- 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

Antivirus scanning: Symantec Mail Security for SMTP. - --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...>

wrote:>

its

> Dear colleagues!

> Let given the triangle ABC. A1B1C1 is the circumcevian triangle of

> 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.

Dear Alexey,

> What is the number of X in ETC?

>

The answer is 22

(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 - --- In Hyacinthos@yahoogroups.com, "Wilson Stothers" <wws@...> wrote:
>

of

> --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@>

> wrote:

> >

> > Dear colleagues!

> > Let given the triangle ABC. A1B1C1 is the circumcevian triangle

> its

and

> > 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

> the

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

> > 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 answer is 22

>

> (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

>

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]

and get the same answer.

Sincerely, Jeff - Please disregard,

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]

and get the same answer.

Sincerely, Jeff