Several revelations:

Given four points, it is not always possible to construct a regular tetahedron and a point inside it such that the four points are the projection of the center point onto the sides. In order for such a thing to exist, for every pair of the original four points, the locus of the points that form angles with them must all intersect at one point, and it's easy to separate two pairs so that the loci in question cannot possibly exist.

Or, more intuitively but much less rigorously, you can count the "dimensions" you need to pick to choose one of these collections. Four points in space means twelve dimensions of choice: three per point. To pick a tetahedron and its point, we can pick the center (three dimensions), one corner (three dimensions), a "spin" for the other three points (one dimension), and a point inside it (three dimensions), which is only 10. Not enough to cover 12 dimensions. Unfortunately as everybody knows and have the same cardinality so to make this rigorous, you'd have to do something about continuous functions not changing the dimension of something. Something really, really, really fancy and really, really, really beyond my range.

Note, though, that the extension of Viviani's to 3D does not require the tetahedron to be regular, only that every face of it have the same area. (Facepalm goes here.) Hopefully this will weaken the requirement enough. Hmm, let's see.

Now, we'll consider, given a triangle, if we can find the right point (in space) that will form such a tetahedron. Okay, so let the triangle be , let denote its side lengths in the usual way, and let's assume its area is . Then every other face must have area . Therefore, if the point we seek is , then the distance from to is , the distance from to is , and the distance from to is .

Conveniently, also happen to be the lengths of the altitudes of . (Well, it's really just what you get with a bunch of equal-area side-sharing triangles.) So let's take 's projection into the plane lies in, call it , and have some fun with barycentrics/trilinears. (I am using these much more often than before I started this blog.) Let 's normalized-to-sum-to-1 barycentric coodrinates be . We have . Next, 's distances from the three sides would be , , . If , we then have , or , and the same thing for the other sides.

Do you feel the urge to stick a or into and watching the radical collapsing in on itself? Preferably, use because the result of the solution would be proportional to and therefore make a nice constant value?

Unfortunately, the previous restriction is holding us back. Nothing useful here, apparently.

Okay, so let's backtrack a little and look at what the problem really wants us to prove. Something about the angles that the center point and the initial tetrahedron would form. Now, since if we're doing this right the tetahedron's points are the projections of the center point onto one of the magical tetahedrons we're studying, the set consisting of the center point, two of the projections, and the point on the edge of the faces on which the projections are that is coplanar with the three other points(,) would be concyclic because two of its opposing angles are right. Anyway, this means that the angles the problem is talking about are supplementary to the face-and-face angles of the hypothetical magical tetrahedron, so we might as well study those.

Okay, let's return to our previous work and study those face-to-face angles. Suppose the angle formed by the triangles and is , and similarly for the other three symmetric angles. Then we have . Surprise! We get to make the radical blow up on itself anyway. Watch.



Yay! And what makes it better is that, since and are both the same if we swap and in all the previous discussion, and the same is true for and and and , and the same quadruple involving , is the same value for any face of the magical tetrahedron! (This is also obvious because the faces of the magical tetrahedron have the same area and the volume of the tetahedron is the same no matter which way you look, but whatever.)

Let's not forget the other condition: or . We sum that over all faces, normalize, subtract, eliminate, and rewrite, and we discover: opposite face-and-face angles are equal! Therefore, since is constant for all edges, opposite edges are equal, therefore:

A magical tetahedron must have congruent faces.

Tada. This helps, but it still doesn't make it obvious that the projections onto the sides of a magical tetahedron from a single point can generate any quadruple of points, which is pretty nasty.

What is the locus of points that form the same angle towards two other fixed pairs of points? I can't even do this in 2D.

Oh well, another time.