## Re: A question regarding table 14 of Bernard Gibert

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• Dear Hauke, What is it exactly you want? Do you demand I make a triangle circumscribe or inscribe a given triangle? If so, I have the similar problem.
Message 1 of 16 , Dec 8, 2008
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Dear Hauke,

What is it exactly you want?
Do you demand I make a triangle circumscribe or inscribe a given
triangle? If so, I have the similar problem. Equilateral triangles are
especially nice ones to begin with.

Sincerely, Jeff
• Dear Nikos, In message #16970 you gave a proof how to construct a triangle A B C inscrided in a given triangle ABC. But now, how to construct A B C
Message 2 of 16 , Dec 8, 2008
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Dear Nikos,

In message #16970 you gave a proof how to construct a triangle A'B'C'
inscrided in a given triangle ABC. But now, how to construct A'B'C'
circumscribed to ABC?

Sincerely, Jeff
• Dear Nikos and friends, Sorry, I meant inscribed triangle A B C . Also, I meant similar triangles A B C . Sincerely, Jeff
Message 3 of 16 , Dec 8, 2008
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Dear Nikos and friends,

Sorry, I meant 'inscribed' triangle A'B'C'. Also, I meant similar
triangles A'B'C'.
Sincerely, Jeff

--- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
>
> Dear Nikos,
>
> In message #16970 you gave a proof how to construct a triangle A'B'C'
> inscrided in a given triangle ABC. But now, how to construct A'B'C'
> circumscribed to ABC?
>
> Sincerely, Jeff
>
• ... are ... Circumscribe an equilateral triangle A B C around ABC such that ABC is a Cevian triangle of A B C with respect to some point O (which coordinates
Message 4 of 16 , Dec 9, 2008
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--- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
>
> Dear Hauke,
>
> What is it exactly you want?
> Do you demand I make a triangle circumscribe or inscribe a given
> triangle? If so, I have the similar problem. Equilateral triangles
are
> especially nice ones to begin with.
>
Circumscribe an equilateral triangle A'B'C' around ABC
such that ABC is a Cevian triangle of A'B'C' with
respect to some point O (which coordinates I desperately
try to compute for the Kimberling triangle but for each
of my NSolve[] tries I get different results. :-O )

Hauke
• Dear Jeff and Hauke, In a triangle ABC we can inscribe infinitely many triangles A1B1C1 that are similar to a given triangle A B C . The method is the
Message 5 of 16 , Dec 9, 2008
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Dear Jeff and Hauke,
In a triangle ABC we can inscribe infinitely
many triangles A1B1C1 that are similar to
a given triangle A'B'C'. The method is the following:
In message #16970 we have seen how to construct the
unique point P such that the pedal triangle XYZ of P
is similar to A'B'C'.
The line PX rotated by an angle w meets BC at the point A1.
The line PY rotated by an angle w meets CA at the point B1.
The line PZ rotated by an angle w meets AB at the point C1.
The triangle A1B1C1 is similar to A'B'C' and w is arbitrary.
If A1B1C1 is a cevian then tan(w) must be the root of
a 3-degree equation. Hence there are 3 such points.

If the triangle A'B'C' is equilateral then P is not unique.
There are two such points X(15), X(16).

If Q is the isogonal conjugate of P then the antipedal triangle
X'Y'Z' of Q wrt ABC is a triangle circumscribed to ABC and
similar to A'B'C'. As previously the rotated QA meets Y'Z' at
a point A2. Similarly we find B2, C2 and the triangle A2B2C2
is a triangle similar to ABC and X'Y'Z' is circumscribed to A2B2C2.
If A'B'C' is an equilateral triangle the points Q are the points
X(13) and X(14).
We can find that the triangles A2B2C2 and X'Y'Z' are
perspective iff tan(w) is the root of a 3-degree equation.
Hence there are 6 points such that Hauke wants to find
and the barycentrics must be complicated.

Best regards

> Dear Nikos and friends,
>
> Sorry, I meant 'inscribed' triangle
> A'B'C'. Also, I meant similar
> triangles A'B'C'.
> Sincerely, Jeff

> > Dear Nikos,
> >
> > In message #16970 you gave a proof how to construct a
> triangle A'B'C'
> > inscrided in a given triangle ABC. But now, how to
> construct A'B'C'
> > circumscribed to ABC?
> >
> > Sincerely, Jeff

> Circumscribe an equilateral triangle A'B'C'
> around ABC
> such that ABC is a Cevian triangle of A'B'C'
> with
> respect to some point O (which coordinates I desperately
> try to compute for the Kimberling triangle but for each
> of my NSolve[] tries I get different results. :-O )
>
> Hauke

>
>
>
>

___________________________________________________________
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• Dear Hauke, I had the same problem with constructing the equilateral circumscribed perspective triangle. Thanks to Jean Pierre Ehrmann who gave me a wonderful
Message 6 of 16 , Dec 10, 2008
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Dear Hauke,

I had the same problem with constructing the equilateral circumscribed perspective triangle.
Thanks to Jean Pierre Ehrmann who gave me a wonderful solution I was able to construct the center at last.
See Hyancinthos message 16394.
With this message I calculated the ETC-coordinate being 1,9094098594 (does it match any of your trials?).
This coordinate does not match any ETC-center.
For as far as I can see there are hardly relations with any other center.
Except that this new point is on the line X16.X370.
It is also situated on the conic ABC-X14-X370.
This is good news for center X370 because sofar X370 did not have any relation with any other ETC-center.

This relationship with the point we are talking about is not strange,
because:X370 = the Equilateral Cevian Triangle Point
The center we are talking about is the Equilateral AntiCevian Triangle Point.

Best regards,

Chris van Tienhoven

----- Original Message -----
From: Hauke Reddmann
To: Hyacinthos@yahoogroups.com
Sent: Tuesday, December 09, 2008 4:47 PM
Subject: [SPAM][EMHL] Re: A question regarding table 14 of Bernard Gibert

--- In Hyacinthos@yahoogroups.com, "Jeffrey Brooks" <cu1101@...> wrote:
>
> Dear Hauke,
>
> What is it exactly you want?
> Do you demand I make a triangle circumscribe or inscribe a given
> triangle? If so, I have the similar problem. Equilateral triangles
are
> especially nice ones to begin with.
>
Circumscribe an equilateral triangle A'B'C' around ABC
such that ABC is a Cevian triangle of A'B'C' with
respect to some point O (which coordinates I desperately
try to compute for the Kimberling triangle but for each
of my NSolve[] tries I get different results. :-O )

Hauke

[Non-text portions of this message have been removed]
• Dear Hyacinthists two little remarks about the equilateral cevian or precevian triangles : If the cevian (or the precevian triangle) of P is equilateral with
Message 7 of 16 , Dec 10, 2008
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Dear Hyacinthists
two little remarks about the equilateral cevian or precevian triangles :
If the cevian (or the precevian triangle) of P is equilateral with
center W, then the line PW goes through the circumcenter.
There allways exist 3 equilateral cevian (or precevian) triangles with
the opposite orientation of ABC; there exists 1,2 or 3 cevian (or
precevian) triangles with the orientation of ABC. The centers of these
triangles (at least 8, at most 12) lie on the Napoleon cubic.
Best regards from France. Jean-Pierre
• Dear Hyacinthists I wrote ... I m sorry. This is true only for the equilateral cevian triangles. With my apologizes. Jean-Pierre
Message 8 of 16 , Dec 10, 2008
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Dear Hyacinthists
I wrote
> two little remarks about the equilateral cevian or precevian
triangles :
> If the cevian (or the precevian triangle) of P is equilateral with
> center W, then the line PW goes through the circumcenter.

I'm sorry. This is true only for the equilateral cevian triangles.
With my apologizes. Jean-Pierre
• Dear Jean-Pierre and Nikos, [ND] Hence there are 6 points such that Hauke wants to find [JPE] There allways exist 3 equilateral cevian (or precevian) triangles
Message 9 of 16 , Dec 11, 2008
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Dear Jean-Pierre and Nikos,

[ND]
Hence there are 6 points such that Hauke wants to find

[JPE]
There allways exist 3 equilateral cevian (or precevian) triangles with
the opposite orientation of ABC; there exists 1,2 or 3 cevian (or
precevian) triangles with the orientation of ABC. The centers of these
triangles (at least 8, at most 12) lie on the Napoleon cubic.

Okay, is it 6, 8 or 12 and what determines this number?

Sincerely, Jeff
• ... (does it match any of your trials?). Yes. That s the one. Thread closed :-) Hauke
Message 10 of 16 , Dec 11, 2008
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--- In Hyacinthos@yahoogroups.com, "van Tienhoven" <van10hoven@...> wrote:
>
> Dear Hauke,
>
> See Hyancinthos message 16394.
> With this message I calculated the ETC-coordinate being 1,9094098594
(does it match any of your trials?).

Yes. That's the one. Thread closed :-)

Hauke
• Dear Jeff, when I spoke, about 6 I was speaking only for the inscribed or the circumscribed and this number is the maximum possible. Jean Pierre spoke about
Message 11 of 16 , Dec 11, 2008
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Dear Jeff,
when I spoke, about 6 I was speaking only
for the inscribed or the circumscribed
and this number is the maximum possible.
Jean Pierre spoke about all and thats why
he gave the number 6+6=12.
More precisely he gave the number 8 as
the minimum.
For example if we want the cevian inscribed
triangles that are similar to ABC,
the maximum number is 3 because there is
a unique point with pedal triangle directly similar
to ABC that is the circumcenter O. If we rotate
around O the perpendiculars from O to the sides
of ABC by an angle w we get as intersections
triangles A'B'C' similar to ABC and these
triangles are cevian triangles iff d=tan(w)
is the zero of a 3-degree equation.
This equation has a real root d = 0.
(because O lies on the Darboux cubic and its
pedal triangle is also a cevian triangle)
The other 2 roots are real only when
the triangle ABC has an obtuse angle.
If I remember well
dd = -S_A.S_B.S_C/(S_A + S_B + S_C).

I don't know in the general case what
determines this number of cevian triangles.

Best regards

> Dear Jean-Pierre and Nikos,
>
> [ND]
> Hence there are 6 points such that Hauke wants to find
>
> [JPE]
> There allways exist 3 equilateral cevian (or precevian)
> triangles with
> the opposite orientation of ABC; there exists 1,2 or 3
> cevian (or
> precevian) triangles with the orientation of ABC. The
> centers of these
> triangles (at least 8, at most 12) lie on the Napoleon
> cubic.
>
> Okay, is it 6, 8 or 12 and what determines this number?
>
> Sincerely, Jeff
>

___________________________________________________________
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• Dear Jeff, I wrote ... Sorry. The correct equation is d.d.S_A.S_B.S_C + S.S.(S_A + S_B + S_C) = 0 Best regards Nikos Dergiades
Message 12 of 16 , Dec 11, 2008
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Dear Jeff,
I wrote

> If I remember well
> dd = -S_A.S_B.S_C/(S_A + S_B + S_C).

Sorry. The correct equation is

d.d.S_A.S_B.S_C + S.S.(S_A + S_B + S_C) = 0

Best regards