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K698

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X(3), X(69), X(74), X(376), X(468), X(3564), X(6090), X(6091), X(7612)

F = X(3)X(895) /\ X(22)X(111) = X(6091)

A' = AX(69) /\ parallel at G to BC, B' and C' likewise

Y = F, X(3564) /\ perpendicular bisector of X(3)X(69)

Q = X(6090), P1, P2 described below

K698 is a focal cubic with singular focus F, the inverse of X(895) in the circumcircle. F = a^2 SA (SA-a^2) / (2 SA-a^2) : : , SEARCH = -2.65808992942898.

Let (L) be the line through O and X(2575) and let (C) be the circle of inversion that swaps G and X(1113). Let f be the transformation which is the (commutative) product of this inversion and the reflection about (L). f can be seen as an isogonal conjugation with respect to an imaginary triangle T whose vertices are O and the circular points at infinity.

K698 is a pivotal isogonal cubic with pivot X(69) with respect to T.

Note that f swaps X(69) and the focus F, also X(74) and X(376).

Q = f(X468) = a^2 (3 a^4+b^4+c^4-4 a^2 b^2-4 a^2 c^2+6 b^2 c^2) : : is unlisted in ETC, SEARCH = 2.40503954322767. It is the intersection of the lines 2,3167 - 3,74 - 6,373 - 25,374...

These points Q, F are now X(6090), X(6091) in ETC (2014-09-27).

Since the intersections P1, P2 of (L) and (C) are clearly invariant, we can see them as two in/excenters of the triangle T.

K698a

 

K698 is the locus of contacts M, N of tangents drawn through F to the circles passing through O and X(69).

M' = f(M) and N' = f(N) lie on the cubic. The following triads are made of collinear points on the cubic :

F, M', N' – X, M, N

X(69), M, M' – X(69), N, N'.

X is the intersection with the real asymptote. It is the pole of F in the pencil of circles above.

f transforms the lines MN', NM' into two circles (CN), (CM) passing through O and another (variable) point S'.

S = f(S') lies on the line MN.

The locus of S' is a rectangular hyperbola (H) with center O, passing through F, X, P1, P2

 

These points solve a problem analogous to Steinhaus' problem raised by Seiichi Kirikami (ADGEOM #1804) as follows.

Given a triangle ABC and a point P, we denote the pedal points of P on BC, CA, AB by D, E, F respectively. Find P such that BPF+CPE=CPD+APF=APE+BPD.

Any solution must be a common point of three concentric rectangular hyperbolas (Ha), (Hb), (Hc) with center O. Furthermore, (Ha) contains the traces on BC of the bisectors at A and the intersections of the circumcircle (O) with the perpendicular bisector of BC. Any point P on (Ha) satisfies the equality CPD+APF=APE+BPD.

These hyperbolas belong to a same pencil and they have two real common points P1, P2 which solve the problem although these points are not always interior to ABC. Note that the pencil contains the hyperbola (H) above and also a decomposed hyperbola which is the union of the axes of the inconic with center O or, equivalently, the parallels at O to the asymptotes of the Jerabek hyperbola. The figures below show a triangle ABC with two interior points.

K698P1 K698P2

P1, P2 are unlisted in ETC with SEARCH numbers 1.828954257033454 and 11.73577153135924.

Another partition of ABC can be found in the paper "On the Thomson triangle". It is related with the center X(5373) and with the cubic K615.