Re: Two isogonal points

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• Dear Quim, in the general case, if P={u,v,w} (barycentric coordinates), then X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w} Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2,
Message 1 of 13 , Aug 4, 2006
Dear Quim,

in the general case, if P={u,v,w} (barycentric coordinates), then

X={a^2*b^2*u*w^2, b^2*c^2*u^2*v, a^2*c^2*v^2*w}

Y={a^2*c^2*u*v^2, a^2*b^2*v*w^2, b^2*c^2*u^2*w}

--- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer" <qcastell@...>
wrote:
>
> Dear friends,
>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of Q
> intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at two
> isogonal points X, Y. When P, Q are G, K the new points X, Y are the
> Brocard points.
>
> Greetings from Bsrcelona,
>
> Quim Castellsaguer
>
´
• Dear Quim. Let PaPbPc, QaQbQc be the cevian triangles of isogonal points P, Q, wrt ABC respectively, in the general case. We denote the six more intersection
Message 2 of 13 , Aug 5, 2006
Dear Quim.

Let PaPbPc, QaQbQc be the cevian triangles of isogonal points P,
Q, wrt ABC respectively, in the general case. We denote the six more
intersection points of their cevians as follows:

- As D, the intersection point of BQb, CPc.
- As D', the intersection point of BPb, CQc.
- As E, the intersection point of CQc, APa.
- As E', the intersection point of CPc, AQa.
- As F, the intersection point of AQa, BPb.
- As F', the intersection point of APa, BQb.

1 - The lines AD, BE, CF, are concurrent at a point so be it, X.
2 - The lines AD', BE', CF', are concurrent at a point sp be it, Y.
3 - The points X, Y, are isogonal wrt ABC.

Another interesting ( I think ) result is that the lines DD',
EE', FF', XY', are concurrent at point so be it, O.

When the points X, Y, are coincided with the Brocard points
( when P, Q, are G, K, respectively ), then the point O coincides
with the midpoint of their segment.

Some notes for the proof about the concurrency point O, have been
posted in message #13310 ( in more general configuration of two
arbitrary isogonal lines through each vertex of a given triangle ).

Best regards.
Kostas Vittas.

--- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
<qcastell@...> wrote:
>
> Dear friends,
>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of
Q
> intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at
two
> isogonal points X, Y. When P, Q are G, K the new points X, Y are
the
> Brocard points.
>
> Greetings from Bsrcelona,
>
> Quim Castellsaguer
>
• Dear Quim. It is clear that the points D, D , are isogonal with respect to the triangle ABC. Similarly about the pairs of points E, E and F, F . So, we can
Message 3 of 13 , Aug 5, 2006
Dear Quim.

It is clear that the points D, D', are isogonal with respect to
the triangle ABC. Similarly about the pairs of points E, E' and F,
F'. So, we can easy to prove the concurrencies at points X, Y and
that these points are also isogonal wrt ABC.

But how about the proof that the points X, Y, coincided with
the Brocard points, when P, Q, are G, K respectively and that the
point O, is the midpoint of their segment?

Best regards.
Kostas Vittas.

--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
wrote:
>
> Dear Quim.
>
> Let PaPbPc, QaQbQc be the cevian triangles of isogonal points
P,
> Q, wrt ABC respectively, in the general case. We denote the six
more
> intersection points of their cevians as follows:
>
> - As D, the intersection point of BQb, CPc.
> - As D', the intersection point of BPb, CQc.
> - As E, the intersection point of CQc, APa.
> - As E', the intersection point of CPc, AQa.
> - As F, the intersection point of AQa, BPb.
> - As F', the intersection point of APa, BQb.
>
> Results mentioned in your message:
>
> 1 - The lines AD, BE, CF, are concurrent at a point so be it, X.
> 2 - The lines AD', BE', CF', are concurrent at a point sp be it, Y.
> 3 - The points X, Y, are isogonal wrt ABC.
>
> Another interesting ( I think ) result is that the lines DD',
> EE', FF', XY', are concurrent at point so be it, O.
>
> When the points X, Y, are coincided with the Brocard points
> ( when P, Q, are G, K, respectively ), then the point O coincides
> with the midpoint of their segment.
>
> Some notes for the proof about the concurrency point O, have
been
> posted in message #13310 ( in more general configuration of two
> arbitrary isogonal lines through each vertex of a given triangle ).
>
> Best regards.
> Kostas Vittas.
>
>
>
> --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
> <qcastell@> wrote:
> >
> > Dear friends,
> >
> > I have noticed this fact:
> > P, Q being two isogonal points, the cevians of P and the cevians
of
> Q
> > intersect at P, Q and six more points. The lines (other than the
> > cevians) joining each vertex with these points are concurrent at
> two
> > isogonal points X, Y. When P, Q are G, K the new points X, Y are
> the
> > Brocard points.
> >
> > Greetings from Bsrcelona,
> >
> > Quim Castellsaguer
> >
>
• Quim, This is particularly interesting when the isogonal points are O and H, in which case the 6 intersections form two triangles antisimilar to ABC. Happy
Message 4 of 13 , Aug 6, 2006
Quim,

This is particularly interesting when the isogonal points are O and
H, in which case the 6 intersections form two triangles antisimilar
to ABC.

Happy geometrizing,

Steve

On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:

>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of Q
> intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at two
> isogonal points X, Y. When P, Q are G, K the new points X, Y are the
> Brocard points.

Steve

Notation:
20web/notation.html

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

[Non-text portions of this message have been removed]
• Dear Steve and Quim and all, I am happy to see this old friend . Please see also: http://groups.yahoo.com/group/Hyacinthos/message/8045 Sincerely, Jeff PS
Message 5 of 13 , Aug 6, 2006
Dear Steve and Quim and all,

I am happy to see this 'old friend'.

http://groups.yahoo.com/group/Hyacinthos/message/8045

Sincerely, Jeff
PS It's nice to see something in a new light.

On Sun Aug 6, 2006 12:58 pm, Steve Sigur wrote:

Quim,

This is particularly interesting when the isogonal points are O and
H, in which case the 6 intersections form two triangles antisimilar
to ABC.

Happy geometrizing,

Steve

On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:

>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of
> Q intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at two
> isogonal points X, Y. When P, Q are G, K the new points X, Y are the
> Brocard points.
• ... Q ... two ... the ... Dear Quim and all my friends. I would like to present a proof, of the result that the isogonals point X, Y, become Brocard s points
Message 6 of 13 , Aug 7, 2006
--- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
<qcastell@...> wrote:
>
> Dear friends,
>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of
Q
> intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at
two
> isogonal points X, Y. When P, Q are G, K the new points X, Y are
the
> Brocard points.
>
> Greetings from Bsrcelona,
>
> Quim Castellsaguer
>

Dear Quim and all my friends.

I would like to present a proof, of the result that the
isogonals point X, Y, become Brocard's points wrt ABC, when the P, Q,
are G, K, respectively, based on the below proposition, as follows:

1  PROPOSITION. A triangle is given and let Ka, Ma be an
arbitrary point and the midpoint of BC, respectively. If K is a point
on AKa, we draw the segment lines BK, CK and we denote as D, E, their
intersection points from the median AMa, respectively. If P is the
intersection point of BE, CD, prove that:

a)  The segment KP is parallel to BC and that it is bisected from
the median AMa.
b)  If K is the Lemmoine point of ABC, then the point P, lies on the
midperpendicular of BC.

2  PROOF. ( in my drawing AB = 10.0, AC = 15.0, BC = 13.00 ).
Let P be, the intersection point of CD, from the line through K and
parallel to BC. So, it is easy to prove that the points B, E, P, are
collinear ( from trapezium BKPC ) and that the median AMa, passes
through the midpoint F, of the segment KP.

We denote as K'a, the intersection point of BC from the segment
line AP and then we have that this point is isotomic of Ka, with
respect to BC. That is, the locus of P, when K is moving on segment
line AKa, is the line AK'a where K'a, is the reflexion of Ka wrt Ma.

As K, in our configuration, is the Lemoine point of ABC, we will
prove that the line through K and parallel to BC, intersects the
median AMa and the midperpenticular of BC, at the points F, P
respectively, such that KF = FP.

We denote as Ba, Ca, the intersection points of sidelines AB, AC
respectively, from the segment line KP. It is well known that the
segment BaCa, is a cord ( one of the three ), of the first Lemoine's
circle (L) of ABC, centered at midpoint L of KO, where O is the
circumcenter of ABC.

The point F ( as intersection point of AMa, BaCa ), is the
midpoint of BaCa, because of BaCa//BC and Ma, is the midpoint of BC.
So, from circle (L) we have that the point F is the orthogonal
projection of L, on BaCa. Hence, the point F is also the midpoint of
KP, because of LF//OP and L, is the midpoint of KO.

That is, we have that KF = FP. If K'a, is the intersection point
of BC, from the segment line AP, we have that KaMa = MaK'a and the
proof is completed.

3  We use now the configuration which is mentioned in message #
13879 and we will prove that the isogonal points X, Y ( when P, Q,
are G, K ), are the Brocard's points of ABC.

( They have been considered the isogonal points G, K, instead of
P,Q and similarly about the traces of their cevians on the sidelines
new drawing: AB = 10.0, AC = 13.0, BC = 15.0 ).

3a  We denote as D the intersection point of BKb, CMc ( Mc is
the midpoint of AC ). So, if D' is the intersection point of BMb ( Mb
is the midpoint of AC ), CKc, we conclude that the point D, D', are
isogonal conjugates wrt ABC. Similarly about the pairs of points E,
E' and F, F' ( where E is the intersection point of CKc, AMa, etc ).

3b  We denote as Y the intersection point of the segment lines
AD', CF' ( F' is the intersection point of AMa, BKb ). So, by
the ''isogonic'' theorem ( see messages #13680, #13685 ), we have
that the lines AF ( where F is the intersection point of AKa, BMb ),
BY, CD, are concurrent. Hence, the line BY passes through the point
E' ( the intersection point of CMc, AKa ).

3c  We conclude now, that the lines AD ( isogonal of AD' ), BE
( isogonal of BE' ), CF ( isogonal of CF' ), are concurrent at a
point so be it X, isogonal conjugate of Y, wrt ABC.

3d  We consider the median BMb and the points F, D', as it's
intersection points, from the symmedians AKa, CKc, respectively. So,
if B' is the intersection point of AD', CF, this point, by the above
already has been proved proposition, lies on the midperpendicular of
AC. Hence, we have that <ACX = <CAB', (1).

<BAX = <CAB',(2).( because of AX, AB' are isogonal wrt angle <A ).

(1),(2), => <BAX = <ACX, (3).

By the same way we can prove that <BAX = <CBX, (4).

From (3),(4), we conclude that the point X, is the first
Brocard's point wrt ABC. Hence the point Y ( as isogonal of X ), is
the second Brocard's point wrt ABC and the proof is completed.

NOTES.

a)  Based on the lemma 1, which has been stated and proved in
message #13310, we can prove that the lines DD', EE', FF', XY, are
concurrent at a point so be it S. I still not have a proof of the
result that the point S, is the midpoint of the segment XY, in this

I have noticed another interesting result in this configuration.
Through K ( Lemoine point ), we draw three lines parallel to BC, AC,
AB and we denote as A', B', C', their intersection points, with the
corresponded midperpendiculars, respectively. The triangle A'B'C', is
antisimilar of ABC and it has as centroid, the centroid G of ABC.

It is clear that the triangles ABC, A'B'C', are perspective at a
point so be it T, isotomic of K with respect to ABC.

Best regards.
Kostas Vittas.
• Dear Quim and all my friends. I have noticed another interesting result, out of our configuration and I don t know if it is already well known. If T, is the
Message 7 of 13 , Aug 7, 2006
Dear Quim and all my friends.

I have noticed another interesting result, out of our
configuration and I don't know if it is already well known.

If T, is the isotomic point of the Lemoine point K, with respect
to a given triangle ABC, then the line connecting the point T, with
the orthocenter H of ABC, is parallel to the line KO, where O is the
circumcenter of ABC.

Best regards.
Kostas Vittas.

--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
wrote:
>
> --- In Hyacinthos@yahoogroups.com, "Quim Castellsaguer"
> > <qcastell@> wrote:
> >
> > Dear friends,
> >
> > I have noticed this fact:
> > P, Q being two isogonal points, the cevians of P and the cevians
> > of Q intersect at P, Q and six more points. The lines (other than
> > the cevians) joining each vertex with these points are
> > concurrent at two isogonal points X, Y. When P, Q are G, K the
> > new points X, Y are the Brocard points.
> >
> > Greetings from Bsrcelona,
> >
> > Quim Castellsaguer
> >
>
>
> Dear Quim and all my friends.
>
> ...
>
> I have noticed another interesting result in this
>configuration. Through K ( Lemoine point ), we draw three lines
>parallel to BC,AC,AB and we denote as A', B', C', their intersection
>points, with the corresponded midperpendiculars, respectively. The
>triangle A'B'C', >is antisimilar of ABC and it has as centroid, the
>centroid G of ABC.
>
> ...
• Dear friends, With these thoughts in mind, can somebody please help me with a problem I am having: Given a known point P, what point or points Q allow
Message 8 of 13 , Aug 8, 2006
Dear friends,

problem I am having:

Given a known point P, what point or points Q allow triangles A1B1C1
and A2B2C2 to be similar with one another?

A1 := BP /\ CQ, A2 := BQ /\ CP,
B1 := CP /\ AQ, B2 := CQ /\ AP,
C1 := AP /\ BQ, C2 := AQ /\ BP;

I have tried numerous times to get the coordinates of Q in terms of
those given by P, but my computer program's kernel keeps shutting
down and I am not getting anywhere fast with hand calculations!

Sincerely, Jeff

>
>
> On Aug 6, 2006, at 4:10 PM, Jeff Brooks wrote:
>
> > I am happy to see this 'old friend'.
> >
> > http://groups.yahoo.com/group/Hyacinthos/message/8045
> >
> > Sincerely, Jeff
>
> An old friend for me too.
>
> Steve
>
• ... Dear Quim, Steve, Jeff, Francois and all. Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary points P, Q, in the plane of a given triangle ABC.
Message 9 of 13 , Aug 10, 2006
On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:
>
> I have noticed this fact:
> P, Q being two isogonal points, the cevians of P and the cevians of
> Q intersect at P, Q and six more points. The lines (other than the
> cevians) joining each vertex with these points are concurrent at
> two isogonal points X, Y. When P, Q are G, K the new points X, Y
> are the Brocard points.
>

Dear Quim, Steve, Jeff, Francois and all.

Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary
points P, Q, in the plane of a given triangle ABC. We denote the
intersection points of the lines connecting the vertices of ABC ( the
other than cevians of P, Q ), with the six , furthermore P, Q,
intersection points of their cevians, as follows:

- As D, the intersection point of BQb, CPc.
- As D', the intersection point of BPb, CQc.
- As E, the intersection point of CQc, APa.
- As E', the intersection point of CPc, AQa.
- As F, the intersection point of AQa, BPb.
- As F', the intersection point of APa, BQb.

RESULTS:
a)- The lines AD,BE,CF, are concurrent at a point so be it, X.
b)- The lines AD',BE',CF', are concurrent at a point so be it, Y.
c)- The lines DD',EE',FF',XY, are concurrent at a point so be it S.

PROOF. Let S be the intersection point of DD', FF'. We consider
the triads of points D, Q, F' and F, P, D', on the segment lines BQb,
BPb, respectively. So, by Pappus theorem, we have that the points E',
S, E, are collinear. Hence the lines DD', EE', FF', are concurrent at
point S.

Because of the concurrency at point S, by Desarques theorem, we
conclude that the triangles E'DF, ED'F', are perspective and so, the
intersection point S', of the segment lines DF, D'F', lies on the
sideline AC of ABC.

From the collinearity of A, C, S', we conclude that the triangles
E'DF, YF'D', where Y is the intersection point of AD', CF', are
perspective and hence, the line E'Y passes through the vertex B of
ABC ( also intersection point of DF', FD' ).

From the collinearity of A, C, S', we also conclude that XDF,
EF'D', where X is the intersection point of AD, CF, are perspective
and hence, the line EX passes through the vertex B of ABC.

From the collinearity of A, C, S', we also conclude that XDF,
YD'F', are perspective and hence, the line XY passes through the
point S ( intersection point of DD', FF' ) and the proof is completed.

It is easy to prove that if the points P, Q, are isotomic
( instead isogonal ) with respect to ABC, then the two new
concurrency point X, Y, are also isotomic wrt ABC.

So, we can play with the special cases of the given pairs of
points P, Q, in two categories of configuration ( isogonal and
isotomic ).

I present a quick result in the isotomic configuration. When the
point Q coincides with the orthocenter of ABC, then the line
connecting the concurrency point S, with the centroid G of ABC,
bisects the segment HP ( where now, P is the isotomic of H wrt ABC ).

Best regards.
Kostas Vittas.
• ... Dear all my friends. As, in message #13935, it already has been proved, we don t need any Lemma for the proof. Best regards. Kostas Vittas.
Message 10 of 13 , Aug 10, 2006
On Aug 7, 2006, at 4:42 PM, Kostas Vittas wrote:
>
> NOTES.
>
> a)  Based on the lemma 1, which has been stated and proved in
> message #13310, we can prove that the lines DD', EE', FF', XY, are
> concurrent at a point so be it S.
>

Dear all my friends.

As, in message #13935, it already has been proved, we don't need
any Lemma for the proof.

Best regards.
Kostas Vittas.
• Dear Steve you wrote: This is particularly interesting when the isogonal points are O and H, in which case the 6 intersections form two triangles antisimilar
Message 11 of 13 , Aug 13, 2006
Dear Steve

you wrote:

This is particularly interesting when the isogonal points are O and
H, in which case the 6 intersections form two triangles antisimilar
to ABC.

*************

These antisimilar triangles (let's call them A1B1C1 and A2B2C2) have
some interesting properties:
1) Their circumcircles intersect at the orthocenter H of ABC and at
a second point S;

Take A' on A1A2 such that A'A1 = k.A2A1 (with k any number) and
define B' and C' similarly

2) The triangles A'B'C' are all antisimilar to ABC and their
circumcircles form a pencil through H and S
3) The triangles A'B'C' and ABC orthological. H is the fixed
orthology center. The other orthology center lies on the
circumcircle of ABC.
4) S is the center of similarity of the triangles A'B'C': a rotation
around S combined with a homothety with center S generates all
triangles A'B'C'

If we take A', B' and C' at the midpoints of A1A2, B1B2 and C1C2

5) The circumcircle of A'B'C' goes through the circumcenter O of ABC
so its circumcenter is X(5) of ABC.
6) The corresponding orthology center on the circumcircle of ABC is
X(74)
7) The medial triangle of ABC is perspective with A'B'C' at X(1147)

The barycentric coordinates of S are quite complex.
If my computations are right the first barycentric should be

(b^2+c^2-a^2)(a^12 - (b^2+c^2)a^10 + (b^4-b^2c^2+c^4)a^8
- 3(b^2-c^2)^2(b^2+c^2)a^6
+ 2(b^2-c^2)(b^6-c^6)a^4 + b^2c^2(b^2-c^2)^4)

Kind regards

Eric
• Dear All, I have been away for a bit, and am just catching up with this thread. As Kostas indicates, the isogonal case is just a special case of a more
Message 12 of 13 , Aug 17, 2006
Dear All,

I have been away for a bit, and am just catching up with this
thread. As Kostas indicates, the "isogonal" case is just a
special case of a more general result. This apears in my paper

http://forumgeom.fau.edu/FG2006volume6/FG200613.pdf

See especially Lemma 11 and Theorems 12, 13, 14. I have
summarized some of the relevant bits below Kostas's result

--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
wrote:
>
> On Aug 4, 2006, at 11:17 AM, Quim Castellsaguer wrote:
> >
> > I have noticed this fact:
> > P, Q being two isogonal points, the cevians of P and the cevians
of
> > Q intersect at P, Q and six more points. The lines (other than
the
> > cevians) joining each vertex with these points are concurrent at
> > two isogonal points X, Y. When P, Q are G, K the new points X, Y
> > are the Brocard points.
> >
>
> Dear Quim, Steve, Jeff, Francois and all.
>
> Let PaPbPc, QaQbQc be the cevian triangles of two arbitrary
> points P, Q, in the plane of a given triangle ABC. We denote the
> intersection points of the lines connecting the vertices of ABC (
the
> other than cevians of P, Q ), with the six , furthermore P, Q,
> intersection points of their cevians, as follows:
>
> - As D, the intersection point of BQb, CPc.
> - As D', the intersection point of BPb, CQc.
> - As E, the intersection point of CQc, APa.
> - As E', the intersection point of CPc, AQa.
> - As F, the intersection point of AQa, BPb.
> - As F', the intersection point of APa, BQb.
>
> RESULTS:
> a)- The lines AD,BE,CF, are concurrent at a point so be it, X.
> b)- The lines AD',BE',CF', are concurrent at a point so be it, Y.
> c)- The lines DD',EE',FF',XY, are concurrent at a point so be it S.
>

Note that much of result c) appears in the glossary for ETC -
the point S is the crosspoint of P and Q.

We also observe that the triangles ABC, DEF, D'E'F' are all
triply perspective - the other perspectors are P and Q in each
case. We have a desmic structure containing A,B,C,D,E,F,D',E',F'
with perspectors X,Y,S. Indeed, any such desmic structure with
triply perspective triangles arises this way from a pair P,Q.

An alternative proof uses barycentrics :
Suppose that P = p1:p2:p3, and Q = q1:q2:q3.
Then X, Y are given by
x1:x2:x3 = 1/p2q3:1/p3q1:1/p1q2 and
y1:y2:y3 = 1/q2p3:1/q3p1:1/q1p2

In fact, we also have that the barycentric products P*Q and X*Y
are equal. If we write W for the common value, then we can say
that P,Q are W-isoconjugates, as are X,Y. W = p1q1:p2q2:p3q3.
This is obvious from the coordinates.

Further S = x1+y1:x2+y2:x3+y3 has the properties in c), and
patently lies on XY.

The point R = x1-x2:x2-y2:x3-y3 is also on XY - it is the
"harmon" of the desmic structure. It can be described as
the intersection of the tripolars of P and Q or as the
perspector of C(R), the circumconic through P and Q.

Alternatively, given R and W, we can recover P and Q as the
intersections of C(R) and the tripolar of R*, the W-isoconjugate
of R. Of course, P and Q may be complex. We can also recover S
as the R*-ceva conjugate of R.

X and Y are the intersections (other than A,B,C and the points
fixed by W-isoconjugation) of the pivotal isocubics pK(W,R)
and pK(W,S).

The points D,E,F,D',E',F' are intersections (other than A,B,C)
of pK(W,S) and nK0(W,R).

That's probably enough for now ....

Regards,

Wilson
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