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Euclidean Spaces

  • Thread starter jgens
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jgens
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Homework Statement



Suppose [itex]n \geq 3[/itex], [itex]x,y\in\mathbb{R}^n[/itex], [itex]||x-y||=d>0[/itex] and [itex]r>0[/itex]. Prove that if [itex]2r>d[/itex], there are infinitely many [itex]z\in\mathbb{R}^n[/itex] such that [itex]||x-z||=||y-z||=r[/itex].

Homework Equations



N/A

The Attempt at a Solution



Well, I figure that no matter how large we choose [itex]n[/itex], it should always be possible to reduce the problem to the case when [itex]n=3[/itex]. Since we just have two points in [itex]\mathbb{R}^n[/itex], these can just be represented by a one dimentional line. Then, we can always cut the line at [itex](x-y)2^{-1}[/itex] with a plane which is perpendicular to [itex]x-y[/itex]. Clearly, any point on this plane is at an equal distance from [itex]x[/itex] and [itex]y[/itex]; moreover, there must be some circle on this plane such that all the points on that circle are a distance of [itex]r[/itex] from [itex]x[/itex].

However, this reasoning is really informal (and not necessarily correct) so I was wondering if I could formalize this idea or if I'm approaching this problem with the completely wrong mindset (I think that I probably am). Any help is appreciated!
 

Answers and Replies

  • #2
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However, this reasoning is really informal (and not necessarily correct)!
You can formalize it by writing explicitly, in Cartesian coordinates, the equations for your orthogonal hyperplane (it does not to be just 2-dimensional at this point). Then you write additional equations for the distance. Then you manipulate with algebra showing that for n>2 these equations always have infinitely many solutions. For this it will be enough to select two linearly independent vectors in the orthogonal hyperplane.
 

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