Is it true that every unit-distance preserving map in the real plane is an isometry?  Is this also true in R^n? (n1) Please mail me directly with any info or references if possible. Thanks in advance.                                                 Joshua Cooper                                                 This e-mail address is being protected from spam bots, you need JavaScript enabled to view it

Is it true that every unit-distance preserving map in the real plane is an isometry?  Is this also true in R^n? (n1) I believe it's true for all n, but I do not have a reference. For n=2, in particular, I vividly remember (15 years ago, back in Thessaloniki, Greece) a German mathematician presenting a *wonderful* proof using only high school Goemetry; unfortunately, I cannot remember his name. George Baloglou

Is it true that every unit-distance preserving map in the real plane is an isometry?  Is this also true in R^n? (n1) Please mail me directly with any info or references if possible. Thanks in advance.                                            Joshua Cooper                                             This e-mail address is being protected from spam bots, you need JavaScript enabled to view it I presume that you mean that if the function is f:R

: Is it true that every unit-distance preserving map in the real plane : is an isometry?  Is this also true in R^n? (n1) : Please mail me directly with any info or references if possible. : Thanks in advance.      Yes, it's true for n 1.  Let f: R^n - R^n have the property that d(x,y) = 1 implies d(f(x),f(y)) = 1 where d is the usual Euclidean metric.  Let z be the origin of R^n, and g(x) = f(x) - f(z). Then g(z) = 0, and |x| = 1 implies |g(x)| = 1, where |x| = d(x,0).  Thus, if we define h by h(z) = 0, and h(x) = |x|*g(x/|x|) when NOT(x = z), then h preserves all lengths, hence all inner products, so h is an isometry.  Further, g(x) = h(x) if |x| = 1.      Now, *for n 1*, each point y with 0 < |y| < 2 in R^n determines an (n-1)-sphere of radius r, 0 < r < 1, contained in |x| = 1 (ie, the intersection of |x| = 1 and d(x,y) = 1) and, given any such (n-1)-sphere S, there is a unique point y, 0 < |y| < 2, so that S is the intersection of |x| = 1 and d(x,y) = 1.      Now g agrees with the isometry h on |x| = 1, so g(S) = h(S) for every (n-1)-sphere S contained in |x| = 1.  If y is the point corresponding to S as described above, clearly g(y) = h(y).  Thus g(x) = h(x) if |x| < 2, and you're off to the races.