The incentral triangle

(with Wilson Strother's help). The incentral triangle and its centers are an extraordinary system, especially when extraversions are taken into account.
The Cevian triangles of all four incenters I_x (x = o, a, b, c) share the same four lines as edges. This means that all four of these triangles, including the incentral triangle, can be described by properties of the complete quadrilateral. In particular the extraversion properties of the strong centers are so explained. The following picture shows this triangle and its Cevian quadrilateral.

For information about properties of the complete quadrilateral, see here for the description from the forthcoming The Triangle Book. See here for an interactive java based complete quadrilateral and its properties. See here for a very thorough and very excellent presentation by Jean-Pierre Ehrman. Paul Yiu has also written a nice paper which uses the complete quadrilateral to generalize the Apollonian circles:

Generalized Apollonian circles,  Journal for Geometry and Graphics, 8 (2004) 225--230

Another important structural feature is that, algebraically, the point coordinates use square roots of the edges of ABC, which are themselves square roots. Hence these centers will not relate well to most triangle points, which are algebraically more simple. Indeed ETC lists only the incenter and centroid of the incentral triangle. This triangle is in fact a great example of the way abstract algebraic properties determine geometric ones.

Notation: Q_P indicates the Q point taken with respect to the Cevian triangle of P. Hence H_Iois the orthocenter of the incentral triangle.

edgelengths

The edgelength can be found by using the distance formula or by the law of cosines.

(1)

where Λ = 5 a b c + 8 s_abc
This can be rewritten as

(2)

where =  is a shape parameter which varies from 0 (for flat triangles) to 1/8 (equilateral).

Important functions of edgelengths

a_Io² is the square of the a-edge of the incentral triangle.

a_Io² + b_Io² + c_Io²

This may be unnecessarily complicated because I have not yet figured out what cocktail of symmetric functions works with this function.

a_Io² – b_Io² + c_Io²

b_Io² (a_Io² – b_Io² + c_Io²)

(a_Io² + b_Io² – c_Io²)(–a_Io² + b_Io² + c_Io²)

Once we know these formulae, we can use   : b(f_c c/(a+b) + f_a  a/(b+c))  : to determine the coordinates with respect to ABC. Here (f_a, f_b, f_c) are the coordinates of a point defined wrt the incentral triangle written in terms of a, b, c, the edgelengths of ABC.  

centers of the incentral triangle

the orthocenters of the incentral triangles

The orthocenter of the incentral triangle and its extraversions are on the Brocard meridian line at its intersection with the four parallels to the Euler line through the incenters, a fact noticed by Wilson Strothers.

ETC has this about this point.

X(500) = ORTHOCENTER OF THE INCENTRAL TRIANGLE
Trilinears      f(a,b,c) = a(b² +c² - a² + bc)[2abc + (b + c)(a² - (b - c)²)]
X(500) lies on these lines: 1,30   3,6   651,943
X(500) = inverse-in-Brocard-circle of X(582)
X(500) = crosspoint of X(1) and X(35)
X(500) = crosssum of X(1) and X(79)

in barycentrics, written symmetrically, I simplify the coordinates to

An even better way to write its b coordinate is

using  λ = as a shape parameter which varies from 0 (flat triangles) to 1/8 (equilateral ones).

which is the definitive way of writing the coordinates of this point and clearly shows that it is on the OK line. To determine the position of this point, we write

where O and K stand for the vectors that represent these points and Ω is the Brocard angle. The position can then be established in relation to O and K, being near O for flat triangles and near K for equilateral ones. H_Io is always between O and K.

Wilson Strothers added an essential piece of this puzzle

Your worries about the orthocentres are unfounded. The four ORTHOcentres are weak, and do lie on OK.  Each is OK intersected with the parallel to the Euler Line of ABC passing through the relevant Ix.
A nice example of four weak points on a strong line!

Nice indeed! Now all I have to do is figure out why this is true.
And I have!

If P is a point and Pa, Pb, Pc its harmonic associates (often called the ex-P points), then the Cevian triangles of these four points involve only 4 lines, the edges of the Cevian triangle of P and the tripolar of P, as shown for the incentral triangle above.

Steiner showed that the orthocenters of a complete quadrilateral are collinear and perpendicular to the Newton line of the quadrilateral.

So consider the complete quadrilateral formed by the edges of the Cevian triangle of I_o and its tripolar, the antiorthic axis. These form four triangles that are also the Cevian triangles of the 4 incenters.

We know that the four orthocenters of a complete quadrilateral are collinear, which in this case becomes the statement that the incentral orthocenter and its extraversions are collinear.

The diagonal triangle of this quadrilateral is ABC. Jean-Pierre showed in his FG paper on the Steiner quadrilateral that the circumcircle of the diagonal triangle is on the orthocenter line.

Both Paul and I have studied this system. The Newton line is the Lemoine latitude line. This and the above proves that the orthocenter line is the Brocard axis. Paul's paper on this is

Generalized Apollonian circles,  Journal for Geometry and Graphics, 8 (2004) 225--230

Here is a picture. Note: H_Io indicates the orthocenter of the Cevian triangle of I_o.

Figure: Three of the four orthocenters are shown in red on the Brocard meridian line. Line parallel to the Euler line of ABC intersect the Brocard line at the four orthocenters. The Cevian and ex-Cevian triangles of I_o, which overlap, are shown in colors.

the circumcenters of the incentral triangles

The circumcenter of the incentral triangle can be computed using the edgelength formulas above. Here we are using the shorthand = 5 abc + 8 s_abc - 4b S_B to both simplify the formula and display essential behavior (which is that the ratio is very important for centers of the incentral triangle.

The four circumcircles of the triangles formed by the edges of a complete quadrilateral concur at the Miquel point of the quadrilateral. The circumcenters are cyclic, the circle being called the Miquel circle of the quadrilateral, which includes the Miquel point (which is mS, the center of the Kiepert hyperbola, a strong point — see below). This tells us the the circumcenter of the incentral triangle and its extraversions lie on a circle.

A tale of strength and weakness: usually when a weak line intersects a strong circle, both of the intersections are weak points. If one is a strong point, the other cannot be, since only one strong point can lie on a weak line. Similarly the four circumcircles intersect the Miquel circle at mS, and must intersect it again at a weak point of which there are 4 versions.

Figure: The Miquel circle goes through the circumcenter of the incentral triangle and its extraversions. The circumcircles concur at the Miquel point.  

Wilson gives more structure, this time for the and its extraversions.

Wilson Strothers notes that (from Cabrii pictures) the Miquel point is the center of the Kiepert hyperbola mS, which is a strong point on the 9 point circle. I take this to be true, as well as his following comments.

Let Cx be the circumcircle of Tx, the cevian triangle of Ix. Cx meets the Nine-point circle at M = mS, the medial Steiner point = center of Kiepert hyperbola (X115), and at one other point Fx, the Feuerbach point. Then the lines IxFx concur at N  (X5). Also, let Nx denote the nine-point center of Tx. Then the lines IxNx concur at O (X3). We can add to this the previous result, let Hx = orthocenter of Tx. Then the lines IxHx concur at the Euler infinity point (X30}.

Figure: In this picture the incircles of ABC are light red. The 9-circle is light blue. The circumcircles of the incentral triangle and its extraversions are black lines. The Feuerbach points are small and green. The small red points are the 9-centers of the incentral triangle and its extraversions. This picture shows the I_x-N_x lines go through O and that the circumcircles of the incentral triangles go through the Feuerbach points.

The circumcircle of the Cevian triangle of I_o

If we assemble Wilson Strother's information above, we see that the circumcircle of the incentral triangle is an unusual circle. It intersects the 9 point circle, a strong circle, twice, once at mS, a strong point, and once at the Feuerbach point F_o, a weak point. This means that all four of these circles intersect at mS and go through the corresponding Feuerbach point, which is on the 9-circle.

The pedal circle of G also intersects the 9-circle at mS, the center of the Kiepert hyperbola, and also at J, the center of the Jerabek hyperbola.

the incenter of the incentral triangle

You can almost figure out this answer in your head.

These coordinates seem to make it difficult to study this point; however they are the barycentric form of a supplementaire of a point having the b-trilinear .

But we do know about the incenters of a complete quadrilateral. Each subtriangle of a quadrilateral has 4 incircles, counting extraversions, so that the complete complete quadrilateral system has 16. Studies of the extraversion of these types of coordinates shows that the incenter has 16 extraversions. These two systems are the same, so we are led to the amazing result that the 16 extraversions of the incenter of the incentral triangle lie, four at a time, on 8 circles which form a coaxal system. Very neat, particularly when I had thought that the coordinates prevented us from knowing anything at all about these points.

Figure: 14  of the 16 incenters are shown (the other two being above the picture). There are four incenters per circle and each incenter is on two circles, one from each family. The incircles are colored so that all four originating from the same triangle share the same color. The only other points shown in this picture are centers of 6 of the 8 circles, all on the perpendicular axes of families of circles.  

The 16 incircles of the incentral triangle are also 4 at a time on 12 lines, 3 lines per point. This is shown by the following picture. This is the dual of a desmic system which has 16 lines through 12 points. So I guess that this must be a projective octahedron to match the projective cube that desmic systems are, but I can't figure out the picture.

Figure: The 16 incenters lie four at a time on 12 lines, each point being on 3 lines. The 4 in/excircles from a given triangle are the same color.

This combination of extraversion with a complete quadrilateral may be uniquely powerful, but it may be more general. Points built from coordinates of half angle sine or cosine functions will have 16 extraversions. Could they form coaxal families as above. An interesting thought.

centroid

This is X(1962) = G_Io

b-centroid

b-extraversion

relations among these points

gi = centroid of the incentral triangle
bxg = b extraversion
bg = b exversion

Not a big deal but these points show some second order relations to incenter and X37.

This point is on a central bisector.

This point is on the b-Cevian of X(37).

This point is the b-Cevian trace of I_o.

This next point is on the b-Cevian of I_o.

This appears to give a desmic system involving I_o and S_o, with harmon and desmon : c + a ± 2b :

isotomic H

for which I can find no interesting properties.

symmedian

Let = 5 a b c+8 s _abc– 4 b S _B. The the symmedian point of the incentral triangle is

Bonus: the Apollonian circles, the isodynamic points, and K

The complete quadrilateral formed by the sides incentral triangle and the antiorthic axis (tripolar of I_o) has diagonals, the diagonal circles of which are the Apollonian circles whose centers are on the Newton line of the quadrilateral, which for this quadrilateral is the Lemoine latitude line. The Lemoine line is the tripolar of K, and the centers of the Apollonian circles are related to K. The intersections of the Apollonian circles are the isodynamic points.

I have always pondered the mystery of the incenter. One assumes that more symmetry is an indication of more importance. While the centroid is a point of high symmetry, the incenter supports no particular symmetry yet it is the source of almost all the quartile points we study. Quartile nature is a result of symmetry, which is part of the answer, but perhaps the picture below is also part of the answer. The Apollonian circles are inversively invariant. Their intersection, the isodynamic points, are inversively invariant, the only points in the triangle that are. So perhaps the amazing structure caused by the Cevian quadrangle formed by the incenter is a part of the anwer.

Figure: This picture shows that the Apollonian circles are the diagonal circles of the incentral quadrilateral for which ABC is the diagonal triangle. The diagonals are the segments of each edge intercepted by the bisectors from the corresponding vertex. The circles described on these segments, with centers at the midpoints between the two points are the Apollonian circles. The Newton line is the Lemoine latitude line, and the axis is the Brocard meridian line.

Created by Mathematica  (October 16, 2005)