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        riddles >> general problem-solving / chatting / whatever >> x + y + sqrt(x^2 + y^2) = 12
        
(Message started by: Ken_Wiley on Oct 31st, 2006, 2:58pm)
Title: x + y + sqrt(x^2 + y^2) = 12
Post by Ken_Wiley on Oct 31st, 2006, 2:58pm
What is the easiest way to get x,y of

xy = 14

x + y + sqrt(x^2 + y^2) = 12
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by towr on Oct 31st, 2006, 3:12pm
My guess would be to start with
sqrt(x^2 + y^2) = 12- (x + y)
and square both sides. (Note that you may get extra solutions, so check which are valid in th eoriginal equations afterwards)
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Michael_Dagg on Oct 31st, 2006, 4:14pm
I think this is one of Diophantus' examples.

Take    x = 1/u  and   y =  14u, then the first equation
is the identitiy 14 = 14 and you can get a quadratic
for the second.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Ken_Wiley on Nov 1st, 2006, 8:14pm
First suggestion nicely eliminates the radical but the second response is something I have never seen before. Please explain/
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Barukh on Nov 2nd, 2006, 3:49am
Are we looking for real solutions?
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by pex on Nov 2nd, 2006, 4:02am
I hope not, since there are none...
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Barukh on Nov 2nd, 2006, 9:17am
Here’s an approach that doesn’t require solving quadratic equations.

Let u = x+y, v2 = x2 + y2. Then, we easily arrive at the following system in variables u, v:

u + v = 12
u2 - v2 = 28

from which it follows u – v = 7/3, and therefore u = 43/6, v = 29/6. Next, we have:

(x-y)2 + 2xy = v2,

therefore x–y = sqrt[(29/6)2 – 28]. This gives the second linear equation in x, y (the first is u).
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by SWF on Nov 2nd, 2006, 6:08pm
Why not just write the first equation as y=14/x and substitute into the 2nd equation. It simplfies considerably to 24*x2-172x+336 =0.  This is similar to why Michael_Dagg suggested without introducing a new variable, u.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Grimbal on Nov 3rd, 2006, 8:54am

on 10/31/06 at 14:58:53, Ken_Wiley wrote:
What is the easiest way to get x,y of

xy = 14

x + y + sqrt(x^2 + y^2) = 12


Post it on a forum where people like to solve this kind of things...
And yes, the answer is x,y = (19 +- i·sqrt(1655))/12

Nice method, Barukh!
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Barukh on Nov 3rd, 2006, 9:14am

on 11/03/06 at 08:54:48, Grimbal wrote:
Nice method, Barukh!

Thanks, Grimbal  :D
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Ken_Wiley on Nov 3rd, 2006, 9:20am

on 11/03/06 at 08:54:48, Grimbal wrote:
Post it on a forum where people like to solve this kind of things...


WHAT? I think you need to look around.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by pex on Nov 3rd, 2006, 9:21am

on 11/03/06 at 08:54:48, Grimbal wrote:
... the answer is x,y = (19 +- i·sqrt(1655))/12


... which gives x + y + sqrt(x2 + y2) = (19 + i*sqrt(647)) / 6. Can you prove that this equals twelve?  ;)

Okay, you probably just made a typo somewhere. I get x,y = (43 +/- i*sqrt(167)) / 12.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Michael_Dagg on Nov 3rd, 2006, 9:29am
You misunderstood Gimbal -- he didn't mean on a forum away from this site but on another forum within this site (another thread have you).
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Sir Col on Nov 3rd, 2006, 10:02am
I believe that Grimbal was being playfully ironic as Ken_Wiley never actually asked anyone to solve the problem; he only asked for the best way to solve it.

For that reason, when I first saw the problem, I was tempted to post
this (http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=advanced&equations=xy=14%0d%0ax%2by%2bsqrt(x^2%2by^2)=12&showReply=true&variables=x%0ay&style=traditional&prefNames=single&prefInput=standard&fontSize=20#reply) link.  ;)

But it was certainly a nice puzzle to share, Ken_Wiley; thanks, and welcome to the forum.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Grimbal on Nov 3rd, 2006, 10:04am

on 11/03/06 at 09:21:37, pex wrote:
... which gives x + y + sqrt(x2 + y2) = (19 + i*sqrt(647)) / 6. Can you prove that this equals twelve?  ;)

Well, ahum ... it comes from
  x,y = (19 +- i·sqrt(4·6·84-19^2))/12
Now, using one of the larger (and seldom used) value of the symbol 19, i.e.19=43, you find:
  x,y = (43 +- i·sqrt(4·6·84-43^2))/12 = (43 +- i·sqrt(167))/12
Which agrees with your result.

:-[
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Icarus on Nov 3rd, 2006, 3:51pm

on 11/03/06 at 09:20:39, Ken_Wiley wrote:
WHAT? I think you need to look around.


What Grimbal means is that the "General problem-solving / whatever" forum is not really intended for posting problems/riddles. Problems like this one are usually posted in either the Medium, Hard, or Putnam forums. I've held off moving this one simply because I decided to wait and see how "hard" it really is. (Usually, for the Hard forum, I prefer problems that require - or provide - some special insight. Putnam problems are those involving advanced mathematics, though I don't mind if less advanced problems get posted there by others.)
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Ken_Wiley on Nov 3rd, 2006, 7:38pm
Fare enough! Thanks for the solutions!
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by Grimbal on Nov 6th, 2006, 4:30am
Actually, my remark was more like a joke.

By saying that the easiest thing is to post it here,
1. I tell you to do what you just did making the advice useless,
2. I don't help you getting a solution at all, making my advice even more dubious.
Title: Re: x + y + sqrt(x^2 + y^2) = 12
Post by srn347 on Sep 8th, 2007, 8:36pm
Are we looking for complex ones? There aren't any either.


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