some time ago, i wrote:
> > Before people go too far looking for such charts, I have a question: is
> > it possible to have 4 points in a plane, no 3 colinear, so that all 6
> > distances are integers? And with the additional constraint that no 2
> > distances are equal?
>
> it is possible to have arbitrarily many points in the plane,
> no three collinear, such that all distances between them are
> integers, and no two such distances are equal.
there hasn't been any response, except for dan's query "why is this true?".
since no one has answered, here is my solution.
rational points on the unit circle x^2 + y^2 = 1 are (rationally)
parametrized by (x, y) = P(t) = ((1 - t^2) / (1 + t^2), 2t / (1 + t^2)) .
let A = P(0) = (1, 0), B = P(oo) = (-1, 0) and O = (0, 0) ; then the
parameter t is tan(<PBA) = tan(<POA / 2) . it is easy to show that
the distances PA and PB are rational if and only if t^2 + 1 is
the square of a rational number, which is the case if and only if
s = tan(<PBA / 2) is rational. we get a new parametrization of the
unit circle by this parameter s :
P(s) = ((1 - 6s^2 + s^4) / (1 + s^2)^2, 4s(1 - s^2) / (1 + s^2)^2)
for rational s , such points have rational distance to both A and B ,
and the points obtained are dense in the unit circle. moreover, by
ptolemy's theorem, any two such have rational distance to each other.
this gives an infinite subset of the plane such that all the pairwise
distances are rational, and one can also include the center of the circle.
i do not know if this is a maximal subset of the plane with this property,
although i think that it is likely.
it remains to get a set with all distances different. this can be done
in several ways. one way is to choose an infinite sequence {P_j} of
points in the above set such that each point is on the top half of the
circle, and such that <P_(j+1)OA < (<P_jOA) / 2 . for an explicit
example, choose s = 2^(-j) , j = 0, 1, 2, ... to get the points
P_j = ((2^(4j) + 6(2^(2j)) + 1) / (2^(2j) + 1)^2 ,
4(2^j)(2^(2j) - 1) / (2^(2j) + 1)^2) .
another way is to choose z to be the complex number corresponding to
P(s) , and then to take the complex numbers z^(2^k) , k = 1, 2, 3, ... .
(this will work as long as s is not 0, 1, -1 or oo , so that z
is not a root of unity.)
for a *finite* set, we can get all distances to be integers simply by
scaling. i do not know if it's possible to have an infinite set, not
all on a line, such that all pairwise distances are integers.
i also do not know about dan's follow-up question:
| And what about the higher-dimensional analogues:
|
| Q3: What is the maximum number of points in R^n, no hyperplane
| (i.e., n-1 dimensional plane) containing n+1 of them, such that all
| interpoint distances are integers?
but i note that even with the weaker condition that no hyperplane contains
all of the points, the question is still non-trivial!
mike
