I recently worked on the problem of finding a sphere tangent to the edges of an irregular tetrahedron. I found that if one of the triangular faces of the tetrahedron is an isosceles triangle, then it is possible to find an infinite number of tetrahedrons that share this face, and each of them will have a unique sphere that is tangent to all its six edges.
My question is, are there instances where a tetrahedron having all faces as scalene triangles can have such a tangent sphere to its edges ?
On
The comment by @achillehui and the reference given by @JeanMarie answered my question. I want to build an arbitrary tetrahedron with a given scalene face. According to Theorem 1 in this paper, a tetrahedron will have an edge-tangent sphere if and only if its edge lengths can be expressed in terms of four positive numbers $x_i$ as follows
$P_i P_j = x_i + x_j ,\hspace{5pt} i,j = 1,2,3,4 , \hspace{5pt} i \ne j $
The $\{P_i\}$'s are the vertices of the tetrahedron. Therefore, given a face $P_1P_2P_3$ of the tetrahedron, I found its incircle, incenter, and the three points where the incircle touches the three edges. The distance from these touching points to the vertices $P_1, P_2, P_3$ on the same side are identified as $x_1, x_2, x_3 $, so that now
$ P_1 P_2 = x_1 + x_2$
$ P_1 P_3 = x_1 + x_3 $
$ P_2 P_3 = x_2 + x_3 $
And the touching points are given by
$\text{On edge } P_1 P_2: \hspace{10pt} Q_1 = P_1 + \dfrac{x_1}{x_1 + x_2} (P_2 - P_1) $
$\text{On edge } P_1 P_3:\hspace{10pt} Q_2 = P_1 + \dfrac{x_1}{x_1 + x_3} (P_3 - P_1) $
$\text{On edge } P_2 P_3:\hspace{10pt} Q_3 = P_2 + \dfrac{x_2}{x_2 + x_3} (P_3 - P_2) $
Now we can choose a "sufficiently large" positive value for $x_4$ , (see comment by @Blue below), then we would have three more edges as follows
$P_1 P_4 = x_1 + x_4$
$P_2 P_4 = x_2 + x_4$
$P_3 P_4 = x_3 + x_4 $
The fourth vertex $P_4$ can be found by solving a system of $3$ quadratic equations in the $3$ unknowns which are the coordinates of $P_4$. These equations are
$ (P_4 - P_i)^T (P_4 - P_i) = (x_i + x_4)^2 ,\hspace{5pt} i = 1,2,3 $
Taking the difference between equations $(1)$ and $(2)$ , and difference between equations $(1)$ and $(3)$ generates a linear system of two equations in the three unknowns. This linear system can be easily solved resulting in
$P_4 = V_0 + t V_1 $
where $V_0, V_1$ are known vectors found by the solver, and $t \in \mathbb{R}$ is to be found. To find $t$, substitute the above form of $P_4$ into any of the three quadratic equations involving $P_4$ and $P_i$'s, and this will give two values for $t$. Either one is good.
To find the sphere itself, we need four points on it, and these points are the edge points of tangency that we already know for the given face, and the fourth point can be generated as follows
$ Q_4 = P_4 + \dfrac{x_4}{x_i + x_4} ( P_i - P_4 ) $
Now if the unknown center is point $C$ then we have
$ (C - Q_i)^T (C - Q_i) = R^2, \hspace{5pt} i = 1, 2, 3, 4$
where $R$ is the (unknown) radius of the sphere. Taking the differences of these equations, gives us
$ -2 C^T (Q_i - Q_1) = Q_1^T Q_1 - Q_i^T Q_i ,\hspace{5pt} i = 2, 3, 4 $
Solving this $3 \times 3$ linear system, gives us the center $C$, and then using this in any of the four equations above gives the radius.
As a side note, the minimum value of $x_4$ that can be chosen to build the tetrahedron, given $x_1, x_2, x_3$ is the positive solution of the following $3 \times 3$ quadratic system
$ (P_4 - P_i)^T (P_4 - P_i) = (x_4 + x_i)^2, \hspace{10pt}, i = 1,2,3 $
where $P_4 = (x, y, 0)$, so that the $3$ unknowns are $x$, $y $ and $x_4$. It is easy to solve these equations by taking differences between equations $(1), (2)$, and equations $(1), (3)$ resulting in linear system of two equations in the three unknowns. Once solved, the form of the solution is
$[x, y, x_4]^T = V_0 + t V_1$
Substituting this into one of quadratic equations, the value of $t$ corresponding to a positive value for $x_4$ can be found, giving the minimum value of $x_4$. Set values of $x_4$ should be strictly above this minimum value to obtain a non-degenerate tetrahedron.
Comment expanded to answer per request.
First we need a lemma
The only if part is trivial. For the if part, since I cannot locate a proof online, I have outlined a proof at the end.
Back to the problem at hand. Start from all $r_i = 1$. If one randomly perturb them by small and distinct amounts. With probability one, one can find a set of $\ell_{ij}$ all different, realizable as edge lengths of tetrahedron. By above lemma, above tetrahedron admits a mid-sphere.
proof of 'if' part of lemma.
Let $r_1,\ldots,r_4 > 0$ and $p_1,\ldots,p_4 \in \mathbb{R}^3$ satisfies $|p_i - p_j | = r_i + r_j$ whenever $i \ne j$.
Given any two points $u,v$, let $(\delta_1,\cdots,\delta_4)$ be the differences of their barycentric coordinates with respect to tetrahedron $p_1p_2p_3p_4$. It is not hard to show:
$$|u-v|^2 = G(\delta_1,\ldots,\delta_4) \stackrel{def}{=} 2 \sum_{k=1}^4 r_k^2 \delta_k^2 - \left(\sum_{i=1}^4 r_k\delta_k\right)^2\tag{*1}$$
Let $\displaystyle\;\rho_1 = \sum_{k=1}^4 \frac{1}{r_k}\;$ and $\displaystyle\;\rho_2 = \sum_{k=1}^4 \frac{1}{r_k^2}\;$.
Let $q$ be the point with barycentric coordinates $(q_1,\ldots,q_4)$ where $\displaystyle\;q_k = \frac{1}{\rho_1^2 - 2\rho_2}\left(\frac{\rho_1}{r_k} - \frac{2}{r_k^2}\right)\;$.
For any edge $p_ip_j$, let $t_{ij}$ be the point on it with $|p_i - t_{ij}| = r_i$ and $|p_j - t_{ij}| = r_j$.
For example, consider the edge $p_1p_2$. The barycentric coordinate of $t_{12}$ is $\left(\frac{r_2}{r_1+r_2},\frac{r_1}{r_1+r_2},0,0\right)$.
Throw this into $(*1)$, we find
$$\begin{align} |q - t_{12}|^2 &= G\left(q_1 - \frac{r_2}{r_1+r_2}, q_2 - \frac{r_1}{r_1+r_2},q_3,q_4\right) \\ &= G(q_1,q_2,q_3,q_4) + \frac{4r_1r_2}{r_1+r_2}(q_3 r_3 + q_4 r_4) \end{align}$$ With a little bit of algebra, one can show that $$G(q_1,q_2,q_3,q_4) = \frac{-4}{\rho_1^2 - 2\rho_2}$$ Together with $$q_3r_3 + q_4r_4 = \frac{2}{\rho_1^2-2\rho_2}\left(\rho_1 - \frac{1}{r_3}-\frac{1}{r_4}\right) = \frac{2}{\rho_1^2 - 2\rho_2}\frac{r_1 + r_2}{r_1r_2}$$ We find $$|q - t_{12}|^2 = r_m^2 \stackrel{def}{=} G(q_1,\ldots,q_4) + \frac{8}{\rho_1^2 - 2\rho_2} = \frac{4}{\rho_1^2 - 2\rho_2} \tag{*2}$$
Since $r_m$ is independent of our choice of edge $t_{ij} = t_{12}$. This means the six point $t_{ij}$ are at a common distance $r_m$ to $q$.
Notice $$|q - p_1|^2 = G(q_1 - 1,q_2,q_3,q_4) = G(q_1,q_2,q_3,q_4) + \underbrace{2r_1(-q_1r_1 + q_2r_2 + q_3r_3 + q_4r_4)}_{X} + r_1^2 $$ and the mess $X$ equals to $\displaystyle\;\frac{2r_1}{\rho_1^2 - 2\rho_2}\left(2\rho_1 + \frac{2}{r_1} - \frac{2}{r_2} - \frac{2}{r_3} - \frac{2}{r_4}\right) = \frac{8}{\rho_1^2 - 2\rho_2}$.
We find
$$|q-p_1|^2 = |q - t_{12}|^2 + r_1^2 = |q - t_{12}|^2 + |t_{12} - p|^2$$
This implies $q - t_{12}$ is perpendicular to $t_{12} - p_1$. By a similar argument, we find $q - t_{ij}$ is perpendicular to $t_{ij} - p_i$ for all $i \ne j$. As a result, the sphere centered at $q$ with radius $r_m$ is tangent to the edges $p_1p_j$ at $t_{ij}$ and the 'if' part of lemma follows.
Upate
One may wonder what happens if $\rho_1^2 - 2\rho_2$ vanishes or even goes to negative. Let $V$ be the volume of tetrahedron, we can compute it using the Cayley-Menger determinant of the edge lengths $\ell_{ij}$:
$$288V^2 = \verb/CM/ \stackrel{def}{=} \left|\begin{matrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & \ell_{12}^2 & \ell_{13}^2 & \ell_{14}^2\\ 1 & \ell_{12}^2 & 0 & \ell_{23}^2 & \ell_{24}^2\\ 1 & \ell_{13}^2 & \ell_{23}^2 & 0 & \ell_{34}^2\\ 1 & \ell_{14}^2 & \ell_{24}^2 & \ell_{34}^2 & 0\\ \end{matrix}\right|$$ Substitute $\ell_{ij}$ by $r_i+r_j$ and with help of an CAS, one find $$288 V^2 = \verb/CM/ = 32(abcd)^2(\rho_1^2 - 2\rho_2) \quad\implies\quad \rho_1^2 - 2\rho_2 = \left(\frac{3V}{r_1r_2r_3r_4}\right)^2$$ For non-degenerate tetrahedron, $V > 0 \implies \rho_1^2 - 2\rho_2 > 0$. When $\rho_1^2 - 2\rho_2 \to 0$, $V \to 0$. The tetrahedron becomes planar and hence the midradius $r_m$ diverges.
If one pick $r_k$ to make $\rho_1^2 - 2\rho_2 < 0$, the corresponding $\verb/CM/ < 0$ and the set of edge lengths $r_i + r_j$ cannot be realized as edge lengths of any tetrahedron.
As a side note, if one compare above results with $(*2)$, we obtain following interesting relation:
$$3Vr_m = 2r_1r_2r_3r_4$$