- Dear Antreas!
>

These triangles are homothetic. It is proved in the paper of L.and T.

>I am wondering what the extraversion of this might be.

>

>That the cevian triangle of I and the Feuerbach triangle

>FaFbFc are perspective?

>

Emelyanov in "Math. Prosveschenie" and the book of Sharygin.

Sincerely Alexey - Dear Antreas,

In Hyacinthos message #10348, you wrote:

>> I am wondering what the extraversion of this

They are not perspective. This is not the way

>> might be.

>>

>> That the pedal triangle of I and the Feuerbach

>> triangle FaFbFc are perspective?

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

[APH]:>> I am wondering what the extraversion of this

[DG]:

>> might be.

>>

>> That the pedal triangle of I and the Feuerbach

>> triangle FaFbFc are perspective?

>They are not perspective. This is not the way

Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3)

>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!

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

I deleted it because I meant to say pedal not cevian

>asked whether the cevian triangle of I and the

>Feuerbach triangle FaFbFc are perspective.

(since the sidelines of the pedal triangle of I

are the polars of A,B,C wrt the incircle)

Antreas

-- - 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:

Actually, you have twelve polars! There

>> Pij := the polar of Aj with respect the

>> circle (Ii),

>> forming C(9,3) (non-ordered) triangles

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

Yes, that's clear to me. But with "cevian"

>> not 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

about extraversion. Finally, extraversion,

when applied properly, can be of great help

in understanding relationships between

theorems from triangle geometry.

Sincerely,

Darij Grinberg - Dear Darij

[APH]:>>> Well... If A1A2A3 is a triangle, and

[DG]:

>>> (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,

[APH]:

>F3. There are 4 Feuerbach points, not 5.

>>> then we have the nine polars:

[DG]:

>>> Pij := the polar of Aj with respect the

>>> circle (Ii),

>>> forming C(9,3) (non-ordered) triangles

>Actually, you have twelve polars! There

Sorry for the stupid errors!

>are four in/excircles, and three vertices

>of triangle ABC. And you thus have C(12,3)

>non-ordered triangles.

(I should have delete it and resend a correct one)

[DG]:>But extraversion is not just replacing

[...]

Thanks for your analysis !

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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