## A question

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• Dear Hyacinthists, 6 points on a circle leads to 60 hexagones and 60 Pascal s lines. These 60 Pascal s lines intersect in 20 Steiner s points. What is the
Message 1 of 10 , Jun 4, 2011
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Dear Hyacinthists,
6 points on a circle leads to 60 hexagones and 60 Pascal's lines.
These 60 Pascal's lines intersect in 20 Steiner's points.
What is the geometric rule when we sta

[Non-text portions of this message have been removed]
• Dear Hyacinthists, 6 points on a circle leads to 60 hexagones and 60 Pascal s lines. These 60 Pascal s lines intersect in 20 Steiner s points. What is the
Message 2 of 10 , Jun 4, 2011
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Dear Hyacinthists,
6 points on a circle leads to 60 hexagones and 60 Pascal's lines.
These 60 Pascal's lines intersect in 20 Steiner's points.
What is the geometric rule when we sta
Dear Hyacinthists,
6 points on a circle leads to 60 cyclic hexagons and 60 Pascal's lines.
These 60 Pascal's lines intersect in 20 Steiner's points which are triple...
Now, what is the geometric rule in order to obtain the Steiner's points?
For example we start withe the cyclic hexagon ABCDEF. How do we obtain geometrically the two other hexagons? I don't wish to use the term of permutation...
Any geometric ideas?
Sincerely
Jean-Louis

[Non-text portions of this message have been removed]
• ... The 3 hexagons with concurrwent Pascal lines are ABCDEF, ADCFEB and AFCBED. You fix ACE and cyclically rotate BDF. Is this what you wanted? -- Barry Wolk
Message 3 of 10 , Jun 6, 2011
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> Dear Hyacinthists,
> 6 points on a circle leads to 60 cyclic hexagons and 60 Pascal's lines.
> These 60 Pascal's lines intersect in 20 Steiner's points which are triple...
> Now, what is the geometric rule in order to obtain the Steiner's points?
> For example we start withe the cyclic hexagon ABCDEF. How do we obtain
> geometrically the two other hexagons? I don't wish to use the term of
> permutation...
> Any geometric ideas?
> Sincerely
> Jean-Louis

The 3 hexagons with concurrwent Pascal lines are ABCDEF,
ADCFEB and AFCBED. You fix ACE and cyclically rotate BDF.

Is this what you wanted?
--
Barry Wolk
• Dear Hyacinthists, what is the nature of the symmedian point of a triangle relatively to his orthic triangle?   Sincerely Jean-Louis [Non-text portions of
Message 4 of 10 , Jul 11, 2012
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Dear Hyacinthists,
what is the nature of the symmedian point of a triangle relatively to his orthic triangle?

Sincerely
Jean-Louis

[Non-text portions of this message have been removed]
• From a quick calculation I find that K is the Mittenpunkt of orthic triangle when ABC is acute. Best regards, Francisco Javier.
Message 5 of 10 , Jul 11, 2012
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From a quick calculation I find that K is the Mittenpunkt of orthic triangle when ABC is acute.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
>
> Dear Hyacinthists,
> what is the nature of the symmedian point of a triangle relatively toï¿½his orthic triangle?
> ï¿½
> Sincerely
> Jean-Louis
>
> [Non-text portions of this message have been removed]
>
• Dear Jean-Louis, The symmedian point of a triangle is the center of homology of triangle and median triangle of his orthic triangle. In attach you find the
Message 6 of 10 , Jul 11, 2012
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Dear Jean-Louis,

The symmedian point of a triangle is the center of homology of triangle and median triangle of his orthic triangle. In attach you find the proof of this theorem.

Best regards,

Catalin

________________________________
From: Jean-Louis Ayme <jeanlouisayme@...>
To: "Hyacinthos@yahoogroups.com" <Hyacinthos@yahoogroups.com>
Sent: Thursday, July 12, 2012 8:11 AM
Subject: [EMHL] A question

Dear Hyacinthists,
what is the nature of the symmedian point of a triangle relatively to his orthic triangle?

Sincerely
Jean-Louis

[Non-text portions of this message have been removed]

[Non-text portions of this message have been removed]
• ... The orthic triangle of the ex-central triangle IaIbIc is the original ABC. So if the P-point of IaIbIc is the Q-point of ABC, then the P-point of ABC is
Message 7 of 10 , Jul 13, 2012
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> From a quick calculation I find that K is the Mittenpunkt of orthic triangle
> when ABC is acute.
>
> Best regards,
>
> Francisco Javier.
>
> Jean-Louis Ayme wrote:
>>
>> Dear Hyacinthists,
>> what is the nature of the symmedian point of a triangle relatively toï¿½his
> orthic triangle?
>>
>> Sincerely
>> Jean-Louis

The orthic triangle of the ex-central triangle IaIbIc is the original ABC. So if the P-point of IaIbIc is the Q-point of ABC, then the P-point of ABC is the Q-point of the orthic triangle of ABC. If P=symmedian, the above "quick calxculation" gives Q=Mitten..And if P=orthocenter then Q=incenter, which is geometrically ttrivial. A few other cases I tried didn't give anything recognizable.
--
Barry Wolk
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