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If we have sphere inscribed in a tetrahedron, and if the distances from the center of the sphere to the edges of the tetrahedron are equal, is it true that this tetrahedron is always regular? I'm looking for hints.

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  • $\begingroup$ Do you mean distances from sphere center to edges as you say, or to the four faces of the tetrahedron? $\endgroup$ – coffeemath Jun 13 '16 at 12:10
  • $\begingroup$ Edges of the tetrahedron $\endgroup$ – marek3487 Jun 13 '16 at 12:13
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Assume that the inscribed sphere is the unit sphere $S^2\subset{\mathbb R}^3$. Let ${\bf u}$ and ${\bf v}$ be two points of contact, and put $\angle({\bf u},{\bf v})=:2\phi\in\>]0,\pi[\>$. The two planes touching the sphere at these points intersect in a line $\ell$, which carries an edge of the tetrahedron $T$. If $r$ is the distance from ${\bf 0}$ to $\ell$ then $\sin\phi={1\over r}$. Since all edges have the same distance from ${\bf 0}$ it follows that all six angles $\angle({\bf u},{\bf v})$ are equal. Assume that ${\bf u}_0$ points into the positive $z$-direction. Then the $z$-coordinates of the remaining three ${\bf u}_i$ are equal, hence these three points are lying in a horizontal plane. It is then easy to see that they form an equilateral triangle.

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  • $\begingroup$ This would appear to go somewhat beyond the "hints" that the OP was seeking. :) $\endgroup$ – John Hughes Jun 13 '16 at 12:55
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Hints:

  1. Think one dimension lower: can you have a circle inscribed in a non-equilateral triangle? When you do, how far is it from the center of the circle to the points of tangency?

  2. One way to inscribe a circle in a triangle is to start with a small circle tangent to two sides, with its center near one vertex, so that the third side of the triangle is completely outside the circle. Now gradually enlarge the circle, keeping it tangent to the two sides. Eventually, it'll touch the third side, and at the moment it does so, it'll be tangent to the third side. Think of the same process, one dimension higher.

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  • $\begingroup$ These are brillant ideas, but how to formalixe them into a proof that this tetrahedron will be regular? $\endgroup$ – marek3487 Jun 13 '16 at 12:10
  • $\begingroup$ I confess, I mis-read the original question, and read "faces" for "edges". or hint 1, the analogy is that of a circle in a triangle, with equal distances to all VERTICES, and then try to show the triangle's equilateral. If we call the tangency point on edge $AB$ by the name $c$, and the circle center $Q$, then $QA^2 = Ac^2 + Qc^2$, by pythagoras. Similar expressions for the other five half-edges show that they all must be equal, and you're done. Now...can you do something analogous in 3D? You'll once again need pythagoras, I suspect. $\endgroup$ – John Hughes Jun 13 '16 at 12:19
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Consider the tetrahedron $T$ with vertices $(0,0,0),\ (0,0,1),\ (0,1,0),\ (1,0,0).$ This is not a regular tetrahedron (three of the six sides are of length $1$ and three of length $\sqrt{2}$). Nevertheless there seem to be two points of the form $(a,a,a)$ at which to center a sphere which has the same distances to the six sides. The symmetry of the coordinates of $T$ make this easier to check. For the edges going "along an axis" we may just look at the $x$ axis (because of the symmetry). Then the squared distance from $(a,a,a)$ to a point $(t,0,0)$ is $(t-a)^2+a^2+a^2,$ and this is minimal at $t=a$ with value $2a^2$ there. For the remaining three edges, of which the edge joining $(1,0,0)$ to $(0,1,0)$ is typical, we can parametrize that edge as $(t,1-t,0)$ and then the squared distance from that to $(a,a,a)$ is given by $(t-a)^2+(1-t-a)^2+a^2,$ which is minimal when $t=1/2$ where its value is $3a^2-2a+1/2.$

So the distances from $(a,a,a)$ to the three axis edges are all $2a^2,$ while the distances from $(a,a,a)$ to each of the remaining edges are all $3a^2-2a+1/2.$ Hence to make all six (squared) distances equal we set these separate values equal, which gives in fact two positive values for $a,$ namely $a=1 \pm 1/\sqrt{2}.$

I don't have the geometry software to make any kind of picture of this result, and also it seems non-intuitive that there should be two such points, let alone one. So if anyone can find an error in my attempt I'd appreciate it, or if someone could somehow plot one of the choices that would be quite interesting.

Major edit: The points $(a,a,a)$ in order to lie "inside" $T$ needs to have coordinate sum at most $1$ since $x+y+z=1$ is the equation of the face not containing the origin. This rules out $a=1+1/\sqrt{2}$ but allows $a=1-1/\sqrt{2}$ which is then the only point I found for $T$ which is inside $T$ and equidistant to the six sides.

A possible objection dealt with:

Consider the regular tetrahedron with vertices $(\pm 1,0,0),\ (0,1/\sqrt{3},0),\ (0,1\sqrt{3},2\sqrt{3}/6).$ The center of this tetrahedron is at $C=(0,1/\sqrt{3},1/\sqrt{6}).$ Now the common distance to each of the six sides is that between $C$ and the origin, i.e. it is $1/\sqrt{2}.$ On the other hand the common distance to each of the faces is that between $C$ and the $xy$ plane, i.e. it is $1/\sqrt{6}.$ So in the usual sense of the term a sphere "inscribed" in the tetrahedron, being one which is tangent to each of the four faces of the tetrahedron, is not the same, even in the regular tetrahedron case, as asking for a sphere centered somewhere inside the tetrahedron which has each of the six sides of the tetrahedron tangent to the sphere. The two spheres have different radii.

So it is not really a drawback in the example above of the tetrahedron $T$ in which a sphere is found with center inside $T$ which has the six edges all tangent to it, but that sphere is not tangent at the same time to the four faces. If it cannot happen that way even for a regular tetrahedron, it would seem too much to want it for a general case. In fact I can't imagine any tetrahedron at all with a sphere centered at a point of its interior which simultaneously was tangent to all four faces and all six edges.

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