## concurrence

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• Dear colleagues! Let given the triangle ABC. A1B1C1 is the circumcevian triangle of its centroid, A2, B2, C2 --- the second common points of its circumcircle
Message 1 of 4 , Mar 1 3:30 AM
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Dear colleagues!
Let given the triangle ABC. A1B1C1 is the circumcevian triangle of its
centroid, A2, B2, C2 --- the second common points of its circumcircle and
three lines passing through its vertex and parallel to opposite sidelines
(i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then A1A2, B1B2
and C1C2 concur. Its common point X is in the Euler line of ABC and the
polar of X wrt the circumcircle is the trypolar of the point isotomic
conjugated to the Lemoine point of ABC.
What is the number of X in ETC?

Sincerely Alexey

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• ... its ... circumcircle and ... sidelines ... A1A2, B1B2 ... the ... isotomic ... Dear Alexey, The answer is 22 (the tripolar of the isotomic conjugate of
Message 2 of 4 , Mar 1 4:20 AM
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--- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...>
wrote:
>
> Dear colleagues!
> Let given the triangle ABC. A1B1C1 is the circumcevian triangle of
its
> centroid, A2, B2, C2 --- the second common points of its
circumcircle and
> three lines passing through its vertex and parallel to opposite
sidelines
> (i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then
A1A2, B1B2
> and C1C2 concur. Its common point X is in the Euler line of ABC and
the
> polar of X wrt the circumcircle is the trypolar of the point
isotomic
> conjugated to the Lemoine point of ABC.
> What is the number of X in ETC?
>

Dear Alexey,

(the tripolar of the isotomic conjugate of Lemoine Point is the
de Longchamps Line. This meets the Euler Line at X(858). The
inverse in the Circumcircle of this is the point you want, and
it is X(22) from ETC)

regards

Wilson
• ... of ... and ... And if this is true we should be able to write: p=N[x22]; q=N[x4]; g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u] T[u_,v_,w_]:=
Message 3 of 4 , Mar 1 4:39 AM
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--- In Hyacinthos@yahoogroups.com, "Wilson Stothers" <wws@...> wrote:
>
> --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@>
> wrote:
> >
> > Dear colleagues!
> > Let given the triangle ABC. A1B1C1 is the circumcevian triangle
of
> its
> > centroid, A2, B2, C2 --- the second common points of its
> circumcircle and
> > three lines passing through its vertex and parallel to opposite
> sidelines
> > (i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then
> A1A2, B1B2
> > and C1C2 concur. Its common point X is in the Euler line of ABC
and
> the
> > polar of X wrt the circumcircle is the trypolar of the point
> isotomic
> > conjugated to the Lemoine point of ABC.
> > What is the number of X in ETC?
> >
>
> Dear Alexey,
>
>
> (the tripolar of the isotomic conjugate of Lemoine Point is the
> de Longchamps Line. This meets the Euler Line at X(858). The
> inverse in the Circumcircle of this is the point you want, and
> it is X(22) from ETC)
>
> regards
>
> Wilson
>

And if this is true we should be able to write:

p=N[x22];
q=N[x4];

g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u]

T[u_,v_,w_]:= (g[v,w,u] - g[w,u,v])

k=1;
M[u_,v_,w_]:= (v-w)^k

f[n_,u_,v_,w_]:= (M[w,u,v] T[v,w,u] w + n M[v,u,w] T[w,u,v] v)/
(M[w,u,v] T[v,w,u] + n M[v,u,w] T[w,u,v])

d1[u_,v_,w_]:=1;

pp1=Chop[N[f[d1[z1,z2,z3],z1,z2,z3],12],10^-6]
pp2=Chop[N[f[d1[z2,z3,z1],z2,z3,z1],12],10^-6]
pp3=Chop[N[f[d1[z3,z1,z2],z3,z1,z2],12],10^-6]

Sincerely, Jeff
• Please disregard, this does nothing but give the inverse in the circumcircle ______________________________________________ And if this is true we should be
Message 4 of 4 , Mar 1 5:35 AM
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this does nothing but give the inverse in the circumcircle

______________________________________________

And if this is true we should be able to write:

p=N[x22];
q=N[x4];

g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u]

T[u_,v_,w_]:= (g[v,w,u] - g[w,u,v])

k=1;
M[u_,v_,w_]:= (v-w)^k

f[n_,u_,v_,w_]:= (M[w,u,v] T[v,w,u] w + n M[v,u,w] T[w,u,v] v)/
(M[w,u,v] T[v,w,u] + n M[v,u,w] T[w,u,v])

d1[u_,v_,w_]:=1;

pp1=Chop[N[f[d1[z1,z2,z3],z1,z2,z3],12],10^-6]
pp2=Chop[N[f[d1[z2,z3,z1],z2,z3,z1],12],10^-6]
pp3=Chop[N[f[d1[z3,z1,z2],z3,z1,z2],12],10^-6]