As noted elsewhere, we may want to restrict attention to orthocentric tetrahedra, which is the family of tetrahedra whose altitudes share a common point that we can call "the orthocenter". Such tetrahedra can be characterized various ways apart from the orthocentricity property; most importantly: each pair of opposite edges determine an orthogonal vectors (hence the alternative name "orthogonal tetrahedra").
This answer to the related question mentions that orthogonal tetrahedra admit a sphere that intersects each face's $9$-point circle; this is the tetrahedron's 24-point sphere. (That's the topic of the Mathematical Gazette article mentioned in the other answer.) Interesting. I've found another analogue, albeit with fewer than twenty-four points.
Place the coordinates of tetrahedron $ABCD$ at
$$A=(0,0,-a) \qquad B=(0,0,b) \qquad C=(h,-c,0) \qquad D=(h,d,0)$$
Here, $h$ is the length of the "edge altitude" between $\overline{AB}$ and $\overline{CD}$. (An edge altitude has its endpoints within two opposite edges, and it determines the unique line perpendicular to both of those edges.) The given coordinates ensure that $\overrightarrow{AB}\perp\overrightarrow{CD}$; we guarantee the orthogonality of the other two pairs of opposing edge-vectors with the relation
$$h^2 = a b + c d$$
The coordinates of the orthocenter are
$$Q = \left(\frac{ab}{h}, 0, 0\right)$$
(Fun fact: Orthogonal tetrahedra are also characterized as those whose three edge altitudes concur. Moreover, the edge altitudes and face altitudes all meet at the same point, $Q$.)
Without too much trouble, one can find the feet of the four face altitudes. These determine a sphere with center $S$ and radius $s$, where
$$S = \left(\frac{a b + h^2}{3 h}, \frac{-c + d}{6}, \frac{-a + b}{6}\right) \qquad
s^2 = \frac{h^2\left( (a+b)^2 + (c+d)^2 \right) - 4 a b c d}{36 h^2}$$
One can verify that this sphere contains
- The centroid of each face: $\frac13(A+B+C)$, etc.
- The one-third-point of the segment from the orthocenter to each vertex: $\frac13(A+2Q)$, etc.
These points are analogues, respectively, of the midpoint of each edge (eg, $\frac12(A+B)$), and the midpoint of the orthocenter-vertex midpoint (eg, $\frac12(A+Q)$), in a triangle's 9-point circle. (Intriguingly, we go from "halves" in two dimensions to "thirds" in three dimensions. Does this pattern continue?)
This sphere does not contain the midpoint of each edge, and therefore does not contain the $9$-point circle of each face; this sphere is distinct from the $24$-point sphere.
I'm currently unaware of any other "interesting" points that may lie on this sphere. (So, for now, it's merely a "$12$-point sphere".) I'll note that the edge intersections aren't particularly pretty; for instance, if $\frac{1}{1+m}(A+mB)$ is on the sphere, then
$$m = \frac{a + b \pm \sqrt{a^2 - 14 a b + b^2}}{4 b}$$
Edit. The appearance of a "1/3" at the (now-deleted) end of my original answer got me thinking a little bit more about the powers of points with respect to various spheres, which I'll denote $\Sigma$ (the circumsphere), $\Sigma_{24}$ (the 24-point sphere), and $\Sigma_{12}$ (the 12-point sphere described above):
$$\begin{array}{rccclr}
\operatorname{pow}(Q,\Sigma_{12}) &=& -\dfrac{a b c d}{3h^2} &=& \phantom{-}\color{red}{\dfrac13}\;\operatorname{pow}(Q,\Sigma_{24}) = \left(\color{red}{\dfrac13}\right)^2 \; \operatorname{pow}(Q,\Sigma) \\[6pt]
\operatorname{pow}(A,\Sigma_{12}) &=& \dfrac{2a(a+b)}{3} &=& \phantom{-}\color{red}{\dfrac13}\;\operatorname{pow}(A,\Sigma_{24}) \color{gray}{\qquad(\operatorname{pow}(A,\Sigma_{\phantom{24}}) = 0)} \\[6pt]
\operatorname{pow}(H,\Sigma_{12}) &=& \dfrac{ab}{3} &=& -\color{red}{\dfrac13}\; \operatorname{pow}(H,\Sigma_{\phantom{24}}) \color{gray}{\qquad(\operatorname{pow}(H,\Sigma_{24}) = 0)}
\end{array}$$
where $H$ is the endpoint in $\overline{AB}$ of the edge altitude $h$. (In our coordinatized tetrahedron, $H$ lies at the origin.) As it happens, all six such endpoints lie on the 24-point sphere, as each is the foot of an altitude in a face. (Importantly, altitudes from neighboring faces meet at these points. For instance, the altitude from $C$ in face $\triangle ABC$, and the altitude from $D$ in face $\triangle ABD$, meet at $H$.)
I'm not sure what (if anything) this is trying to tell me, but it's pretty neat.
Best Answer
I've significantly re-worked this answer. For a previous version, see the Edit History.
"The" nine points of a triangle fall into three types:
These points naturally determine three concyclic right triangles:
By Thales' Theorem, each hypotenuse is necessarily a diameter of the nine-point circle. Each diameter has a triangle midpoint and an ortho-midpoint as its endpoints. The left-over points are the altitude feet, which determine the original triangle's orthic triangle. These elements happen to be related in an interesting way:
The figure shows the situation for diameter $\overline{PP'}$ and orthic triangle side $\overline{Q''R''}$:
For proof, we note that, since $\angle BQ''C$ and $\angle BR''C$ are right angles, Thales tells us that $Q''$ and $R''$ live on a circle with diameter $\overline{BC}$ (and, therefore, with center $P$). Similarly, with $G$ the orthocenter of $\triangle ABC$, we have right angles $\angle AQ''G$ and $\angle AR''G$, so that $Q''$ and $R''$ live on a circle with center $P'$. Thus, $\overline{Q''R''}$ is a chord common to the two circles, so that it must be perpendicular to, and bisected by, the line $\overline{PP'}$ connecting the centers. $\square$
The Fun Fact provides the "only if" portion of this characterization of viable nine-point sets:
For the "if" part, we provide a construction a quartet of solution triangles. To begin, one can show (as @mathlove did in this answer)
The four triangles determined by the orthocentric system share a common nine-point set, namely: the three altitude feet ($P''$, $Q''$, $R''$), as well as the three midpoints of $\overline{AB}$, $\overline{BC}$, $\overline{CA}$ (the midpoints of $\triangle ABC$), and the three midpoints of $\overline{GA}$, $\overline{GB}$, $\overline{GC}$ (the ortho-midpoints of $\triangle ABC$). The roles of some points change for $\triangle GBC$, $\triangle AGB$, $\triangle ABG$, but the point-set itself remains the same.
By the Fun Fact, the midpoints and ortho-midpoints determine diameters perpendicular to the sides of $\triangle P''Q''R''$. Such diameters are unique, so they must coincide the the Theorem's assumed pairs $\{P,P'\}$, etc, so that the midpoints and ortho-midpoints themselves coincide with the points $P$, $P'$, $Q$, $Q'$, $R$, $R'$. Thus, the given nine points are "the" nine points of $\triangle ABC$. $\square$
So, given nine points, the Theorem tells us when the set is viable, and the Lemma tells us how to construct an orthocentric system that yields exactly four solution triangles.
Note that nine points can contain four diametric pairs, raising the possibility of additional solutions. What happens then?
In the case of four diametric pairs, there the altitude-feet triad must contain one of them (so that the other three remain in tact), making the orthic triangle a right triangle (by Thales, yet again). Consequently, we have two midpoint/ortho-midpoint diameters perpendicular to the legs, which makes them perpendicular to each other, while the third midpoint/ortho-midpoint diameter is perpendicular to the hypotenuse diameter. In other words: Those four diameters comprise two mutually-perpendicular pairs.
I'll leave it as an exercise to the reader to show that there is only one choice of altitude-foot triad, unless the diameters make angles of $30^\circ$ and $60^\circ$; in that case, there are two symmetric choice of triad, leading to two orthocentric systems, for a total of eight solution triangles.