- 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 - [Eric Danneels]:
>X(1358) is defined in the current ETC as the Brisse-transform

Dear Eric

>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.....)

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

> 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

>Greetings from Bruges & good luck on sunday

Thank you. Your wishes are very much appreciated.

>

>Eric Danneels

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 !

--