## Re: [EMHL] Re: Perspector

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• Dear Antreas, ... They are not perspective. This is not the way extraversion goes! Under extraversion, the triangle formed by the polars of the points A, B, C
Message 1 of 6 , Sep 1, 2004
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Dear Antreas,

In Hyacinthos message #10348, you wrote:

>> I am wondering what the extraversion of this
>> might be.
>>
>> That the pedal triangle of I and the Feuerbach
>> triangle FaFbFc are perspective?

They are not perspective. This is not the way
extraversion goes! Under extraversion, the
triangle formed by the polars of the points A,
B, C with respect to the A-excircle, B-excircle,
C-excircle, respectively, is "mapped" to the
triangle formed by the polars of the points A,
B, C with respect to the incircle, C-excircle,
B-excircle, respectively. And not all with
respect to the incircle!

Actually, in a message you have deleted, you
asked whether the cevian triangle of I and the
Feuerbach triangle FaFbFc are perspective. And
in Hyacinthos message #10349, Alexey answered
that "These triangles are homothetic".
Actually, they are perspective but NOT
homothetic. This matter is a real pitfall: In
the paper

Lev Emelyanov and Tatiana Emelyanova, "A
Note on the Feuerbach Point", Forum
Geometricorum, 1 (2001) pp. 121-124,

http://forumgeom.fau.edu/FG2001volume1/FG200117index.html

it has been proven that the cevian triangle
of I and the Feuerbach triangle FaFbFc are
1) perspective and 2) similar (even directly
similar). It took me a long time until I
realized that two triangles can be both
perspective and directly similar without being
homothetic! Actually, the quickest way to see
that the cevian triangle of I is (in general)
NOT homothetic to the Feuerbach triangle
FaFbFc is to show that their corresponding
sides are not parallel. In fact, the
corresponding sides of both triangles pass
through the feet of the exterior angle
bisectors of triangle ABC, and hence they
can't be parallel for a generic triangle ABC.

Sincerely,
Darij Grinberg
• Dear Darij ... Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3) are the in/excircles and F0,F1,F2,F3,F4 the Feuerbach Points, then we have the nine
Message 2 of 6 , Sep 1, 2004
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Dear Darij

[APH]:
>> I am wondering what the extraversion of this
>> might be.
>>
>> That the pedal triangle of I and the Feuerbach
>> triangle FaFbFc are perspective?

[DG]:
>They are not perspective. This is not the way
>extraversion goes! Under extraversion, the
>triangle formed by the polars of the points A,
>B, C with respect to the A-excircle, B-excircle,
>C-excircle, respectively, is "mapped" to the
>triangle formed by the polars of the points A,
>B, C with respect to the incircle, C-excircle,
>B-excircle, respectively. And not all with
>respect to the incircle!

Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3)
are the in/excircles and F0,F1,F2,F3,F4 the Feuerbach Points,
then we have the nine polars:
Pij := the polar of Aj with respect the circle (Ii),
forming C(9,3) (non-ordered) triangles
and also we have C(4,3) triangles with vertices Fi

Which ones are perspective? And which are the perspectors?

[DG]:
>Actually, in a message you have deleted, you
>asked whether the cevian triangle of I and the
>Feuerbach triangle FaFbFc are perspective.

I deleted it because I meant to say pedal not cevian
(since the sidelines of the pedal triangle of I
are the polars of A,B,C wrt the incircle)

Antreas
--
• Dear Antreas, ... ^^^^^^^^^^^^^^ Well, you probably wanted to say F0, F1, F2, F3. There are 4 Feuerbach points, not 5. ... Actually, you have twelve polars!
Message 3 of 6 , Sep 1, 2004
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Dear Antreas,

In Hyacinthos message #10353, you wrote:

>> Well... If A1A2A3 is a triangle, and
>> (I0),(I1),(I2),(I3) are the in/excircles
>> and F0,F1,F2,F3,F4 the Feuerbach Points,
^^^^^^^^^^^^^^

Well, you probably wanted to say F0, F1, F2,
F3. There are 4 Feuerbach points, not 5.

>> then we have the nine polars:
>> Pij := the polar of Aj with respect the
>> circle (Ii),
>> forming C(9,3) (non-ordered) triangles

Actually, you have twelve polars! There
are four in/excircles, and three vertices
of triangle ABC. And you thus have C(12,3)
non-ordered triangles.

But extraversion is not just replacing
*some* incircle by *some* excircle. If you
extravert with respect to, say, A, then
you replace

incircle --> A-excircle;
A-excircle --> incircle;
B-excircle --> C-excircle;
C-excircle --> B-excircle.

This is a strict rule, and if you have a
theorem, then its extraversion with
respect to A is uniquely defined. You can
also extravert with respect to B or to C
(just analogously to the A-extraversion
explained above). If you have a triangle
formed by the polars of the points A, B, C
with respect to the A-excircle, B-excircle,
C-excircle, respectively, then you can
never extravert it into a triangle formed
by the polars of A, B, C with respect to
the incircle, since an extraversion
"transforms" (I put this word in quotes
since extraversion is not a real
geometrical transformation) different
in/excircles into different in/excircles,
and thus it cannot "transform" all three
excircles into one and the same incircle.

>> Which ones are perspective?

Hmm, well, this is a difficult question.
Of course, you have the perspector of the
triangles P11P22P33 (meaning: the triangle
formed by the lines P11, P22 and P33) and
F1F2F3. This is the perspector found by
Juan Carlos. You also have the three
extraversions of this perspector, namely

- its A-extraversion, i. e. the perspector
of triangles P01P32P23 and F0F3F2.
- its B-extraversion, i. e. the perspector
of triangles P31P02P13 and F3F0F1.
- its C-extraversion, i. e. the perspector
of triangles P21P12P03 and F2F1F0.

[I hope I have done no typos or mistakes
in extraverting.]

But concerning the other combinations of
polars and Feuerbach points - I don't know.
Maybe somebody could write a computer
program to check all possible cases...

>> I deleted it because I meant to say pedal
>> not cevian

Yes, that's clear to me. But with "cevian"
it's correct (it's not a correct
extraversion, but the triangles are
perspective), while with "pedal" it's wrong.

Please forgive me that, as I'm just noting,
the above comments sounded rude. They were
actually by no means meant in this way. I
wanted to clear up some fundamental things
when applied properly, can be of great help
in understanding relationships between
theorems from triangle geometry.

Sincerely,
Darij Grinberg
• Dear Darij ... Sorry for the stupid errors! (I should have delete it and resend a correct one) ... [...] Thanks for your analysis ! Now, how about a
Message 4 of 6 , Sep 1, 2004
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Dear Darij

[APH]:
>>> Well... If A1A2A3 is a triangle, and
>>> (I0),(I1),(I2),(I3) are the in/excircles
>>> and F0,F1,F2,F3,F4 the Feuerbach Points,

[DG]:
>Well, you probably wanted to say F0, F1, F2,
>F3. There are 4 Feuerbach points, not 5.

[APH]:
>>> then we have the nine polars:
>>> Pij := the polar of Aj with respect the
>>> circle (Ii),
>>> forming C(9,3) (non-ordered) triangles

[DG]:
>Actually, you have twelve polars! There
>are four in/excircles, and three vertices
>of triangle ABC. And you thus have C(12,3)
>non-ordered triangles.

Sorry for the stupid errors!
(I should have delete it and resend a correct one)

[DG]:
>But extraversion is not just replacing
[...]

Now, how about a generalization of JCS's problem?

I think that a possible one is this:

Let ABC be a triangle, P a point, A'B'C'
the antipedal triangle of P, and A'aA'bA'c,
B'aB'bB'c, C'aC'bC'c the pedal triangles
of A',B',C' resp.

Let Pa, Pb, Pc be the points of contact of the
circle which touches externally the pedal circles
of A', B', C' (ie the circles (A'aA'bA'c),
(B'aB'bB'c), (C'aC'bC'c)) .

Which is the locus of P such the triangle PaPbPc and
the triangle bounded by the lines A'bA'c, B'cB'a, C'aC'b
are perspective?

PS: Probably are interesting the cases
(1) ABC, PaPbPc (2) ABC, bounded by (A'bA'c, B'cB'a, C'aC'b)
be perspective.

Greetings

Antreas

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