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K859

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

on both curves :

A, B, C, X(2) with tangent the Euler line

X(11078), X(11092), infinite points of the Steiner ellipses

on K859a :

X(13), X(299), X(532), X(619), X(3180), X(11118), X(11119), X(11122)

Ea (see below)

on K859b:

X(14), X(298), X(533), X(618), X(3181), X(11117), X(11120), X(11121)

Eb (see below)

K859a = pK(X11078, X13) and K859b = pK(X11092, X14) are two pivotal cubics, one being the isotomic transform of the other, with poles the barycentric quotients X(11078) = X(13) ÷ X(14) = X(16) ÷ X(15) and X(11092) = X(14) ÷ X(13) = X(15) ÷ X(16) respectively.

These poles lie on the two cubics, on the line X(30)X(74) and on the conic (C) which is its isotomic transform.

K859a and K859b are both tangent at G to the Euler line hence the tangentials Ea, Eb of G are two other points on the Euler line. Peter Moses observes that :

• Ea = X (471) x X (621), on the lines {232, 5978}, {264, 3642}, {340, 532}, {533, 648}, {619, 6116}, {1968, 9988}, SEARCH = 0.61554816872172722664.

• Eb = X (470) x X (622), on the lines {232, 5979}, {264, 3643}, {340, 533}, {532, 648}, {618, 6117}, {1968, 9989}, SEARCH = 3.5882743534311368912.

• The midpoint of Ea, Eb is X(297).

• Ea = SB SC (Sqrt[3] SA - S) (S SA + Sqrt[3] SB SC) : : and Eb = SB SC (Sqrt[3] SA + S) (S SA - Sqrt[3] SB SC) : : .

Ea, Eb are now X(11093), X(11094) in ETC (2016-12-07).

Moreover Fa = X(30)X(74) /\ X(11118)Ea and Fb = X(30)X(74) /\ X(11117)Eb lie on K859a and K859b respectively. These points are rather complicated with SEARCH numbers 0.00783861310604469 and 4.06649036110565.

Obviously, their isotomic conjugates tFa, tFb are the sixth common points of (C) and K859b, K859a respectively.

***

K859a and K859b generate a pencil of cubics which contains the nodal cubic K185 = cK(#X2, X523) and K860 = pK(X2, X30), the third pK of the pencil. All these cubics contain the nine following points :

• A, B, C,

• X(2) counted twice since the tangent is the Euler line, except for K185 since X(2) is a node,

• the infinite points of the Steiner ellipses,

• X(11078), X(11092).

This pencil is very similar to a pencil of circular cubics where the circular points at infinity would be replaced with the infinite points of the Steiner ellipses. The point analogous to the singular focus would then be the intersection of the tangents at these points. It can be seen that this point lies on an ellipse homothetic to the Steiner ellipse, with center X(9979), passing through X(648), X(671). The two centers of homothety are rather complicated.

***

Barycentric products and quotients of K859a and K859b

K278 = K859a x X(14) = K859b x X(13),

K390 = K859a x X(15) = K859b x X(16),

pK(X7799, X2) = K859a ÷ X(13) = K859b ÷ X(14),

pK(X76 x X94, X94) = K859a ÷ X(16) = K859b ÷ X(15),

K859a x X(11092) = K859b and K859b x X(11078) = K859a.