circlesgeometrytriangles
In $\Delta ABC$, $AD, BE, CF$ are the altitudes and $\Delta A'B'C'$ is the medial triangle. $K, L, M$ are the midpoints of $AH, CH, BH$. Consider the nine-point circle with centre $G$ (not to be confused with the centroid) and diameters shaded in yellow. Prove that $G$ is the midpoint of the Euler line $HO$.
This, frankly, is an amazing result. The fact that the nine-point circle exists at all is amazing in itself. But, I haven't seen a convincing proof for this fact yet.
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
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.
After some symbol-bashing in Mathematica, here are barycentric coordinates of key points:
$$\begin{align} A' &= (0 : a + b - c : a - b + c) \\[0.75em] A'' := \tfrac12(B'+C') &= \left(a +\frac{(a-b+c)(a+b-c)}{-a+b+c} : b : c\right) \\[0.75em] A''' &= (2 a + b + c : b : c ) \\[0.75em] O_A &= (2 a^3 : a^3 - b^3 + c^3 + a^2 b - a^2 c - 3 a b^2 - a c^2 - b^2 c + b c^2 - 4 a b c \\ &\phantom{(2 a^3\;}\quad : a^3 + b^3 - c^3 - a^2 b + a^2 c - a b^2 - 3 a c^2 + b^2 c - b c^2 - 4 a b c) \end{align}$$ The coordinates of the points where, say, $\bigcirc O_A$ (the one passing through incenter $X_1$) meets the side-lines of the circle are a little messy, so I'll omit them. Nevertheless, the six prescribed points are indeed cyclic, and the equation of their common circle has this barycentric form (in $u:v:w$ coordinates): $$\begin{align} 0 &= u^2 b c (-a + b + c) \cos A + v^2 c a (a - b + c) \cos B + w^2 a b (a + b - c) \cos C \\[0.75em] &\quad+ v w (-a^3 + b^3 + c^3 - 2 a^2 b - 2 a^2 c - b^2 c - b c^2) \\[0.75em] &\quad+ w u (-b^3 + c^3 + a^3 - 2 b^2 c - 2 b^2 a - c^2 a - c a^2) \\[0.75em] &\quad+ u v (-c^3 + a^3 + b^3 - 2 c^2 a - 2 c^2 b - a^2 b - a b^2) \end{align}$$
Converting to trilinear form (in $\alpha:\beta:\gamma$ coordinates), the equation can be manipulated into this: $$(\lambda \alpha + \mu\beta+\nu\gamma)(a\alpha+b\beta+c\gamma)+\kappa(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda &:= (-a + b + c)\cos A \\ \mu &:= (\phantom{-}a-b+c)\cos B\\ \nu &:= (\phantom{-}a+b-c)\cos C\\ \kappa &:= -2(a+b+c) \end{align}$$
As $\lambda:\mu:\nu$ are the trilinear coordinates for Kimberling center $X(219)$ ("$X(8)$-Ceva Conjugate of $X(55)$"), the circle is the Central Circle of that point. This answers the "alternatively" aspect of OP's question.
I don't know what Kimberling centers may lie on the circle. I don't even know how one would go about searching for them. Someone with more fluency in triangle-center lore may have some insights.
Incidentally, for the $\bigcirc O_A$ passing through $B''$, $B'''$, $C''$, $C'''$, etc, the corresponding six-point circle has trilinear form $$(\lambda'\alpha+\mu'\beta+\nu'\gamma)(a\alpha+b\beta+c\gamma)+\kappa'(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$ where $$\begin{align} \lambda' &:= a (-a + b + c)^2 \\ \mu' &:= b (\phantom{-}a - b + c)^2 \\ \nu' &:= c (\phantom{-}a + b - c)^2 \\ \kappa' &:= -8 a b c \end{align}$$ This circle is therefore the Central Circle of Kimberling's $X(220)$ ("$X(9)$-Ceva conjugate of $X(55)$").
I have not investigated whether there is a general connection between OP's construction, Ceva-conjugates, and/or $X(55)$.
Best Answer
Behold!
(MSE doesn't like such concise answers, so here are some extra characters.)