Re: Near-isometries of the plane

In article <du5o4m$ok0$1@xxxxxxxxxxxxxxxxx>, Dave Rusin
<rusin@xxxxxxxxxxxxxxxxxxxxx> wrote:

In article <13217929.1141259578413.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Valeriu Anisiu <vanisiu@xxxxxxxxxxx> wrote:

Suppose f : R^2 --> R^2 has the property that

dist( P, Q ) = 1 implies dist( f(P), f(Q) ) = 1

Is f an injection?

dave


Yes, f is actually an isometry. This is valid in R^n for n>1 (it is easy to
see that it is false for n=1).
It is proved in Intelligencer 10(4)(1988) by S. Kolodziej.

Thank you . I should have said that this IS homework -- homework that I
assigned! (I intended the students to prove that f is an isometry,
as you note, but in order to get the proof rolling I intended the students
to show f also preserves distances when dist(P,Q) = sqrt(3), by
consideration of some equilateral triangles. That first step falters
unless there is an easy way to show that f(P) <> f(Q) in that case.)

Once one knows that f carries the hexagonal lattice isometrically,
and can barely increase any distances (by the triangle inequality
applied to polygonal paths of integer lengths) it is not hard to
show f is an isometry, and it is clear the proof generalizes to R^n.

Oh, lovely--what course is this for? And where do you get students you
could expect to solve it? (I might have a couple--but I wouldn't
expect a significant fraction of a CLASS to do this.)

I was vaguely familiar with the result, although I didn't have the
reference at my fingertips. I think it's been a topic on sci.math
before.

--Ron Bruck
.

Relevant Pages

  • Re: Near-isometries of the plane
    ... Dave Rusin wrote: ... (I intended the students to prove that f is an isometry, ... but in order to get the proof rolling I intended the students ...
    (sci.math)
  • Re: Near-isometries of the plane
    ... f is actually an isometry. ... to show f also preserves distances when dist= sqrt, by ... consideration of some equilateral triangles. ...
    (sci.math)