Certainly not off topic for this member! Of interest in connexion with The Inquisitive Problem Solver (Vaderlind, Guy & Larson) Problem 234. There, we're concerned with the perimeters of the triangles formed by a side and two of the distances, and the relevant equation is a sextic, I believe I recall, though we don't bother our readers with that. When you move into higher dimensions you must make sure that all dimensions are represented. As you observe, in three dimensions, the four corners of one face won't do, because the point doesn't know that it's outside that plane. Of course, in general, four distances won't be compatible. There's also a connexion with an unsolved problem, now possibly describable as clasical (see D19 in Unsolved Problems in Number Theory): Is there a point, in the plane of a rational sided square, at rational distances from all four corners of the square? Let me know if any Hyacinthian or Math Funster solves this! R. On Tue, 25 Mar 2003, mjc_w wrote:

Recently, inspired by a thread in sci.math, I worked out the solution to determining the size of a square given the distances from a point to 3 of its vertices. I know that this has been done many times before, but I found some of my results interesting.

To summarize, if the distances to 3 consecituve vertices from a point are a, b, and c, with b in the middle, then the side s has its square t=s^2 satisfying the quadratic

0 = 2*t^2 - 2*t*(a^2+c^2) + (a^2-b^2)^2 + (c^2-b^2)^2

So, the side is clearly constructable.

I noticed that the discriminant of this quadratic can be factored as

-( (a-c)^2 - 2*b^2 )*( (a+c)^2 - 2*b^2 )

so that, since this must be positive to have a real root, the condition for the square to exist is

|a-c| <= b*sqrt(2) <= a+c.

I have extended this to the case of determining the side of an n-dimensional cube from the distances of a point to n+1 of the vertices. It turns out that this can be done if the ordinates of the vertices do not omit any of the ordinates of the sides of the cube (e.g., for a 3-d cube, having all 4 vertices on one face does not work). In particular, having n vertices all adjacent to the n+1-st vertex works.

The resulting equation is, again, a quadratic with roots, when they exist, being the square of the side of the n-dimensional cube.

If anyone is interested, I would be glad to show details.

BTW, if this if off topic for this group, my apologies.

Martin Cohen

------------------------ Yahoo! Groups Sponsor ---------------------~--> Your own Online Store Selling our Overstock. http://us.click.yahoo.com/rZll0B/4ftFAA/46VHAA/6HYolB/TM ---------------------------------------------------------------------~->

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/