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K1094

too complicated to be written here. Click on the link to download a text file.

X(2), X(20), X(69), X(1992), X(3543), X(31886), X(31887)

infinite points of K002

Geometric properties :

Let P be a finite point and Q the midpoint of GP. Any spK(P, Q) passes through the vertices Q1, Q2, Q3 of the Thomson triangle. See CL055.

The tangents at Q1, Q2, Q3 concur (at X) if and only if P lies on K1094.

With P = X(2), X(20), X1992) we obtain K002, K615, K1093 with X = X(5646), X(154), X(6) respectively.

K1094 is a central cubic with center X(2) and asymptotes parallel at G to those of the Thomson cubic K002. These asymptotes are the altitudes of the Thomson triangle Q1Q2Q3.

K1094 and K002 meet again at six finite points lying on a rectangular hyperbola (H) passing through X(2), X(76), X(193), X(3413), X(3414) hence homothetic to the Kiepert hyperbola.

If P' and P" are two points on K1094 symmetric about X(2) then the corresponding points X' and X" are collinear with X(6).

• if P' = X(20) and P" = X(3543), we find X' = X(154) and X" = X(6)X(25) /\ X(23)X(3066), SEARCH = -0.372397625789591, barycentric coordinates : a^2 (5 a^4+2 a^2 b^2-7 b^4+2 a^2 c^2+14 b^2 c^2-7 c^4) : : .

• if P' = X(1992) and P" = X(69), we find X' = X(6) and X" = X(3)X(6) /\ X(20)X(141), SEARCH = 8.71218102723990, barycentric coordinates : a^2 (3 a^4+2 a^2 b^2-5 b^4+2 a^2 c^2-6 b^2 c^2-5 c^4) : : .

When P' and P" traverse K1094, these points X', X" lie on a cubic which is the pK with pivot X(6), isopivot X" just above with respect to the Thomson triangle. See figure below.
K1094a

P69 and P3543 are the two points X" mentioned above.

P69 = X(31884) now in ETC (2019-03-31).

(C) is the polar conic of P69 passing through X(6), X(376) and the vertices of the Thomson triangle Q1Q2Q3.

JT is the Thomson-Jerabek hyperbola.

The cubic meets the circumcircle again at the same points as nK0(Ω, X6) where Ω lies on the line X(32)X(184) with first barycentric a^4(5a^2+11(b^2+c^2)).

Ω = X(31885) now in ETC (2019-03-31).