- [APH]
>GENERALIZATION

that is,

>

>Let ABC be a triangle P a point and A'B'C' the cevian triangle

>of P.

>

>Let A#BC, B#CA, C#AB be three fixed triangles erected on the sides

>of ABC (out / inwardly).

>

>Ac := AC# /\ CC'

>

>Ab := AB# /\ BB'

>

>A* := BAc /\ CAb. Similarly B*, C*

>

>Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc

>

>Now, the triangles ABC, A*B*C* are perspective for all P, if

>

>

> cotA + cot(A,C)

> ------------------------ * CYCLIC = 1

> cotA + cot(A,B)

cotB + cot(B,C)

------------------------ * CYCLIC = 1

cotC + cot(C,B)

that is, ABC, A#B#C# are perspective.

So we have the

THEOREM

Let ABC, A#B#C# be two triangles and P a point.

Denote

Ac := AC# /\ CP, Ab := AB# /\ BP

A* := AcB /\ AbC. Similarly B*,C*.

ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective

for all Ps

Synthetic proof??

Antreas

-- - --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"

<xpolakis@...> wrote:>

Hey Antreas!

> [APH]

> >GENERALIZATION

> >

> >Let ABC be a triangle P a point and A'B'C' the cevian triangle

> >of P.

> >

> >Let A#BC, B#CA, C#AB be three fixed triangles erected on the sides

> >of ABC (out / inwardly).

> >

> >Ac := AC# /\ CC'

> >

> >Ab := AB# /\ BB'

> >

> >A* := BAc /\ CAb. Similarly B*, C*

> >

> >Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc

> >

> >Now, the triangles ABC, A*B*C* are perspective for all P, if

> >

> >

> > cotA + cot(A,C)

> > ------------------------ * CYCLIC = 1

> > cotA + cot(A,B)

>

> that is,

>

> cotB + cot(B,C)

> ------------------------ * CYCLIC = 1

> cotC + cot(C,B)

>

>

> that is, ABC, A#B#C# are perspective.

>

>

> So we have the

>

> THEOREM

>

> Let ABC, A#B#C# be two triangles and P a point.

>

> Denote

>

> Ac := AC# /\ CP, Ab := AB# /\ BP

>

> A* := AcB /\ AbC. Similarly B*,C*.

>

>

> ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective

> for all Ps

>

>

> Synthetic proof??

>

Have you given any more thought to this theorem?

Sincerely, Jeff - Waiting patiently for a synthetic proof, Jeff.
>

sides

> --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"

> <xpolakis@> wrote:

> >

> > [APH]

> > >GENERALIZATION

> > >

> > >Let ABC be a triangle P a point and A'B'C' the cevian triangle

> > >of P.

> > >

> > >Let A#BC, B#CA, C#AB be three fixed triangles erected on the

> > >of ABC (out / inwardly).

> > >

> > >Ac := AC# /\ CC'

> > >

> > >Ab := AB# /\ BB'

> > >

> > >A* := BAc /\ CAb. Similarly B*, C*

> > >

> > >Denote fixed angles C#AB = (A,B), B#AC = (A,C) etc

> > >

> > >Now, the triangles ABC, A*B*C* are perspective for all P, if

> > >

> > >

> > > cotA + cot(A,C)

> > > ------------------------ * CYCLIC = 1

> > > cotA + cot(A,B)

> >

> > that is,

> >

> > cotB + cot(B,C)

> > ------------------------ * CYCLIC = 1

> > cotC + cot(C,B)

> >

> >

> > that is, ABC, A#B#C# are perspective.

> >

> >

> > So we have the

> >

> > THEOREM

> >

> > Let ABC, A#B#C# be two triangles and P a point.

> >

> > Denote

> >

> > Ac := AC# /\ CP, Ab := AB# /\ BP

> >

> > A* := AcB /\ AbC. Similarly B*,C*.

> >

> >

> > ABC, A#B#C# are perspective ==> ABC, A*B*C* are perspective

> > for all Ps

> >

> >

> > Synthetic proof??

> >

>

> Hey Antreas!

>

> Have you given any more thought to this theorem?

>

> Sincerely, Jeff

>