X(3), X(4), X(64), X(6523)
infinite points of the altitudes
A', B', C' : common points of the Orthocubic K006 and the circumcircle
midpoints of ABC
pedals of H with respect to A'B'C' = feet of the altitudes of A'B'C'
foci of the inconic with center O, perspector X(69)
more generally, every point of the Orthocubic with respect to A'B'C' and in particular the in/excenters of A'B'C' whose reflections about O lie on the Stammler hyperbola
The two triangles ABC and A'B'C' share the same orthocenter, circumcenter hence the same Euler line and centroid. The orthocubic of A'B'C' is K376, the Orthocubic's sister. See other analogous cubics in Table 58.
The tangents at A', B', C' and H pass through O. The polar conic of O is the rectangular hyperbola (H) passing through X(3), X(4), X(110), X(155), X(1351), X(1352), X(2574), X(2575) and obviously A', B', C'.
The tangents at A, B, C, H pass through X(25) and the polar conic of X(25) is the rectangular circum-hyperbola through X(378).
See also another characterization and a generalization at Q098.
From Table 54, it can be seen that K376 belongs to the pencils generated by :
K376 also belongs to the pencils generated by :
• K006 and the union of the circumcircle (O) and the Euler line (E),
More generally, K376 and one cubic (K) = pK(X6, P) of the Euler pencil generate a pencil of cubics spK (see CL055) which always contains a circular cubic (Kc), a stelloid (Ke), a central cubic (Ko) with center O, a psK (Ks) tangent at A, B, C to the symmedians. With K002, all the cubics of the pencil are psKs.
More precisely, taking P on the Euler line such that HP = T HG, every cubic of the pencil is F(t) = (1 - t) K376 + t (K) where t and T are two real numbers or infinity.
• F(t) is a circular cubic (Kc) if and only if t = 1/4 and then, it passes through X(3), X(4), X(30), X(74). The singular focus F lies on the line X(3), X(74), X(110), etc, such that OF = T / (T - 3) OX(110). F(t) is a proper focal cubic if and only if T = 3/2 giving K187. With T = 0, F(1/4) = (E) ∪ (O) and with T = 3, F(1/4) = (L) ∪ (J). These two decomposed cubics generate the pencil, stable under isogonal conjugation, which contains K001 and several other cubics. See table in Table 54, line Q = X20.
• F(t) is an equilateral cubic (Ke) if and only if t = 1 / (2T - 2) in which case it is a McCay stelloid with radial center X on the Euler line such that HX = 1 / (3 - T) HO. The pencil is generated by (L) ∪ (J) and K003 obtained with T = 3/2 hence t = 1. See table in Table 54, column P = X3 for other examples.
• F(t) is a central cubic (Ko) with center O if and only if t = 1 / (4 - T) in which case its points at infinity are those of pK(X6, S) such that OS = 3(2 - T)/T OH.. See table in Table 54, column P = [X20] for other examples. The pencil is generated by (E) ∪ (O) and K004 obtained with T = 3.
• F(t) is a psK other than K376 (for t = 0), and (K) (for t = 1) if and only if t = 1 / (3T - 2) in which case its pseudo-isopivot is X(6), its pseudo-pole lies on (J), its pseudo-pivot lies on the circum-conic with perspector X(525). Note that these two latter points are collinear with X(69). See table in Table 54, column P = aQ for other examples. The pencil is generated by K002 and K028 obtained with T = 0.
Each line in the table below shows the mentioned cubics in the pencil generated bt K376 and (K).