## THEOREM (was : Locus and generalization)

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• [APH] ... that is, cotB + cot(B,C) ... 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
Message 1 of 4 , Jun 24, 2005
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[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??

Antreas

--
• ... Hey Antreas! Have you given any more thought to this theorem? Sincerely, Jeff
Message 2 of 4 , Mar 2, 2006
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--- 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 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??
>

Hey Antreas!

Have you given any more thought to this theorem?

Sincerely, Jeff
• Waiting patiently for a synthetic proof, Jeff. ... sides
Message 3 of 4 , Mar 3, 2006
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Waiting patiently for a synthetic proof, Jeff.
>
> --- 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
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??
> >
>
> Hey Antreas!
>
> Have you given any more thought to this theorem?
>
> Sincerely, Jeff
>
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