What is the most symmetric arrangement of points in 3-D space so that they are all equidistant?

Question

By symmetric arrangement, what I mean is that when the points are viewed in x-y or x-z or y-z planes, they look the same in the 3 planes.

Imagine placing small marbles in space. We want each neighboring marble to be equidistant. And when we view those marbles, they appear to have a symmetric pattern.

Points placed on vertices of a cube are the most symmetrical arrangement since they look alike in whatever plane you look. However all neighboring points are not equidistant. They will all be either x apart or sqrt(2)x apart. We want all points to be equidistant.

Dutch_prof has given a v detailed answer and i suppose i should give 10 pts---but I disagree with the answer. Because a regular tetrahedron is not 100% space filling (3-D space). When stacked the arrangement of tetrahedrons will turn out to be assymetric because of the fact that tetrahedrons can not be completely stacked without overlapping or gaps--IMHO. Correct me if I am wrong...

Answer 1

If you want all points to be equidistant, the largest arrangement you can possibly make is a regular tetrahedron, with four points.

Now you want a proof :)

Let the first point be P1 = (0,0,-1) and the second point be P2 = (0,0,1). [The coordinate system can always be translated and scaled to make this happen.] The distant between the points is 2.

Any additional points P3 should also have distant 1 to each of the given points. All points with this property lie on a circle with radius sqrt 3 around the origin, in the xy-plane. Proof: because d(P1,P3) = d(P2,P3) = 2

x^2 + y^2 + (z+1)^2 = 4 ==> x^2 + y^2 + z^2 + 2z = 3
x^2 + y^2 + (z-1)^2 = 4 ==> x^2 + y^2 + z^2 - 2z = 3

It follows immediately that 2z = -2z, so z = 0. The resulting formula is x^2 + y^2 = 3, which describes the circle.

Let's pick the point P3 on the circle. By rotating the coordinate system around the z-axis we can always make this point P3 (sqrt 3, 0, 0).

A fourth point P4 must not only lie on the circle (x^2 + y^2 = 3), but also have distance 2 to point P3. This gives

(x - sqrt 3)^2 + y^2 + z^2 = 4
x^2 + y^2 + z^2 - (2 sqrt 3) x = 1

Using z = 0 and x^2 + y^2 = 3, we get

-(2 sqrt 3) x = -2
x = 1/sqrt 3
(1/sqrt 3)^2 + y^2 = 3 --> y^2 = 8/3 --> y = +/- 2/3 sqrt 6

For point P4 we have two choice: (1/sqrt 3, +/- 2/3 sqrt 6, 0). However, only one point can be added to our set because the distance between the two possible points is 4/3 sqrt 6, which is not equal to 2.

Answer 2

There are two answers.

A. Equidistant points- They can be any two points in a space, three points forming vertices of an equilateral triangle or four points forming vertices of equilateral triangular pyramid. You can't have five or more points, all of which are equidistant from each other.

B. Arrangement looking same from all 3 directions. Either one point or Eight points forming vertices of a cube so that one can see just four points from any of these three directions, the rest of four being hidden behind the first four. Or 27 points forming an arrangement of 3 by 3 by 3. Or 64 points......... in that way.

Answer 3

Its a tetrahedron. All of the four corner points are cental symmetric and equidistant one from an other.

Answer 4

if you put a marbel in each corner of a cube, you'll have it !!!. I mean, only the marbels will be the arrangement u look for !!

in each plane u'll see something like

o o
o o

Answer 5

1.ax+ay+az
2.-ax-ay-az
3. 0

Answer 6

simple, a sphere.