Ex-extra Points

Introduction

Each point P in the triangle plane has 3 harmonic associates, which are the vertices of its pre-Cevian triangle, notated as A^P, B^P, C^P, ofen called the ex-P points. The relationship between these four points is projectively invariant. Many points, particularly if they are constructed using the incenter or incircle, have 3 extra-versions. The extraversions come from a symmetry, discovered by John Conway and independently by Georgey Kapetis, in the manner of construction of the point. The original version of a point P is notated P_o, and the extraversions as P_a, P_b, and P_c. The surprise is that the 3 ex-P points and the three extraversions are often in perspective, as shown in the figure using P = G_o the Gergonne point. Since the perspector comes from two forms of G's we call it the GG_o point, the "o" indicating that it is the "original" member of a set of 4.

Figure: This shows the construction of GG_o from the extraGergonne points G_a, etc., and the ex-Gergonne points A^Go, etc.

This topic was a wonderful ride: the discovery of a new type of point and its construction; the clarification of what points this constuction works for; and finally the amazing connection to cubic curves. There was the mysterious discovery that the ex-extra operation is equivalent to the Cevain quotient operation, a purely projective operation. It was also a collaborative effort of the Swarthmore and Hyacinthos groups with Barry Wolk and John Conway making significant contributions.

The next surprise is that the traces of the perspectrices on the sides are themselves in perspective with ABC and that this perspector is the pivot of a cubic on which the P and its desmic mates, if they exist, reside. Here is the picture that shows this perspector for the Gergonne points. Perhaps even more surprising is that this operation is inversive, and forms the second generator (with conjugation) of points following the group law on the cubic.

Figure: the traces of the GGo perspectrices are themselves in perspective to ABC. This perspector is the pivot of the cubic on which Gx (and GGx) reside.

Exposition

The extra-Nagel points are N_a N_b N_c and the theorem is that this is in perspective with A^No, B^No, C^No. We call the perspector the "Nagel-Nagel point" NN_o, and use similar terminology in other cases.

John Conway was the first to find the coordinates of one of the points.

[John Conway] Here's my short proof. We have

      N_o = ( s_a : s_b : s_c)  and so

    B^No = ( s_a : –s_b : s_c ),  and by extraversion

      N_b = ( s_c : –s_o : s_a ).

Now the sum of these two vectors is

 ( s_o–s_b : –s_o–s_b : s_o–s_b ) ~ ( 1/s_b – 1/s_o : –1/s_b – 1/s_o : 1/s_b – 1/s_o )

and their difference is

 ( s_a–s_c : s_a+s_c : s_c–s_a ) ~ ( 1/s_c–1/s_a : 1/s_c+1/s_a : 1/s_a–1/s_c )

and the sum of the two second vectors here is the desired answer.

Nag-nag: NN_o = ( : 1/s_a – 1/s_b + 1/s_c – 1/s_o : )

[JC] I couldn't find a similarly simple proof for  GG_o. However, I did find something nice, namely that  NN_a, NN_b, NN_a, NN_o  forms one tetrad of a desmic triple with  A,B,C,D , the third tetrad being formed from the isoNagNags; and a similar result holds for the GerGers and their isotomic conjugates.

Figure: This shows the desmic relationship between the NagNag points and their isotomic conjugates.

More ex-extra perspectors

For many points the precivian triangle of a point is in perspective to the triangle made from its extraversions. Here I list some of these points.

We noted that, for the Nagel point, the precevian triangle A^No B^No C^No and the Nagel triangle N_aN_bN_c are in perspective. So too for the Gergonne point.

I have tested this for many points. For most, but not all, the corresponding triangles are in perspective and its Ex-Extra point exits.

By all rights these perspectors should not be there; I comment on this at the end.

Chart of perspectors

Point                    Perspector for triangles A^PB^PC^P and P_aP_bP_c
_____                    ___________________________________________

Nagel N_o = : s_ac :            Nag-nag: NN_o = ( : 1/s_a – 1/s_b + 1/s_c – 1/s_o : )
 (This is John's rather silly name, which I rather like).
                   
Gergonne  G_o = : s_ac :       Ger-ger = more complicated, similar to the above
Mittens   M_o = :b s_b:             incenter I_o
Mittenmates gM_o = :b s_ac:     : -2 s_abc + b S_B :
proNagel  pG_o = :b² s_ac:     proNag-nag pNN_o = : b² (1/s_a – 1/s_b + 1/s_c - 1/s_o) :)
Feuerbach F_o = : (a+c)² s_ob :  :(c²–a²)(-abc s_abc+2(ab+bc+ac)s_oabc-4ac s_oabc+abc b S_B):
Feuerbach mate E_o= :(a-c)² s_ac:    :(a+c)² s_ac (10th degree polynomial) :
isogonal Nagel gN_o = : b²/s_b :    : b² s_ca (–S_ABC–2abc(2s_cba+abc)+4(ab+bc-ac) s_oabc+2abc b S_B):
Shifflers  H_o = :b s_b/(a + c):        does not exist, presumably because the H_x are colinear.
Shiffler mates = :b /(c –a) s_b:    : b² S_B/ (s_b (c-a)) : = proShiffler mate
X37     : b (c + a):             proSpieker = pSo = :b²(c+a):
Spieker  = S_o  = : c+a:            : a c (a + c) :  = retroX37
iso and coSpiekers              does not exist

Recapping, let P = P_o be any of the above points. We ask whether the triangle of extra-P points (P_a, P_b, P_c) is centrally perspective to the triangle of ex-P points (A^Po, B^Po, C^Po).

Barry Wolk's work

Barry added much creativity to this investigation and supplied many proofs.

[from Barry Wolk] Here are the results of a lot of computer calculation. YES means this perspectivity is true, and I give the coordinates of its center, whenever I was able to reduce those coords to something simple. NO means this perspectivity does not hold.

 (c+a)²/sb    YES
 (c-a)² sb Feuerbach   YES
 b/(c-a)    YES (: b²/(c-a) :)
 b/(c+a)    NO -- Pa,Pb,Pc are collinear
 b(c-a)    NO -- Pa,Pb,Pc are collinear
 b(c+a)    YES (: b²(c+a) :)
 b²(c+a)    YES (: b(c+a) :)
 1/b    trivial
 b²(c+a)²    YES
 b²(c-a)²    YES
 1/b(c-a) Steinerian    YES (: 1/(c-a) :)
 1/b(c+a)    NO -- Pa,Pb,Pc are collinear
 b²/(c-a)    YES (: b/(c-a) :)
 b²/(c+a) coSpieker    NO -- Pa,Pb,Pc are collinear
 1/(c+a)²    YES
 1/(c-a)²    YES
 c-a    NO -- Pa,Pb,Pc are collinear
 c+a Spieker    YES (: (c+a)/b :)
 (c+a)/b    YES (: c+a :)
 b incenter    trivial
 sb Nagel  YES (: aSA-bSB+cSC-abc :)
 (c+a)²    YES
 (c-a)²    YES
 1/(c-a)    YES (: 1/b(c-a) :)
 1/(c+a) isoSpieker    NO -- Pa,Pb,Pc are collinear
 b sb Mitten    YES (: b :)
 1/sb Gergonne    YES (: (bc/SA-ca/SB+ab/SC-1)/sb :)
 b/sb coMitten    YES
 b sb/(c+a)  Schiffler   NO -- Pa,Pb,Pc are collinear

Since January I have found some generalizations, thus explaining a few of these many unexpected concurrences. The proper setting for this generalization is a quadrangle PoPaPbPc, where these points may not have anything to do with extraversion. However, the order of the four points is significant. Recall that if P = (x:y:z), then the ex-versions (or pre-Cevians or harmonics) of P are A^P = (–x:y:z), B^P = (x:–y:z), C^P = (x:y:–z). Write X—Y for the line XY.

Definition:
The quadrangle Q_oQ_aQ_bQ_c is the ex-mate of quadrangle P_oP_aP_bP_c if (1)  the 3 lines P_a--A^Po, P_b--B^Po, P_c--C^Po are concurrent at Q_o
(2)  the 3 lines P_o--A^Pa, P_c--B^Pa, P_b--C^Pa are concurrent at Q_a
(3)  the 3 lines P_c--A^Pb, P_o--B^Pb, P_a--C^Pb are concurrent at Q_b
and (4) the 3 lines P_b--A^Pc, P_a--B^Pc, P_o--C^Pc are concurrent at Q_c

Note: In the original case where P_aP_bP_c are the extraversions of P_o, the concurrences (2) (3) (4) all follow from the concurrence (1), by applying the three extraversion operations.

Steve's first example used P_o = the Nagel point, so P_aP_bP_c were the extra-Nagel points, and A^Po, B^Po, C^Po are the ex-Nagel points. And it was the perspectivity of these two triangles that led to all this stuff.

My first rather easy result was: If Q_oQ_aQ_bQ_c is the ex-mate ofP_oP_aP_bP_c, then P_oP_aP_bP_c is the ex-mate of Q_oQ_aQ_bQ_c . (Sketch of proof: Show that the four points P_o, Q_b, B^Qo, B^Pb are collinear [as shown in the figure for the case where P_o = N_o, the Nagel point].  This justifies using the term "mate" for this property.

Now, why do these concurrences exist so often? There are two interesting cases.

Theorem 1: If the quadrangle P_oP_aP_bP_c is desmic, then it has an ex-mate, and this ex-mate is also desmic.

Theorem 2: If the quadrangle P_oP_aP_bP_c is skew-desmic, with skew-desmic mate P'_oP'_aP'_bP'_c , then the following are equivalent:
   (1) P_oP_aP_bP_c has an ex-mate
   (2) The four points P'_o P'_a P'_b P'_c are collinear

Notes to theorem 1, the desmic case:
1.1   Recall that desmic means there are coordinates such that P_o = (u:v:w), P_a = (u:V:W), P_b = (U:v:W), P_c = (U:V:w)
and the desmic mate of P_oP_aP_bP_c is
       P'_o = (U:V:W), P'_a=(U:v:w), P'_b=(u:V:w), P'_c=(u:v:W)

1.2   Let P_oP_aP_bP_c be desmic, with desmic mate P'_oP'_aP'_bP'_c. Then the ex-mate of P_oP_aP_bP_c exists and is desmic, and the same for the ex-mate of P'_oP'_aP'_bP'_c, but these two ex-mates are not desmic mates of each other.

Notes to theorem 2, the skew-desmic case:
2.1   Recall that skew-desmic means there are coordinates such that
       P_o = (u:v:w), P_a=(u:V:-W), P_b=(-U:v:W), P_c=(U:-V:w)
and the skew-desmic mate of P_oP_aP_bP_c is
       P'_o = (U:V:W), P'_a=(U:v:-w), P'_b=(-u:V:w), P'_c=(u:-v:W)

2.2   Let PoPaPbPc be skew-desmic, with skew-desmic mate P'oP'aP'bP'c. If the ex-mate QoQaQbQc of PoPaPbPc exists, then QoQaQbQc is also skew-desmic. And it follows from these results that the skew-desmic mate of QoQaQbQc will then consist of four collinear points.

2.3   As vectors, P'a + P'b + P'c = P'o. So if any three of these four points are collinear, then all four are collinear (and then PoPaPbPc has an ex-mate). This condition is just
     UVW+Uvw+uVw+uvW=0
or equivalently
     (u-U)(v-V)(w-W) = (u+U)(v+V)(w+W)
which is a form that is quite useful in doing calculations.

2.4  If the product uvwUVW is not equal to zero, then each of the four concurrences needed for the existence of an ex-mate reduces to this same equation given in 2.3. The degenerate case where uvwUVW=0 might be interesting, but I haven't explored it at all.

The "friendly points" that Steve and I investigated create quadrangles by extraversion, and each one is either desmic or skew-desmic. The desmic case is basically settled by Theorem 1. In the skew-desmic case, the main questions remaining are:

Why does this collinearity (of P'o P'a P'b P'c ) hold so often? When this collinearity holds, so the ex-mate of PoPaPbPc exists, why are the coordinates of this ex-mate generally so simple?

Partial answer to first question: The collinearity seems to be true for many small values of the exponents involved, but not for larger values. And these small exponents correspond to the interesting points, the ones that have been given names.

Harmonic equivalent desmic triples and cubics

Notation: A^P is the a ex-P, a precevian vertex of P in ABC.

I have been making a chart of known desmic triples. In doing so I noticed that every desmic triple with desmon D and harmon H, has a switched triple with desmon H and harmon D.

Here is the usual configuration with desmon (u+U:v+V:w+W) and harmon (u-U:v-V:w-W)

(u+U:v+V:w+W)	A	B	C
(U:V:W)	(U:v:w)	(u:V:w)	(u:v:W)
(u:v:w)	(u:V:W)	(U:v:W)	(U:V:w)
(u-U:v-V:w-W)	(U-u:v-V:w-W)	(u-U:V-v:w-W)	(u-U:v-V:W-w)

where the blue points are the perspectors and are colinear. The red points is the harmonic quadrangle around the Harmon. The yellow oints are three triangles in mutual perspective at the perspectors.

We will name these as follows

       

D	A	B	C
P	P1	P2	P3
Q	Q1	Q2	Q3
H	H1	H2	H3

Since    A A^P1 , B B^P2 , C C^P3 have the same concurrence as AP1, BP2, CP2, we can replace Px and Qx with their corresponding harmonics.

In this case the chart looks like

(u-U:v-V:w-W) = H	A	B	C
(-U:-V:-W) = P	(-U:v:w) = AP1	(u:-V:w) = BP2	(u:v:-W) = CP3
(u:v:w) = Q	(u:-V:-W) = AQ1	(-U:v:-W) = BQ2	(-U:-V:w) = CQ3
(u+U:v+V:w+W) = D	(-u-U:v+V:w+W) = AD	(u+U:-v-V:w+W) = BD	(u+U:v+V:-w-W) = CD

This desmic triple has desmon H and harmon D and the roles of Desmon and Harmon are reversed. The two perpectors are the same. Each desmic triple has an associated cubic.

Here's how to construct this new desmic triple.

First note that the points B = (0:1:0), (U:-v:W), B_P = (U:0:W), (U:v:W) are harmonically conjugate pairs.



                  
                                                             
                                                          

The triangles made from the extraversions of a quartile point P_o is often in perspective to its precevian triangle A^Po, B^Po, C^Po. In TB these will be called the harmonic points of P_o. We call this perspector the ex-extra perspector and notate it as PP_o.

Barry Wolk has conjectured that if the quadrangle P_x (x = o,a,b,c) has a desmic mate, then the ex-extra perspector will exist. If it does exist, it will be quartile. Barry calls this quadrangle the ex-mate.

A nice example comes when P_o = N_o, the Nagel point. In this case the ex-extra perspetor, which John calls the Nag-Nag point is given by a very simple formula.

                 NN_o = : –2s_abc + b S_B :

so that this point lies on the G—W_o line. This line will occur many times in this post.

Here I am using John's shorthand notation s_abc = s_a s_b s_c, where s_b = s_o – b = (c+a–b)/2, and S_B is the analogous strong fromula S_B = (c²+a²– b²)/2.

A very nice thing now happens: the NN_x quadrangle is desmic with its isotomic conjugates (John pointed this out), its desmon being :S_B:, currently called the Desmon point. Further NN_o itself has an ex-extra perspector which cycles back to the Nagel point

         NNN_o = :s_b: = N_o

I believe that Barry was the one who pointed out that this would happen.

The Gergonne situation is almost as nice

        G_o = : s_ca :     GG_o = :(bc S_BC–ca S_CA + ab S_AB –S_ABC)/s_b:
        GGG_o = :s_ca: =  = G_o

The Mitten mates cycle back again as well

         gM_o = : b/s_b :    gMgM_o = b(–2s_abc+bS_B)/s_b
          gMgMgM_o = :b/s_b: = gM_o

Some other patterns work out nicely (here n can be positive or negative)

    : bⁿ s_ob :   -->   bⁿ (2s_abc – bS_B) if n even;  --> : bⁿ s_ob :
                -->   bⁿ if n odd

For n = odd, the extra-versions and the ex-versions are the same so the chain stops there.

I am finding this stuff interesting and fun.


Desmic quadrangles of other than quartile points.

Cubics organize points into sets of 4. The group law tells us which sets of points are desmicly related. The group laws of three important cubics are reproduced below. Notice that the rows alternate between strong and weak points. It is the strong points that interest me now.

Looking at the first cubic, we find three strong points H,G,O that are colinear. Combined with the other points in their rows we have a desmic triple of quadrangles with perspectors H and O and desmon G.

Now what use can I make of one of these desmic quadrangles? Well, let's test our theory that a desmic quadrangle has an ex-extra perspector. We do this as follows. I will use the L A B C quadrangle from the Darbeau cubic. The last three members of the row take the place of the extraversions, while the harmonics of the perspector, the points A^L, B^L, C^L are, as before, the "ex" triangle. So we ask whether the two triangles

            A    B    C
            A^L  B^L  C^L
           
are in perspective. Of course they are with perspector L. Most of these strong "ex-extra" perspectors exist either because, as above, the triangles are trivially in perspective or because any precevian triangle is perspective to any cevian triangle.

Following Barry Wolk, I use the term "ex-mate" to represent the quadrangle formed from the ex-extra operation. If PoPaPbPc is a quandrangle then its main ex-extra perspector is the concurrence of the lines Pa--A^Po, Pb--B^Po, Pc--C^Po, where A^Po is Conway's notation for the a-precevian vertex of Po.

Barry has presented a theorem that if the Px (x = o,a,b,c) are a desmic quadrangle, then the ex-extra perspector exists.

I want to show how far this idea goes by applying it to cubics.

All 12 points of a desmic quadrangle lie on a cubic. Barry has given us the equation of the cubic. Thought of the other way, each cubic determines many desmic quadrangles. So let us look at a cubic equation. Here I choose the Lucas cubic which has pivot dK, the symmedian point of the antimedial triangle.

Here is its group table.

       cubic for which cevian of Q = pedal of P
                    isotomic
       
            o    a    b    c
           -------------------
    -V     GG_o GG_a GG_b GG_c
    -IV    L    *    *    *
    -III   NN_o  NN_a  NN_b  NN_c
    -II    dK   A    B    C
    -I     N_o  N_a  N_b  N_c
    0      G    dA   dB   dC
    I      G_o G_a G_b G_c
    II     H    A_dK  B_dK  C_dK
    III    tNN_o  tNN_a  tNN_b  tNN_c
    IV     tL    t*  etc
    V      tGG_o tGG_a tGG_b tGG_c

        pivot      dK
        center     G
        constant   dK

Here N_x are the Nagel pts, G_x the Gergonne pts, dA a vertex of the antimedial triangle, A_dK is the cevian foot of dK and NN_x is the ex-extra perspector of Nx.

Every row in the group table forms a quadrangle whose ex-extra perspector IS ON THE CUBIC!

This seems to me a big deal and both tells us a lot about cubics and the ex-extra operation.

Following Barry Wolk, I use the term "ex-mate" to represent the quadrangle formed from the ex-extra operation. If PoPaPbPc is a quandrangle then its main ex-extra perspector is the concurrence of the lines Pa--A^Po, Pb--B^Po, Pc--C^Po, where A^Po is Conway's notation for the a-precevian vertex of Po.

Barry Wolk's theorem is that if a quadrangle is "desmic," it has an ex-extra perspector. I will comment on this below.

Barry has presented a theorem that if the Px (x = o,a,b,c) are a desmic quadrangle, then the ex-extra perspector exists.

I want to show how far this idea goes by applying it to cubics.

All 12 points of a desmic quadrangle lie on a cubic. Barry has given us the equation of the cubic. Thought of the other way, each cubic determines many desmic quadrangles. So let us look at the group table of a cubic equation. Here I choose the isogonal cubic which has pivot G, the centroid.
______________

  precevian of R = pedal of P      
        isogonal                 
                                                            
        o   a   b   c       
       -------------       
-V                           
-IV                          
-III   sM'M'o  :sM'M'b:      
-II    H   ***               
-I     M'o M'a M'b M'c       
0      K   AG  BG  CG        
I      Io  Ia  Ib  Ic        
II     G   A   B   C         
III    Mo  Ma  Mb  Mc        
IV     O   H+A H+B H+C       
V      M'M'o :M'M'b:        

pivot      G                    
center     K       
constant   O   

Notation:  Mo = Mittenpunkt, M'o = mate of Mo = isogonal conjugate of Mo, AG = A_G = a-trace of G , H+A = midpt of HA.

*** - the points in this row are the third meets of lines AO, BO, CO with the cubic.

The I row are the incenters, which are self conjugate. Corresponding rows above and below row I are isogonally conjugate.

-------

Now the new results: the ex-extra operation, if defined for a row, is an operation on this cubic, taking a particular row to another row.

For example row 2 goes to itself. The lines Pa--A^Po = A--A^G concur at G.
Another example: row III --  row I since the ex-extra of Mo = Io.

Here is a table showing what rows go to what rows followed by the period of the operation

-III   ?     
-II    --  : (bbSB - SAC)/SB:  do not yet know what row this is             
-I     --  V       period 2
0      --  IV      period 2  
I      not defined since A^Io = Ia       
II     --  II      period 1  
III    --  I       not defined
IV     --  0       period 2
V      --  -I      period 2

I have done this for 4 cubics; the ex-extra operation gives a new point on the cubic. I noticed this as I was computing group tables for most cubics.

[Barry Wolk] On June 20, in message 1049 I said:
Theorem 1: If the quadrangle PoPaPbPc is desmic, then it has an ex-mate, and this ex-mate is also desmic.

The ex-mate of a desmic quad actually does lie on that cubic.

The general desmic structure contains 12 points and 16 lines, and contains three related quadrangles. However we are only looking at the special case where one of these three quadrangles is of the form (D,A,B,C), where A,B,C is the triangle of reference, and D is the desmon. In this case each of the other two quadrangles is called a desmic quadrangle. Each desmic quadrangle determines an entire desmic structure, which then determines a cubic curve that passes through all twelve of its points. That curve will be called the cubic determined by the desmic quadrangle.

Theorem 1.4 revised: The ex-mate of a desmic quadrangle always lies on the cubic determined by that quadrangle.

Proof: Use the formulas I gave in message 1072 to calculate the ex-mate of a desmic quadrangle, and plug them into the equation of the cubic.

This is getting more and more interesting. It proves that, in the group tables for cubic curves that Steve has been posting, the ex-mate of each row will be another row of the same table. I now suspect that, for the special case of a desmic quadrangle, this ex-mate construction is a generalization of some other construction that is already known for cubic curves. It would be nice to remove the requirement that the cubic passes through A,B and C.

After writing all the above, I just saw Steve's latest messages. He wrote: Now the new results: the ex-extra operation, if defined for a row, is an operation on this cubic, taking a particular row to another row.
[snip] I have done this for 4 cubics; the ex-extra operation gives a new point on the cubic. I noticed this as I was computing group tables for most cubics. I will be working on a proof while I am hiking.

Barry again about cubics

Steve, I can prove it. First, if four points on a cubic form one of the three quadrangles of a desmic structure, and if all 12 points of that structure lie on the cubic, then those first four points are all in the same row of the group table of the cubic.

Partial proof: Use k for the constant of the cubic. Then
Io + IIo + IIIo = k
Io + IIa + IIIa = k
Ia + IIo + IIIa = k
Ia + IIa + IIIo = k
Adding and cancelling, Io + Io = Ia + Ia. This means that Io and Ia are in the same row of the group table.

Combine this with my calculations for arbitrary desmic quads and their ex-mates. This calculation shows that the ex-mate is desmic and (after I redid that calc correctly) also lies on the cubic.

Also, almost every row of the group table is such a desmic quad.
Just use the D A B C row as the second quad, and find the third quad.

   The rows of these group tables are all things of the form

    P, P+x, P+y, P+z

where  x,y,z  are the three elements of order two in the group. Also, three points  P,Q,R  of the cubic are in line just if

    P+Q+R  is the particular element I like to call "line",

and it then follows that their three rows form a desmic system (provided they're distinct).  In particular, the "desmic mate" of row  P  is always another row, namely that of 

    line - P.

   I'm very intrigued indeed by Barry's result that the ex-mate of a row is always another row. What is the relation between the corresponding group elements?  The fact that the ex-mate relation is symmetric suggests that it has the form

      P + Q  might be a certain constant, which I'll call "wolk".

If this is the case, then   P + Q + (line - wolk) = line,  so that the line  PQ  passes through a fixed point 

    R = line - wolk

on the cubic.  Is this in fact the case, and if so, what is this fixed point  R?  Is it isogonally symmetric to  D,  the fourth point of the row  A B C D ?


(BW)  Write X--Y for the line XY.
 
  Definition:
  The quadrangle QoQaQbQc is the ex-mate of quadrangle PoPaPbPc if
  (1)  the 3 lines Pa--A^Po, Pb--B^Po, Pc--C^Po are concurrent at Qo
  (2)  the 3 lines Po--A^Pa, Pc--B^Pa, Pb--C^Pa are concurrent at Qa
  (3)  the 3 lines Pc--A^Pb, Po--B^Pb, Pa--C^Pb are concurrent at Qb
  and (4) the 3 lines Pb--A^Pc, Pa--B^Pc, Po--C^Pc are concurrent at Qc
 
This definition is a generalization of the existence of the  extra-ex-perspectors, and reduces precisely to that case when the quadrangle PoPaPbPc is obtained from Po by the extraversion operators.

(John Conway)     I'm very intrigued indeed by Barry's result that the ex-mate of a row is always another row.  What is the relation between the corresponding group elements?  The fact that the ex-mate relation is symmetric suggests that it has the form
 
        P + Q  might be a certain constant, which I'll call "wolk".

The problem is that the general definition of ex-mate involves points  such as A^Po which do not lie on the cubic curve that is associated  with the desmic case. So why should it be related to the group of that  cubic? However, I have now worked this out, and your last statement is  correct. The constant is related to twice the row number of the  A B C D line. The formula is Po + Qo = D + D. So any row PoPaPbPc of  the group table, and its ex-mate row QoQaQbQc, are equidistant from the D A B C row.

     (And, like all rows of the group table, we know
     Po+Qo=Pa+Qa=Pb+Qb=Pc+Qc,
     Pa=Po+A-D, Pb=Po+B-D, Pc=Po+C-D
     A+A=B+B=C+C=D+D)

Proof: In the general case, we have
(1)  the 3 lines Pa--A^Po, Pb--B^Po, Pc--C^Po are concurrent at Qo plus three other conditions. In the desmic case, Po = (u:v:w),  Pa = (u:V:W), Pb = (U:v:W), Pc = (U:V:w), and A^Po = (-u:v:w), etc.  The point (0:v+V:w+W) lies on the Pa--A^Po line (and also on the  lines Po--A^Pa, Pc--A^Pb, Pb--A^Pc). And this new point is the A-trace (or A-cevian) of the desmon D = (u+U:v+V:w+W).

Now D and its three traces AD, BD, CD all lie on the cubic. So  Qo can be described as the point of concurrence of the lines   Pa--AD, Pb--BD, Pc--CD, and this construction of Qo involves only points that are on the cubic.  When these concurrences defining  QoQaQbQc are described in terms of the group operations, the result  is as stated above, namely Po+Qo = D+D.   End of proof.

Incidentally, Qo is also the point of concurrence of the lines  AD--A^Po, BD--B^Po, CD--C^Po. And this description is exactly that  of the Cevian quotient of D and Po. Similarly, Qx is the Cevian  quotient of D and Px, for x=o,a,b,c. That is an amusing connection  with another topic that showed up here on Hyacinthos.

I can now restate some formulas I gave for the ex-mate. Quoting  from message 1072:
 Case 1: Take Po Pa Pb Pc desmic, so
       Po=(u:v:w),    Pa=(u:V:W),    Pb=(U:v:W),    Pc=(U:V:w)
 
  Let x,y,z,t be the polynomial solution of the three linear equations
       x(-u,V,W) + y(U,-v,W) + z(U,V,-w) = t(u,v,w)
  Specifically,
       x=u v w + 2U v w - u V W + U v W + U V w
       y=u v w + 2u V w + u V W - U v W + U V w
       z=u v w + 2u v W + u V W + U v W - U V w
       t=-u v w + 2U V W + u V W + U v W + U V w
 
  Then
        Qo=(ux:vy:wz), Qa=(ut:Vz:Wy), Qb=(Uz:vt:Wx), Qc=(Uy:Vx:wt) is the ex-mate of PoPaPbPc. And rewriting these as       (utx:vty:wtz), (utx:Vxz:Wxy), (Uyz:vty:Wxy), (Uyz:Vxz:wtz) makes it obvious that QoQaQbQc is also desmic.

To get another version of this formula for the ex-mate, just scale  the x,y,z,t factors by dividing each of them by (u+U)(v+V)(w+W).  This leads to the new definitions
     x = -u/(u+U) + v/(v+V) + w/(w+W)
     y =  u/(u+U) - v/(v+V) + w/(w+W)
     z =  u/(u+U) + v/(v+V) - w/(w+W)
     t = 2 - u/(u+U) - v/(v+V) - w/(w+W) = 2-x-y-z
which certainly look better than the previous formulas. However,  even in the simplest case of the Nagel points, these formulas give  answers that need to be simplified.

Finally, one more comment. In message 1049 I said:
(BW)  1.3   I tried the sequence (1)=desmic PoPaPbPc, (2)=ex-mate of (1), (3)=desmic mate of (2), (4)=ex-mate of (3), etc, looking for a short cycle, but nothing interesting happened, andl the coordinates eventually got hopelessly complicated.

This is now easily explained in terms of the group table. There is  a constant k such that, for all n, the desmic mate of the ex-mate  of row n is row n+k. So I was just hopping along the group table  in steps of length k, and would never find a cycle.

 (JC)  I'mvery intrigued indeed by Barry's result that the ex-mateof   a row (P) is always another row (Q) [suggests that]             P + Q  might be a certain constant, which I'll call "wolk".

[Barry]:       The formula is Po + Qo = D + D.

 
  Now D and its three traces AD, BD, CD all lie on the cubic. So  Qo can be described as the point of concurrence of the lines   Pa--AD, Pb--BD, Pc--CD, and this construction of Qo involves only  points that are on the cubic.  When these concurrences defining  QoQaQbQc are described in terms of the group operations, the result is as stated above, namely Po+Qo = D+D.   End of proof.

    This is very nice indeed!  So at last we have a "cubic" definition of the "ex-mate".