Is it possible to have four equidistant points?

Answer 1

Yes, you simply need to go from 2 dimensional thinking into 3. The shape is called an equilateral tetrahedron - a polyhedron composed of four triangular faces, three of which meet at each vertex.

Answer 2

To have n equidistant points, you must have n-1 dimensions.

2 equidistant points on a line. Of course, 2 points are always equidistant from each other, so as long as you have room enough for 2 points - meaning 1 dimension - then you can find 2 equidistant points.

3 equidistant points in a plane - equilateral triangle.
http://en.wikipedia.org/wiki/Equilateral_triangle

4 equidistant in 3D - a regular tetrahedron.
http://en.wikipedia.org/wiki/Tetrahedron

5 equidistant in 4D -the 4D analog of those other figures, called a pentatope.
http://mathworld.wolfram.com/Pentatope.html

The interesting pattern here is:
Start with a point.
The ends of a line segment are 2 points.

The sides of an equilateral triangle are 3 line congruent segments.

The faces of a regular tetrahedron are 4 congruent equilateral triangles.

The facets of a pentatope are 5 congruent tetrahedra.

This pattern continues forever. In general, these are called simplexes.
http://mathworld.wolfram.com/Simplex.html

The defining property of an n dimensional simplex is that it has n+1 vertices that are all equidistant. They also have Schläfli Symbol {3,3,3....,3} where there are n-1 3s.
http://mathworld.wolfram.com/SchlaefliSymbol.html

So anyway, you need to be thinking in 3D (or more) to get 4 equidistant points.

To see that this is impossible in 2D, consider 3 equidistant points (equilateral triangle), and then try to find a fourth. You'd draw a circle from each (radius = side length of the triangle) and they'd have to all three meet in one place. They don't:
http://math.colgate.edu/~kellen/interspace/4pts.gif

You can actually use this to construct a "cut out" of a regular tetrahedron, by connecting those six points where the circles intersect and then "making them" concurrent by cutting out the figure and folding it together so that the 3 extraneous points in the previous diagram become one point - equidistant from the other 3. That will give you a regular tetrahedron. See:
http://math.colgate.edu/~kellen/interspace/4pts2.gif

Answer 3

Yes, but not in 2-dimensional space, as on a sheet of paper.

The shape you would be looking for is a tetrahedron, which looks a little like a three sided pyramid (if you discount the base). Each side is an equilateral triangle, which means each of the four vertices would be the same distance apart from each other.

Another way to look at the problem would be to cut out two equilateral triangles, say ABC and DCB and place them together so that they now share a common side, BC. Clearly the points A and D are too far apart, but if you fold this new shape along BC so that A and D approach each other then at some point, the distance along AD will be the same as AB and AC, etc., giving four points equidistant from each other

Answer 4

Yes. The vertices of a regular tetrahedron are equidistant. See my source for more details.

Answer 5

Four points equidistant from each other? I think not.

Answer 6

If you are in a 2-dimensional plane you cannot, 3-dimensional yes because one point would be on the third demonsion with the other three on the 2nd dimension, shaped like a triangular-based pyramid.

Answer 7

not at all..you can have a million of them ...I think