One for the geniuses out there...

[A$$Bandit]

You are given only this information:

x + y + z = 4
x² + y² + z² = 6
xyz = 3

You are required to determine a numerical value for:

x³ + y³ + z³

No fancy stuff, just algebraic manipulation... I couldn't figure it out so anyone who can point me in the right direction, please relieve me of this.

[mmiill]

that is so easy!!!

(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=16

x^2+y^2+z^2=6

=> xy+yz+xz=5

now

(xy+yz+zx)*(x+y+z)=x^2y+xyz+zx^2+xy^2
+y^2z+zxy+xyz+yz^2+z^2x =
x^2y+zx^2+xy^2+y^2z+yz^2+z^2x+3*xyz=20

A=x^2y+zx^2+xy^2+y^2z+yz^2+z^2x=11
A=11
now

(x^2+y^2+z^2)(x+y+z)=
x^3+y^3+z^3 +A =24

and
x^3+y^3+z^3 =13

SEE YA

[NotLKH]

I might have made a stupid mistake somewhere in here, but I can't find one.
Anyways, this should help you along:

[bugzpodder]

how did u do these images, NotLKH?

[A$$Bandit]

Yeah thanks a lot, as soon as I read your first lines I realized I'd missed the obvious way of doing it, hehe.

Now for part 2:

Given the drawing, find X:
I had a 5 minute stab at this and couldn't be arsed to try any harder...

[phanbachloc]

its easy to do,hehe

[bugzpodder]

what are these lines? medians? if its random lines i doubt it can be solved.

[A$$Bandit]

They are in fact, random. You are given EXACTLY what I am. The question gives that exact diagram and says "Find the area x." I think it can be done but I don't have the time to try and work it out for myself... also I can't be arsed

[NotLKH]

It looks to me a bit trig intensive, but definetely do-able.
Too bad I don't have the time.

Here's an angle map of the problem I'd use if I did have time:

[Gandalf_Grey_]

is this an equalateral, scalene or isocolese triangle? or do you not have that information?

[A$$Bandit]

You only get EXACTLY what I posted: that drawing and the quote that goes with it. You would have to assume that it's scaline I suppose...

[A$$Bandit]

I just spent 5 minutes thinking and worked this out....
So now I need the area A

[bugzpodder]

solution is in this thread: http://www.vbforums.com/showthread.p...hreadid=194972