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June 6th, 2012, 06:12 PM   #1
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Equation...

Find all groups of three real numbers (x,y,z), such that

(x-1)/y = (y-1)/z = (z-1)/x
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June 6th, 2012, 11:35 PM   #2
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Re: Equation...

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June 7th, 2012, 02:38 PM   #3
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Re: Equation...

Quote:
Originally Posted by karthikeyan.jp
Your solution is simply x = y = z ? 0.

Are there solutions where they are not all equal.
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June 7th, 2012, 02:48 PM   #4
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Re: Equation...

Quote:
Originally Posted by mathman
Quote:
Originally Posted by karthikeyan.jp
Your solution is simply x = y = z ? 0.

Are there solutions where they are not all equal.
Since it is symmetric on x,y,z, without loss of generality let x<=y<=z.
If we assume x<y<=z or x<=y<z or x<y<z; then which is impossible since both are equal to m.
The only remaining alternative is x=y=z, which obviously holds.
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