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zetafunc.
2012-10-14 01:58:41

Bump. I still need help on this.

zetafunc.
2012-10-04 05:32:21

Haven't done much work on this puzzle, but I applied the *correct* version of the AM-GM inequality and got this result:

or

so

and since x, y, z > 0 (all positive), then xyz > 2, right?

But, again, this looks pointless, seeing as xyz = 2 + x + y + z, so of course xyz > 2.

Can anyone give me a push in the right direction for this? This question seems to look like it is designed to use AM-GM but I am not seeing what kind of upper or lower bound I can get on xyz using it.

bobbym
2012-10-02 04:47:28

Hi;

Oh, that is it! I was assuming only integers.

zetafunc.
2012-10-02 04:43:59

Why are there 1728 possibilities? I can see where that comes from (12^3) but aren't there more since x, y, z can be real and positive...?

I posted the exact wording of the problem as shown on my problem sheet -- I am assuming by 'distinct' they mean x, y and z must have different values to each other.

bobbym
2012-10-02 04:40:48

I tried out all 1728 possible answers. Only 2,2,2 fits. I am not suggesting that you solve the problem like that although in practice the computer solution gets the nod over a math one! I am just saying that something is wrong here. What do they mean by distinct?

zetafunc.
2012-10-02 04:38:02

Did you use a computing program to do it? Whilst I agree I would probably use one in any other scenario, this problem is from an olympiad, with only pen and paper allowed... I'd be interested in hearing what you used to find that out though.

bobbym
2012-10-02 04:32:56

There are other ways to solve math problems.

zetafunc.
2012-10-02 04:29:01

How come? The question says that there are...

bobbym
2012-10-02 04:27:40

I do not think there are any other solutions than 2,2,2.

zetafunc.
2012-10-02 04:16:44

The problem:

"

Find a solution of the above simultaneous equations, in which all of x, y and z are positive, and prove that it is the only such solution.

Show that a solution exists in which x, y and z are real and distinct."

I haven't really made much progress on this problem. I divided the first equation by the second and got;

or

if we do this:

I notice that the LHS is the arithmetic mean of {2, x, y, z} and the RHS is the harmonic mean of {x, y, z}. They are equal iff x = y = z = 2, so one solution is x = 2, y = 2, z = 2. BUT it appears that they want x, y and z to be distinct.

Can anyone help me here?

Thanks.

Wait... on second thought, what I wrote is wrong. The two sets {2,x,y,z} and {x,y,z} aren't identical. Looks like I am back to square one.