Distance between points on a regular tetrahedron

[Sorcerer]

I suck at geometry, but I have this intuitive notion that the points on the corners of a regular tetrahedron are all equidistant. How do I go about proving this true (or false, if I'm wrong)? Note that the highest geometry class I've taken is high school, but I'm okay with any undergraduate math.

The shape is made of nothing but equilateral triangles, and the points obviously for each one are equidistant, and just thinking about it there are several axes upon which the shape has rotational symmetry for three flips. Four, I think.



Also, as an aside, if there is a generalized version of this for higher dimensions I'd be interested to see it. My guess is that a regular tetrahedron is the shape that arises when four points are equidistant, and I suspect it is not possible to add a fifth point and have all five equidistant in three dimensions. But this is all intuitive. How do I verify this?


Thanks!

 

[mfb]

That is the definition of a regular tetrahedron.

The shape is made of nothing but equilateral triangles

Of the same side length. These side lengths are the distances between the corners.

My guess is that a regular tetrahedron is the shape that arises when four points are equidistant, and I suspect it is not possible to add a fifth point and have all five equidistant in three dimensions. But this is all intuitive. How do I verify this?

That is right and you can find a 4-dimensional shape with five equidistant points. You can prove that a line, a triangle and a tetrahedron are the only options to arrange 2,3,4 points such that all pairs have the same distance, and it is easy to show that you cannot add a fifth point in three dimensions with the same property.

 

[Sorcerer]

That is the definition of a regular tetrahedron.Of the same side length. These side lengths are the distances between the corners.That is right and you can find a 4-dimensional shape with five equidistant points. You can prove that a line, a triangle and a tetrahedron are the only options to arrange 2,3,4 points such that all pairs have the same distance, and it is easy to show that you cannot add a fifth point in three dimensions with the same property.

Thanks. I can easily see this in two-dimensions with just an equilateral triangle. Any other point on the 2D plane will be closer to one or some points but further from others. It's also easy to see on the one dimensional number line (assuming you can't have two or more points at the same location, of course).

Now, I'm making a guess here, but if I could figure out a way to translate the geometry into "math language," would induction be a good way to prove that the number of points that can be equidistant from each other is one more than the number of dimensions? I.e., if D = 3, then the maximum number of points that can be equidistant is D + 1 = 4. You know, first prove it true for D = 1, then prove that if it's true for D = n then it's true for D = n + 1? Again, I'm throwing darts blindly, and I'm probably way off, but since there seems to be a +1 relationship between dimension and number of equidistant points I'm wondering if that's the way to go.

 

[mfb]

Induction is a good approach, yes. Show that there are no points with the same distance to all existing points in the same dimension. Show that there is no third point on a line. The third one has to be in a plane. For a given plane there are just two options for the place of the third point. A fourth point cannot be at either of them (one distance is too large or too small) - but you can add it if you go to three dimensions. And so on.