
#1
Jan2608, 05:46 AM

P: 2,268

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
Jan2608, 05:58 AM

P: 13

But in the case of a reflection the transformation of a midpoint is still a midpoint, no?




#3
Jan2608, 06:06 AM

P: 2,268

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. 



#4
Jan2608, 12:17 PM

P: 139

f preserves midpoint?
Are we in R^n?




#5
Jan2608, 03:47 PM

P: 13

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.




#6
Jan2608, 07:13 PM

P: 2,268

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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