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K1180

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X(3), X(4), X(5), X(54), X(14979), X(32423), X(38539), X(38542)

imaginary foci of the MacBeath inconic

other points below

Geometric properties :

K1180 is an isogonal focal nK, the locus of foci of inconics with centers on the line X(5), X(54). Also, K1180 is spK(X32423, X5) as in CL055.

Its singular focus is X(14979). It is a member of CL062 hence it is invariant under both Cundy-Parry transforms. See CL037.

K1180 is also globally invariant under another quadratic transformation 𝚿OH very similar to Psi and Psi_P described here.

𝚿OH is the commutative product of the reflection in the Euler line and the inversion in the circle (C) with diameter OH and center X(5).

K1180 is then the locus of M such that X(54), M and 𝚿OH(M) are collinear.

Points on K1180

• K1180 must contain the singular points of 𝚿OH (namely X(5) and the circular points at infinity) and its fixed points (namely O, H).

• K1180 meets (C) again at two points S1, S2 on the perpendicular at X(54) to the Euler line.𝚿OH obviously swaps these points.

• K1180 meets the sidelines of ABC again at A', B', C' which lie on the trilinear polar (L) of the root X(24978). (L) is the tangent at X(24977) to the MacBeath inconic.

• K1180 passes through A1 = 𝚿OH(A) and A2 = 𝚿OH(A') which are isogonal conjugates. The points B1, B2, C1, C2 are defined cyclically.

The triangles ABC and A1B1C1 are perspective at X(54) and the circumcircle (C1) of A1B1C1 passes through X(54). Its center is X(14130) , the midpoint of X(15062). It follows that A1, B1, C1 are the projections of X(15062) on the cevian lines of X(54).

The triangles ABC and A2B2C2 are perspective at X(5) and the circumcircle (C2) of A2B2C2 passes through X(5). Its center is Ω2 and the reflection Q2 of X(5) in Ω2 are unlisted in ETC. It follows that A2, B2, C2 are the projections of Q2 on the cevian lines of X(5).

• More generally, if M and N are two isogonal conjugates on K1180 then the Cundy-Parry transforms OM /\ HN and HM /\ ON are two other isogonal conjugates on K1180. For example, with M = X(14979) on (O) and N = X(32426) at infinty, we find X(38539) and X(38542).

With M = A and N = A', we find Oa = OA /\ HA' and Ha = HA /\ OA' on the cevian lines of O and H.

K1180a

These triangles A1B1C1 and A2B2C2 are very similar to the 2nd and 4th Brocard triangles obtained when the transformation Psi mentioned above is used instead of 𝚿OH. In the same way, (C1) can be seen as a Brocard circle and the circle B1C1A2X(5) as a McCay circle, this latter being the 𝚿OH-image of the sideline BC.