Figure: A portion of the group table of the Lucas cubic with the Steiner foci and its isotomic conjugates added in. The blank entries are known from the usual group table, and are not repeated here. The red row is the constant to which the indices of all rows must add. F1—dA refers to the point where the line F1—dA meets the cubic.
From the group table we conclude that tF1--tF2 goes through L and we later lean that it is parallel to the Steiner major axis, and both F1--tF1 and F2--tF2 both go through D (= dK = tH), the pivot for this cubic.
The major picture
Master Figure: This picture shows the Seiner circumellipse and inellipse, the circumcircle, the Brocard circle, the director circle of the circumellipse.
The foci and points related to the foci
Most of these results are due to Peter Moses, Bernard Gibert, and Wilson Strothers. Many of these results are shown in the Master Figure.
[PM] Call the foci of the Steiner circumellipse F+ & F-
Call the foci of the Steiner inellipse f+ & f-, note that F+ and f+ are on opposite sides of G.
gF+, tF+ and F- are colinear.
gF-, tF- and F+ are colinear.
These lines meet on the circumellipse at P = (SA2 – SBC – SA Q :: ) ~ ( SA2– – SA Q ::), a meet of GK with the circumellipse.
gf+ = f–, tf+ and P are colinear.
gf- = f+, tf- and P are colinear.
Circumconic through tF+, tf+, G, tF- and tf+ passes through P with center: midpoint P and G, on the Steiner inellipse. [this is the isotomic of the major Steiner axis]
Circumconic through gF+, f-, f+, gF- passes through P, 1380. [ this one defined from the colinearity of rF+ = tgF+, tf-, tf+, rF– = tgF- , tP; since tP is the infinite point on the Steiner axes, this line must be parallel to an axis]
[WS]
Fact 1: The vertices of the cevian triangles of tF+ and tF- lie in a circle C whose center L is on line tF+ tF-.
Fact 2
The inconic I(L) with centre L is bitangent with C above with contact points on tF+ tF-, so this is the axis containing the real foci G+, G– (isogonal conjugates of course, and on the Darboux Cubic) The perspector is tdL the cevapoint (whatever that is) of tF+ and tF–. The inconic has axes parallel to those of the Steiner ellipses.
f+--H and f- —O meet at G–
and f- —H, f+ —O at G+
Is it relevant that f+, f-, G+,G-, O,H are isogonal conjugates? [SS] Yes there is more to this story than we have found (and we have found so much more than I ever thought was there!). G+, G-, O, H are on the Darboux cubic, which should give more relations. f+, f-, G, O, H are on Thomson cubic (at least I think they all are), so this will give more.
Conjecture 2
The lines F+G+1 and F-G- meet on the Euler line.
The lines F+G- and F-G+ meet on the Euler line.
One of the points of intersection is the anticomplement of L (so reflection of L in H) . The other cuts LH in the ratio 2:3(a strange ratio)
Fact 3
The duals of F+ and F- are perpendicular to the major axis of the Steiner ellipses
[SS] This one is known already. Peter (I think) credited this fact to Jean-Pierre, who showed that the dual of any point on a Steiner axis is perpendicular to the axis.
Conjecture 3
These tripolars (duals) are the directrices of the Steiner Inellipse.
[BG] all you need to know is that the tripolar of P is the polar of mtP in the Steiner in-ellipse.
[SS] If G+ and G+ are on the Darboux cubic, its two operations are the isogonal conjugate and reflection in the circumcircle, which we call o.
Both the group table and observation says that oG+ — oG- goes through H.
The tripolars of points on the non-focal axis is an inparabola with perspector the intersection of GK with Steiner ellipse and Focus the OK intersection with the circumcircle.
The dual of this inparabola is a circumhyperbola with perspector an end of a Steiner axis.
[PM] Wonderful results Wilson ..
Various lines concerning the Steiner foci ...
f +/-, foci of the Steiner Inellipse
F +/-, foci of the Steiner Circumellipse
g = isogonal
t = isotomic
I+ = Steiner major axis at infinity = g X(1379)
I- = Steiner minor axis at infinity = g X(1380)
P = r X(1379)
[SS] I am adding things to this for my benefit. I am adding the new results as well as why we know the statements are true. G+ is taken to be in the same direction as F+. the "o" operation represents reflection in the circumcircle. It is one of the two generating operations of the Darboux cubic.
{tF-, rtF-, rF+}.
{tF+, rtF+, rF-}.
Don't know why this is true.
{D, F-, tf+, tF-, and tG+}
{D, F+, tf-, tF+, and tG-}
(D, F-, tF-) comes from cubic properties.
{f-, pf-, X(1379)}.
{f+, pf+, X(1379)}.
p is the projective operation that takes G to K and the Steiner ellipse to the Circumcircle. This operation is ubiquitous in this schema.
{pF+, f+, gF+, O, oG-}
{pF-, f-, gF-, O, oG+}
Dont know why.
{gF-, pF+, pgF-, X(1379)}.
{gF+, pF-, pgF+, X(1379)}.
dont know why
{F+, P, tF-, gF-}
{F-, P, tF+, gF+} p X1379 = P these lines are p of above lines.
{tK, tf-, tf+, tgF-, tgF+, I+}.
this very interesting results follows from two things. Jean-Pierre's (I think) demonstration that the tF+ , tF- line is parallel to focal axis. Also that p of this line is the focal axis.
{pF-, pf-, K, pF+, pf-, I+}, this is the tripolar of X(1380).
gtG = K, this line is the gt of the focal axis and parallel to it (perhaps)
{L, I+ ,tF-, tF+} add G+, G-
From group properties for two cubics
{G, X(1341), X(1348), X(2542), F+, F-, f+, f-, I+} .. major axis .. and infinite point
The x pts here are centers of similitude with the Brocard circle and other circles.
Some others with the new points
G- tG- f+ H
G+ tG+ f- H
oG+ oG- H
f- D G- gF-
f+ D G+ gF+
G+ F+ dL
G- F- dL
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Some Properties of Points on a Steiner axis
[BG] If P is a point, gP, tP, sP its isogonal, isotomic, reflection in G then gP, tP, sP are colinear iff P lies on the axes of the Steiner ellipses, hence your property is also true for the in-foci and for the non-real foci.
when P lies on the focal axis, the line passes through X524 + Q X2, which is the points called P above by Peter.
when P lies on the non-focal axis, the line passes through X524 – Q X2, which is the Steiner antipode of the above.
[Note: X524 is the infinite point on the GK line.]
[SS] The dual of a point on a Steiner axis is perpendicular to that axis. In particular the dual of the point at infinity on one axis is the other axis. This property allows us to find other, rational points on the Steiner axis (done below)
The isotomic conjugates of two points on a Steiner axis lie on a line parallel to that axis. [Jeff Brooks says that this was proved by Jean-Pierre Ehrman]
Peter's rectangles
[PM] The line P -- X(1379) -- X(99) is parallel to the Steiner minor axis.
(line t g Q -- Q passes through X(99) for Q on the circumcircle)
On the circumcircle, X(99), X(1379), X(98) & X(1380) form a rectangle parallel to the Steiner axes.
Another rectangle, this time on the circumellipse is X(99), P = tgX(1379), X(671), tgX(1380) [Note: tg = r is the "retro" projective transformation of The Triangle Book].
Centers of Similitude
[PM] Forgot to mention that
X{2,1341,1348,2542} are on the Steiner Focal axis and
X{2,1340,1349,2543} on the other one.
Note:
X(1341) = EXSIMILICENTER(CIRCUMCIRCLE, BROCARD CIRCLE)
X(1346) = INSIMILICENTER(NINE-POINT CIRCLE, ORTHOCENTROIDAL CIRCLE)
X(2542) = INSIMILICENTER(BROCARD CIRCLE, (X(4),2R))
Indeed, the similitudes of the Brocard circle and a circle centered on the Euler line at a distance k |GO| from G with radius k R lie on the axes of the Steiner ellipse.
Constructions of the Steiner axes.
These facts led John Conway to propose the easiest construction of the Steiner and Kiepert axes:
Begin with the Euler line and construct the circle on GH, which John calls "the shield" and create its center U (for "umbo," Greek for center of a shield -- John uses these names as pneumonic devices, he does not write new discoveries down, rather he finds a way to put them into memory so that they will be remembered forever). This circle plays an important role in John's conception of the triangle because both the incenter and symmedian points (and many others) will be inside it.
Next draw U—K, the Fermat line f. Because K must be inside the shield, this line will meet the circle twice. The lines from G to these two intersections are the Steiner axes.
Now draw a circle with center U through mS, the medial Steiner point. The Euler line is a diameter and will meet this circle twice. Connect mS to these two meets. These will be the Kiepert axes.
Figure: Construction of the Steiner and Kiepert axes. The circle on GH is the "shield" circle, U its center. The second circle is through mS, the Kiepert center, and centered at U.
The axes of the Steiner ellipse
General Principles
For P in the triangle plane, there are two natural lines: G—P and ~P where the "~" symbol means "dual." I prefer the affine invariant dual to the projectively invariant tripolar. The dual of (l:m:n) is lx + my + nz = 0. Note: "natural" in this context means "leading to structure which is affinely preserved."
The isotomic conjugates of the infinite points on these two lines are antipodal on the Steiner ellipse. This allows us to associate a particular point with antipodal points on the Steiner ellipse.
P (l:m:n) -> P2- (l2–mn : m2–nl : n2–lm) is inversion in the Steiner ellipse. Both points are on G--P.
G—P and ~P intersect at mP2–, the medial (complementary) of P2-.
E+ = P2– + Z P and E– = P2– – Z P . where Z = √( l4 + m4 + n4 – m2n2 – n2l2 – l2m2) are the intersections of G--P with the Steiner ellipse.
The tripolars of E± go through G, so that their duals are at infinity. The dual of E+ is the infinite point on E–. These are the infinite points on the axes of the conic that is the isotomic of G—P. This statement can be checked by computation. It is the only statement whose verification is not trivial (although I figure that there must be an easy argument out there somewhere).
The dual of the isotomic of E+ is the line through G with the correct infinite point. The axes through the conic center will be parallel to these lines. The particular case we are interested in has P = K, the symmedian point in which case Z is Q = √( a4 + b4 + c4 – a2b2 – b2c2 – c2a2). The relation between Q and S, twice the area, is Q2 = SW2 – 3 S2 .
These are all results of my affine theory that I have been expounding here.
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Now let P = K, then the points named above become (here •infinity refers to the intersection of a line with infinity.)
~K•infinity = : c2 – a2 : the isotomic Steiner point tS
G—K•infinity = : c2 + a2 – 2 b2 : X(524)
t ((~K)•infinity) = : 1/ (c2–a2) : = S, the Steiner point.
t( (G—K)•infinity) = :1/(c2 + a2 – 2 b2 ): , antipodal to the above, X(671).
E+ and E– are the isotomic conjugates of X530 and X531, the infinite points on the Kiepert hyperbola because the Kiepert hyperbola is the istomic of the line G—K
The duals of X530 and X531 are the Steiner axes since they have the same directions as the Kiepert hyperbola asymptotes.
The equations of the axes thus can easily be found as ~tE1 and ~tE2
My form of the axes of the Steiner ellipse are
... + y / (b4–c2a2 + b2Q) + .... = 0,
with Q being oppositely signed for the other axis.
The endpoint of this line is : 1/ (b4 – c2a2 – b2Q) : which has the opposite sign as the axis so that the endpoint of an axis is the dual of the other axis. This is not easily verified without a computer. Since the line goes through the centroid, the point must be on the line at infinity, which is the dual of the centroid.
The foci of the Steiner circumellipse
The focus is now, in principle, easy to find, but fining it requires a computer to do the computation.
The end point s∞ on a Steiner axis is as above so that s∞ ± Z
G, where Z is computed in terms of s∞, is one of the meets of the
ellipse with an axis . To find the focus we then contract this point by
3√Q/√(SW2 + Q2) which values
were obtained from the basic Steiner ellipse picture above.
from Bernard Gibert:
[BG] I have computed the Steiner foci again from scratch trying to get symmetric formulas.
After a very painful and long work, I have found the following results :
Let :
S2 = a2+b2+c2,
S4 = a4+b4+c4,
S6 = a6+b6+c6,
Q = sqrt(a4+b4+c4 – b2c2 – c2a2 – a2b2), as above
X2 = 1:1:1 (centroid)
X543 = 3(a4+2b2c2) – S22 : : ,
X524 = b2+c2 – 2a2 : : ,
E530 = X543 + Q X524
E531 = X543 – Q X524
these are the infinite points of the asymptotes of the Kiepert hyperbola i.e. the infinite points of the axes of the Steiner ellipsea (E530 for the axis containing the real foci)
Then :
Q1 = 2 S42 – 6 a2b2c2 S2 – 7 (b2c2(b2 – c2)2 + cyclic )
Q2 = 6 S6 – S23 + 21 a2b2c2
R = sqrt(Q1 – 2 Q2 Q)
S = sqrt(Q1 + 2 Q2 Q)
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Foci of the Steiner inellipse :
(real) E530 + R X2, E530 – R X2,
(imaginary) E531 + S X2, E531 – S X2,
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Foci of the Steiner circumellipse :
(real) 2 E530 + R X2, 2 E530 – R X2,
(imaginary) 2 E531 + S X2, 2 E531 – S X2.
[Note: it turns out that the real and imaginary foci are backwards. We have always said that the Steiner axes are parallel to the Kiepert ones; it is more true that they are perpendicular, so that the Kiepert asymptote that corresponds to the Steiner focal axis is the perpendicular one.]
Remember that the 4 in-foci lie on the Thomson cubic and on K018; the 4 other on the Lucas cubic.
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from Peter Moses
(with his Z changed to Q to conform to the above usage)
For the Steiner foci I arrive at (barycentrics)
QQ +/- k (b2 – c2) (a4 - b2 c2 – a2 Q) ::
Where QQ = Sqrt[2 (a2 b2 c2 Z3) – 2 (b4 c4 (2 S2 - a2 SW) + a4 c4 (2 S2 – b2 SW) + a4 b2 (2 S2 – c2 SW))]
and Q = sqrt(a4 + b4 + c4 – b2c2 – c2a2 – a2b2)
k = 1 for inellipse and 2 for circumellipse.
[Note: X524 is the infinite point on GK.]
[BG] where Q = ± S √( cot2 W – 3)
these two points are antipodes on the Steiner circum-ellipse and lie on the line GK, P being on the same side as K.
Note that you find an equivalent property when sP is the reflection of P in any point on the the Steiner hyperbola, the diagonal rectangular hyperbola with equation (b2 – c2) x2 + cyclic = 0 and center S = X(99).
New Identities
This circumstance gives quite a few new identities. There is a lot to chew over here
(B2– + B Q )(b4– + b2 Q) = ± (B-C)(B-A) (S4 + Q S2)
(S4 + S2 Q)(S4 – S2 Q) = 4S2 S22
(b4 - c2 a2 – b2 Q)(b4 - c2 a2 + b2 Q) = –(a2 – b2) (b2 – c2) S22
(a2 – 2b2 + c2 – Q)(a2 – 2b2 + c2 + Q) = –3(a2 – b2) (b2 – c2)
: (c2 – a2) (a2 – 2b2 + c2 + Q) : dot :(c2 – a2) (a2 – 2b2 + c2 – Q) :
