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Homework Help: Structure R^k and midpoints of vectors (rudin's PMA chapter 1 problem 16)

  1. Sep 8, 2007 #1
    First this is NOT a homework problem. I am undertaking a self-study of mathematical analysis by following Rudin's Principles of Mathematical analysis. I have done a course in analysis before but this is a high-powered review of sorts.

    So I'm currently on the first chapter's problem set and I've gotten stuck on problem 16, which asks:

    Let x,y in R^k, k>/=3 (at least 3-space), norm[x-y] = d>0 prove:

    a)If 2r>d There are infinitely many z in R^k s.t. norm[z-y]=norm[z-x]=r
    b)If 2r=d there is exactly one such z.
    c)If 2r<d there are no such z.

    That is the question as stated, to clarify the norm I speak of is the tradition euclidean k-space norm (i.e. root of sum of squares of components).

    Part a) is the one I've made the least progress on, b) I'm half done and c) is a simple proof by contradiction using the triangle inequality.

    Geometrically (for part a)) consider the line between x and y, there is a perpindicular plane at the midpoint (x+y)/2 (perpendicular to the line connecting x and y) and the set of infinite z that a) asks for is the circle of radius (r^2 - (d/2)^2)^(1/2) which is nonzero b/c of the hypothesis lying on the tangent plane centered at the midpoint (x+y)/2. The thing is, I presume rudin wants me to construct a general z that admits infinitely vectors, but I've found this very difficult to do using the definitions and theorems given in the chapter. Any ideas?

    Part b) the only such z is x+y=2 but I can't for the life of me prove that it is the ONLY solution with rigour.

    Help would be much appreciated.

    Siddharth M.

    PS: obviously the problem I'm having is providing a clean and neat proof strictly using definitions and theorems as is required of an analyst but such a solution has thus far escaped me.
  2. jcsd
  3. Nov 25, 2010 #2
    $|x-z|+|z-y| =2r \Rightarrow |x-y| \leq 2r \forall z$ by the triangle equality, the rest pretty much falls in your lap.
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