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• Dear Hyacinthians, X(1358) is defined in the current ETC as the Brisse-transform of X(101). I would like to present a new remarkable property of this point Let
Message 1 of 4 , Jul 1, 2004
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Dear Hyacinthians,

X(1358) is defined in the current ETC as the Brisse-transform
of X(101).

I would like to present a new remarkable property of this point

Let A'B'C' be the medial triangle of triangle ABC.
Consider the circle, different from the ninepointcircle, through B'
and C' touching the incircle. Let A* be the point where both circles
touch. Define B* and C* similarly.

X(1358) is the perspector of the triangles ABC and A*B*C*
Its barycentrics are ( (b-c)^2/(b+c-a) : cyclic.....)

If we replace the medial triangle by the orthic triangle we get a
new point S with similar barycentrics

S = ( (b-c)^2.(b+c-a)^3 : cyclic.....)

Greetings from Bruges

Eric Danneels
• ... Dear Eric How about if we replace the medial triangle with the Euler triangle? (ie the triangle formed with the midpoints of AH, BH, CH) ? Also, how about
Message 2 of 4 , Jul 1, 2004
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[Eric Danneels]:
>X(1358) is defined in the current ETC as the Brisse-transform
>of X(101).
>
>I would like to present a new remarkable property of this point
>
>Let A'B'C' be the medial triangle of triangle ABC.
>Consider the circle, different from the ninepointcircle, through B'
>and C' touching the incircle. Let A* be the point where both circles
>touch. Define B* and C* similarly.
>
>X(1358) is the perspector of the triangles ABC and A*B*C*
>Its barycentrics are ( (b-c)^2/(b+c-a) : cyclic.....)
>
>If we replace the medial triangle by the orthic triangle we get a
>new point S with similar barycentrics
>
>S = ( (b-c)^2.(b+c-a)^3 : cyclic.....)

Dear Eric

How about if we replace the medial triangle with the Euler triangle?
(ie the triangle formed with the midpoints of AH, BH, CH) ?

Also, how about if we replace the incircle with the three excircles?

Antreas (waiting the great victory of Greece !)

--
• Dear Antreas, you wrote ... triangle? ... My sketches show that the Euler triangle does not produce a perspector. However I think I ve found another
Message 3 of 4 , Jul 3, 2004
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Dear Antreas,

you wrote

> How about if we replace the medial triangle with the Euler
triangle?
> (ie the triangle formed with the midpoints of AH, BH, CH) ?

My sketches show that the Euler triangle does not produce a
perspector.

However I think I've found another generalization leading to a new
perspector.

Let A'B'C' be the cevian triangle of a point P wrt triangle ABC.
Through B' and C' we can draw 2 circles tangent to the incircle.
Let A1 and A2 be the touchpoints.
Similarly define the touchpoints B1-B2, and C1-C2 corresponding to
C'A' and A'B'.
It seems that A1A2, B1B2 and C1C2 go through 1 point.
I could not find a proof of it yet.

For the orthic triangle and the medial triangle this new perspector
if of course the Feuerbach point.

Greetings from Bruges & good luck on sunday

Eric Danneels
• Dear Eric ... Thank you. Your wishes are very much appreciated. After your nice problems with incircle, I have this locus problem: Let ABC be a triangle, P, P*
Message 4 of 4 , Jul 3, 2004
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Dear Eric

>Greetings from Bruges & good luck on sunday
>
>Eric Danneels

Thank you. Your wishes are very much appreciated.

After your nice problems with incircle,
I have this locus problem:

Let ABC be a triangle, P, P* two isogonal points, and
A'B'C', A"B"C" their pedal triangles.

Consider the [finite] circle passing through A',A"
and touching the incircle at A*.

Similarly B*,C*.

Which is the locus of P such that ABC, A*B*C* are perspective?

Greetings from Athens

Antreas

SHKWSE TO TO G...MENO
DEN MPORW NA PERIMENW !
--
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