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F preserves midpoint?

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f is an isometry that fixes O (origin). Prove f preserves midpoints of line segments.


    3. The attempt at a solution
    Geometricallly, f could be a reflection in which case it would not preserve the mid point of any line segment that does not intersect the origin anywhere.

    So I don't see a proof at all and infact sees a mistake.
     
  2. jcsd
  3. Jan 26, 2008 #2
    But in the case of a reflection the transformation of a midpoint is still a midpoint, no?
     
  4. Jan 26, 2008 #3
    That is true. I was thinking along the wrong lines (no pun intended) in that I was thinking that f maps midpoint to the exact same mid point.

    Everything makes geometric sense. The only problem is to prove it algebraically. Can't see how to do it.
     
  5. Jan 26, 2008 #4
    Are we in R^n?
     
  6. Jan 26, 2008 #5
    If f is a isometry, u.v=f(u).f(v) holds. So the vector norm (u.u)^1/2 and distance stays the same.
     
    Last edited: Jan 26, 2008
  7. Jan 26, 2008 #6
    I have worked out the quesion in the OP. I now need to show that f(ru)=rf(u) with the same conditions given in the OP.
     
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