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Topic: x + y + sqrt(x^2 + y^2) = 12 (Read 944 times) |
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Ken_Wiley
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x + y + sqrt(x^2 + y^2) = 12
« on: Oct 31st, 2006, 2:58pm » |
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What is the easiest way to get x,y of
xy = 14
x + y + sqrt(x^2 + y^2) = 12
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towr
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #1 on: Oct 31st, 2006, 3:12pm » |
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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)
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Michael Dagg
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #2 on: Oct 31st, 2006, 4:14pm » |
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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.
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Regards,
Michael Dagg
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Ken_Wiley
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #3 on: Nov 1st, 2006, 8:14pm » |
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First suggestion nicely eliminates the radical but the second response is something I have never seen before. Please explain/
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Barukh
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #4 on: Nov 2nd, 2006, 3:49am » |
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Are we looking for real solutions?
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pex
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #5 on: Nov 2nd, 2006, 4:02am » |
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I hope not, since there are none...
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Barukh
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #6 on: Nov 2nd, 2006, 9:17am » |
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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).
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SWF
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #7 on: Nov 2nd, 2006, 6:08pm » |
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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.
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Grimbal
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #8 on: Nov 3rd, 2006, 8:54am » |
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on Oct 31st, 2006, 2:58pm, Ken_Wiley wrote:What is the easiest way to get x,y of
xy = 14
x + y + sqrt(x^2 + y^2) = 12 |
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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!
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| « Last Edit: Nov 3rd, 2006, 9:56am by Grimbal » |
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Barukh
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #9 on: Nov 3rd, 2006, 9:14am » |
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on Nov 3rd, 2006, 8:54am, Grimbal wrote:
Thanks, Grimbal 
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Ken_Wiley
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #10 on: Nov 3rd, 2006, 9:20am » |
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on Nov 3rd, 2006, 8:54am, Grimbal wrote:
Post it on a forum where people like to solve this kind of things...
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WHAT? I think you need to look around.
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pex
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #11 on: Nov 3rd, 2006, 9:21am » |
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on Nov 3rd, 2006, 8:54am, Grimbal wrote:| ... the answer is x,y = (19 +- i·sqrt(1655))/12 |
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... 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.
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Michael Dagg
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #12 on: Nov 3rd, 2006, 9:29am » |
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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).
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Regards,
Michael Dagg
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Sir Col
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #13 on: Nov 3rd, 2006, 10:02am » |
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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 link.
But it was certainly a nice puzzle to share, Ken_Wiley; thanks, and welcome to the forum.
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| « Last Edit: Nov 3rd, 2006, 10:18am by Sir Col » |
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Grimbal
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #14 on: Nov 3rd, 2006, 10:04am » |
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on Nov 3rd, 2006, 9:21am, pex wrote:| ... which gives x + y + sqrt(x2 + y2) = (19 + i*sqrt(647)) / 6. Can you prove that this equals twelve? |
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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.
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Icarus
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #15 on: Nov 3rd, 2006, 3:51pm » |
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on Nov 3rd, 2006, 9:20am, Ken_Wiley wrote:
WHAT? I think you need to look around.
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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.)
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When their digits are reversed? " - Anonymous
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Ken_Wiley
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #16 on: Nov 3rd, 2006, 7:38pm » |
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Fare enough! Thanks for the solutions!
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Grimbal
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #17 on: Nov 6th, 2006, 4:30am » |
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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.
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srn437
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Re: x + y + sqrt(x^2 + y^2) = 12
« Reply #18 on: Sep 8th, 2007, 8:36pm » |
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Are we looking for complex ones? There aren't any either.
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