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• ... As Francisco pointed out: this follows inmediately from Desargues theorem for any ABC and A B C We can use it in the reference triangle ABC to get
Message 1 of 4 , May 1, 2010
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--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC, A'B'C' be two homothetic triangles
> and X,Y,Z three collinear points.
>
> Are the triangles bounded by the lines
> (AX, BY, CZ) and (A'X, B'Y, C'Z) perspective?

As Francisco pointed out: "this follows inmediately from Desargues
theorem for any ABC and A'B'C'"

We can use it in the reference triangle ABC to get points,
from two given triangles and three collinear points.

Here is an example:

Inside ABC exists a unique point P such that:
Three Congruent (equal) circles (A'),(B'),(C') concur at P,
and circle (A') touches the sides of the angle A,
(B') of angle B and (C') of angle C.
[A', B', C' are the centers of the circles]

Now, let Lp a line passing through P and
intersecting the circles (A'),(B'),(C') at
X,Y,Z resp. (other than P).

The triangles bounded by the lines
(AX, BY, CZ) and (A'X, B'Y, C'Z) are perspective.

Which is the locus of the perspectors
as L moves around P?

APH
• ... I get two such points P. If your 3 circles have radius r*R/(R+r) then P=X(55). And if their radius is r*R/(R-r) then P=X(56). -- Barry Wolk
Message 1 of 4 , May 6, 2010
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Antreas wrote:
> Here is an example:
>
> Inside ABC exists a unique point P such that:
> Three Congruent (equal) circles (A'),(B'),(C') concur at P,
> and circle (A') touches the sides of the angle A,
> (B') of angle B and (C') of angle C.
> [A', B', C' are the centers of the circles]

I get two such points P. If your 3 circles have radius r*R/(R+r)
then P=X(55). And if their radius is r*R/(R-r) then P=X(56).
--
Barry Wolk
• ... inside triangle (that s what I had in mind) APH [Non-text portions of this message have been removed]
Message 1 of 4 , May 6, 2010
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On Thu, May 6, 2010 at 11:18 PM, Barry Wolk <wolkbarry@...> wrote:

>
>
> Antreas wrote:
> > Here is an example:
> >
> > Inside ABC exists a unique point P such that:
> > Three Congruent (equal) circles (A'),(B'),(C') concur at P,
> > and circle (A') touches the sides of the angle A,
> > (B') of angle B and (C') of angle C.
> > [A', B', C' are the centers of the circles]
>
> I get two such points P. If your 3 circles have radius r*R/(R+r)
> then P=X(55). And if their radius is r*R/(R-r) then P=X(56).
> --
> Barry Wolk
>
>
> Indeed. Two points P. But I think only one point with circles centers
inside triangle (that's what I had in mind)

APH

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