-(C/2)]=cos(C/2)

 

[fortune_cookie]

Re: HSC 2012 Marathon

A+B+C =180 (angle sum triangle)

A+B = 180-C

(A+B)/2 = 90-(C/2)

Take sin of both sides and remember that cos(90-x) = x

 

[fortune_cookie]

Re: HSC 2012 Marathon

edit: should have been sin(90-x) = cos(x).

Slow internet is really pissing me off.

 

[bleakarcher]

Re: HSC 2012 Marathon

I love that question Carrotsticks posted before on the volume between the ball of unit radius and y=abs(x), I was just working on it. Missed out on this marathon because I was moving houses and only recently got the internet back.

 

[Sy123]

Re: HSC 2012 Marathon

Nice work both of you, now someone post a question!

 

[bleakarcher]

Re: HSC 2012 Marathon

Prove that the acceleration of a body a is given by a=v(dv/dx) where x and v are the displacement and velocity of the particle after a time t.

 

[bleakarcher]

Re: HSC 2012 Marathon

Just one off the top of my head lol.