Good Text Books (2 Viewers)

[0o0]

What is a good maths text book with the HARDEST questions, for 2/3/4 unit?

i dont need explanations or walkthroughs, i just want the hardest possible questions so when I walk into the HSC, all the questions would be shit easy.

thanks for help ppl

 

[enak]

You should probably just do question 8's from past papers if that's what you want, and listen to dr buchanan and attempt to prove the riemann hypothesis

 

[Affinity]

scan through this board, most questions here are here because someone found it hard.

 

[0o0]

"... and attempt to prove the riemann hypothesis "

i already have, back when i was in yr 9. It is very difficult, but I managed.

honestly, i dont think i need study for 4unit, I got 97% in the 2001 HSC when i was yr 9. (doing the 4U paper for fun after it was released)

 

[Affinity]

hmm you skipped a year? I was in year 10 in 2001

 

[freaking_out]

Originally posted by 0o0
"... and attempt to prove the riemann hypothesis "

i already have, back when i was in yr 9. It is very difficult, but I managed.

honestly, i dont think i need study for 4unit, I got 97% in the 2001 HSC when i was yr 9. (doing the 4U paper for fun after it was released)

so why didn't u just sit the hsc in 2002??

 

[0o0]

although i could get 100% for 3 and 4 unit maths

thats another 6 required units i have left....

im sure you will be able to work this one out

 

[ND]

Originally posted by 0o0
"... and attempt to prove the riemann hypothesis "

i already have, back when i was in yr 9. It is very difficult, but I managed.

So what did you do with the million dollars?

 

[enak]

Originally posted by ND
So what did you do with the million dollars?

That's exactly what I was thinking. Let's wait for dr buchanan.

 

[freaking_out]

Originally posted by 0o0
im sure you will be able to work this one out

no i can't...would like to do that for us, "o wise one"

 

[ND]

Ok here's a question for you:

Prove:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab) >= 1

Hope it's not too easy for you.

 

[freaking_out]

Originally posted by ND
Ok here's a question for you:

Prove:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab) >= 1

Hope it's not too easy for you.

actually u know, if u can't do the above question, u shouldn't do 4u, in my opinion.

 

[Fosweb]

HAHAHA!!!

Another fraud?

For the answer: Just call Richard Lee, or refer to the other thread...

 

[Fosweb]

Originally posted by ND
a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab) >= 1

I think that if i see this question in the 2003 4U HSC, then i'll probably have to laugh.

 

[Rahul]

Originally posted by Fosweb
For the answer: Just call Richard Lee, or refer to the other thread...

hahahahha!

Originally posted by Fosweb
I think that if i see this question in the 2003 4U HSC, then i'll probably have to laugh.

why? [as you can tell i have not attempted this, nor does it seem likely that i will, i just want in on the joke]

 

[turtle_2468]

I love frauds. They make my day especially when I feel down..

 

[0o0]

firstly morons, if someone says theyve proved the reimann zeta whateverthefuck, chances are, they are lying. especially if they were in year 9.

am i the idiot for lying? or are you the idiots for taking me seriously?

the answer is you are all idiots.

 

[0o0]

Originally posted by ND
Ok here's a question for you:

Prove:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab) >= 1

Hope it's not too easy for you.

....

i know how to solve pyfagarus, is that good enuf?

 

[0o0]

Originally posted by ND
Ok here's a question for you:

Prove:

a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab) >= 1

Hope it's not too easy for you.

you had me going, i was panicing, i even started to sweat



a/sqrt(a^2+8bc) + b/sqrt(b^2+8ac) + c/sqrt(c^2+8ab)
>= 1

for all positive real numbers a, b and c.
We sue the convexity 1/sqrt(x): if sum (x)=1,
then sum (x/sqrt(y))>=1/sqrt(sum(x*y)).
Since this inequality is
homogeneous, we may assume a+b+c=1, so
left part is
>=1/sqrt(a*(a^2+8bc)+b*(b^2+8ac)+c*(c^2+8ca))=1/sqrt(a^3+b^3+c^3+24abc)

It is enough to show that
1>=a^3+b^3+c^3+24abc. But 1=(a+b+c)^3=a^3+b^3+c^3+
6abc+3b(a^2+c^2)+3a(c^2+b^2)+3c(b^2+a^2)>=
a^3+b^3+c^3+6abc+3b*2ac+3a*2cb+3c*2ba=a^3+b^3+c^3+24abc.



thankyou

 

[ND]

Yeh and i suppose you posted that exact solution here too:

http://groups.yahoo.com/group/imo-problems/message/230