## Re: New point related to Feuerbach-point

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• Dear Eric ... and ... triangle ... Your point is the reflection of X(40) (reflection of I wrt O) wrt X(100) (anticomplement of F); he lies on the line
Message 1 of 8 , Nov 3, 2003
Dear Eric
> I wonder if the following property is known
>
> -------------------------------------------------
> Let A', B' and C' be the midpoints of BC, CA and AB
> Let Ia, Ib and Ic be the A-, B- and C-excenter
> Let F be the Feuerbach-point
> Consider the perpendiculars Pa from Ia to FA', Pb from Ib to FB'
and
> Pc from Ic to FC'
>
> ==> the lines Pa, Pb and Pc are concurrent
>
> or in words
>
> The perpendiculars from each excenter to the line joining the
> Feuerbach point with the corresponding vertex of the medial
triangle
> are concurrent.

Your point is the reflection of X(40) (reflection of I wrt O)
wrt X(100) (anticomplement of F); he lies on the line I-F-NPcenter
and on the circle (I,2R).
Friendly. Jean-Pierre
• ... Dear Eric Your theorem rephrased: The perpendiculars from each vertex of the antipedal triangle of I to the line joining the Feuerbach point with the
Message 2 of 8 , Nov 5, 2003
[Eric Danneels]:

>The perpendiculars from each excenter to the line joining the
>Feuerbach point with the corresponding vertex of the medial triangle
>are concurrent.

Dear Eric

The perpendiculars from each vertex of the antipedal triangle of I
to the line joining the Feuerbach point with the corresponding vertex
of the pedal triangle of O are concurrent.

The "dual" (by replacing the points I,O):

The perpendiculars from each vertex of the antipedal triangle of O
to the line joining the Feuerbach point with the corresponding vertex
of the pedal triangle of I are concurrent.

Is it true??

Antreas
--
• Antreas wrote ... vertex ... vertex ... I don t think so (if I may believe my sketches) I also tried anticevian triangles instead of antipedal triangles
Message 3 of 8 , Nov 5, 2003
Antreas wrote

>
> The perpendiculars from each vertex of the antipedal triangle of I
> to the line joining the Feuerbach point with the corresponding
vertex
> of the pedal triangle of O are concurrent.
>
> The "dual" (by replacing the points I,O):
>
> The perpendiculars from each vertex of the antipedal triangle of O
> to the line joining the Feuerbach point with the corresponding
vertex
> of the pedal triangle of I are concurrent.
>
> Is it true??
>
----------------------------

I don't think so (if I may believe my sketches)

I also tried anticevian triangles instead of antipedal triangles
(because IaIbIc is the antipedal and anticevian triangle of I) but
it doesn't work either

I also tried replacing one of the excenters by the incenter and
replacing the Feuerbach point by the tangent point of the
ninepointcircle and the selected excircle but without success

PS

I would like to thank Paul for the coordinates of this point
and Jean-Pierre for some nice properties of it and I'm sorry for
the typing errors in the formulae in my original mail

Greetings from Bruges

Eric Danneels
• ... [...] ... What the extraversions of this problem are? Antreas --
Message 4 of 8 , Nov 5, 2003
[ED]:
>Antreas wrote
>
>>
>> The perpendiculars from each vertex of the antipedal triangle of I
>> to the line joining the Feuerbach point with the corresponding
>> vertex of the pedal triangle of O are concurrent.
[...]
>I also tried replacing one of the excenters by the incenter and
>replacing the Feuerbach point by the tangent point of the
>ninepointcircle and the selected excircle but without success

What the extraversions of this problem are?

Antreas
--
• ... of I ... Sorry, I don t understand your last question Meanwhile I discovered that using my original statement we get a new concurrency if we replace the
Message 5 of 8 , Nov 5, 2003
--- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis"
<xpolakis@o...> wrote:
> [ED]:
> >Antreas wrote
> >
> >>
> >> The perpendiculars from each vertex of the antipedal triangle
of I
> >> to the line joining the Feuerbach point with the corresponding
> >> vertex of the pedal triangle of O are concurrent.
> [...]
> >I also tried replacing one of the excenters by the incenter and
> >replacing the Feuerbach point by the tangent point of the
> >ninepointcircle and the selected excircle but without success
>
> What the extraversions of this problem are?
>
-------------------
Sorry, I don't understand your last question

Meanwhile I discovered that using my original statement we get a new
concurrency if we replace the excenters by the vertices of the pedal
triangle of I

so

The perpendiculars from the vertices of the intouch triangle on the
lines joining the Feuerbach-point with the corresponding vertices of
the medial triangle are concurrent

Greetings from Bruges

Eric Danneels
• ... Extraversists (JHC et al) understand. JHC? Antreas --
Message 6 of 8 , Nov 5, 2003
>> [ED]:
>> >Antreas wrote
>> >
>> >>
>> >> The perpendiculars from each vertex of the antipedal triangle
>of I
>> >> to the line joining the Feuerbach point with the corresponding
>> >> vertex of the pedal triangle of O are concurrent.
>> [...]
>> >I also tried replacing one of the excenters by the incenter and
>> >replacing the Feuerbach point by the tangent point of the
>> >ninepointcircle and the selected excircle but without success
>>
>> What the extraversions of this problem are?
>>
>-------------------
>Sorry, I don't understand your last question

Extraversists (JHC et al) understand.

JHC?

Antreas

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