## Re: [EMHL] Re: Morley's double miracle

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• Dear Gilles You are right. Sketchpad and perhaps Cabri (I don t know it) are very good programs as visual constructors of geometric figures driven by the mouse
Message 1 of 10 , Feb 24, 2002
Dear Gilles

You are right. Sketchpad and perhaps Cabri
(I don't know it)
are very good programs as visual constructors
of geometric figures driven by the mouse but
the precision I think is very bad.
I would like to have a program that would be
a combination of Sketchpad and Maple V that
has very good precision and ability of symbolic
(not only float point) calculations.

You are right because I have I a proof that the
opposite incenters I1, I'1and the B'C' side of Morley's
triangle are not perpendicular. Only in the case that
cos(x/2)sin(60+x) = cos(y/2)cos(z/2) where
x=A/3 y=B/3 z=C/3 it is true, if I don't have an error
in my calculations.

The same I think (I have not a proof) for the lines
I1I'1, I2I'2, I3I'3. I think they are not concurrent.

Best regards
-----������ ������-----
���: Boutte Gilles <g.boutte@...>
����: Hyacinthos@yahoogroups.com <Hyacinthos@yahoogroups.com>
����������: ������� 6, 2002 11:27 PM
����: Re: [EMHL] Re: Morley's double miracle

Sorry, I don't have a proof, only the values of the angles with Cabri.

Regards,

Gilles

> > I've made an accurate sketch with Cabri, which answers : lines through
> > opposite incenters and sides of Morley's triangle are not perpendicular.
> > I've drawn some triangles, for which the angles of theses lines are
> from
> > 89.8 to 90.1 !!!
>
> I didn't think of measuring the angle. I was just looking at it by eye.
> I tried the same thing with Sketchpad and can confirm that the
> angle varies from 89.8 to 90.1.
>
> I also took the 3 points of intersection of pairs of lines
> joining opposite incenters and measured the area of the triangle
> they form. To 3 decimal places, I get 0.000 all the time.
> So the 3 lines _do_ look concurrent. Do we have a proof yet?

[Non-text portions of this message have been removed]
• Sorry, I don t have a proof, only the values of the angles with Cabri. Regards, Gilles
Message 2 of 10 , Mar 6, 2002
Sorry, I don't have a proof, only the values of the angles with Cabri.

Regards,

Gilles

> > I've made an accurate sketch with Cabri, which answers : lines through
> > opposite incenters and sides of Morley's triangle are not perpendicular.
> > I've drawn some triangles, for which the angles of theses lines are
> from
> > 89.8 to 90.1 !!!
>
> I didn't think of measuring the angle. I was just looking at it by eye.
> I tried the same thing with Sketchpad and can confirm that the
> angle varies from 89.8 to 90.1.
>
> I also took the 3 points of intersection of pairs of lines
> joining opposite incenters and measured the area of the triangle
> they form. To 3 decimal places, I get 0.000 all the time.
> So the 3 lines _do_ look concurrent. Do we have a proof yet?
• ... The maple program you are looking for might be found at http://www.math.rutgers.edu/~zeilberg/PG/gt.html
Message 3 of 10 , Mar 8, 2002
> I would like to have a program that would be
> a combination of Sketchpad and Maple V that
> has very good precision and ability of symbolic
> (not only float point) calculations.

The maple program you are looking for might be found at
http://www.math.rutgers.edu/~zeilberg/PG/gt.html
• ... Dear Nikos and Hyacinthians: Even if they are, I still grope for the proof. Let s just summarize the configuration as I see it with regard to several
Message 4 of 10 , Mar 8, 2002
>
> The same I think (I have not a proof) for the lines
> I1I'1, I2I'2, I3I'3. I think they are not concurrent.

Dear Nikos and Hyacinthians:

Even if they are, I still grope for the proof. Let's just
summarize the configuration as I see it with regard to several
present or plausible concurrencies in that configuration.

There's triangle ABC and Morley's triangle A'B'C'. In each
of the six smaller triangles we choose a point - the incenter
or the circumcenter - and connect them to the opposite one.
Let's then denote the points in triangles that share a single
vertex with ABC as O_A, O_B, O_C, and in the triangles that
share a single vertex with A'B'C' as Q_A, Q_B, Q_C. Then we
have two more triangle: O_A O_B O_C (shortened to OOO) and,
similarly, QQQ. So there are four triangles, and the question
is which ones are perspective. The question could be extended
from Morley's to a more general configuration of isogonal lines.

#| triangle pair | perspective | isogonal | support
| | | generalization |
---------------------------------------------------------------
| | | |
1| ABC/A'B'C' | yes | yes | (*)
| | | |
---------------------------------------------------------------
Incenter case
---------------------------------------------------------------
| | | |
2| ABC/OOO | yes | yes | :-)
3| ABC/QQQ | yes | yes | (*)
4| A'B'C'/OOO | yes | ? |
5| A'B'C'/QQQ | ? | ? |
6| OOO/QQQ | ? | ? |
| | | |
---------------------------------------------------------------
Circumcenter case
---------------------------------------------------------------
| | | |
7| ABC/OOO | no | no | Sketchpad, Cabri, Java
8| ABC/QQQ | no | no | Sketchpad, Cabri, Java
9| A'B'C'/OOO | yes | ? | :-)
10| A'B'C'/QQQ | yes | ? | ;-)
11| OOO/QQQ | yes | ? | N. Dergiades
| | | |
---------------------------------------------------------------

(*) This is a well known fact, see for example,
http://www.cut-the-knot.com/triangle/Morley/Morley.html

There is of course the fifth triangle A"B"C", that Antreas mentioned
in one of the recent posts, which is perspective to ABC and A'B'C',
and is an out standing case of the isogonal generalization.

All the best,
Alexander Bogomolny
• Dear Alexander, Thanks for your table. I prefer figures rather than tables :-) A B C = Morley Triangle of ABC. A / / / / / C B / A
Message 5 of 10 , Mar 8, 2002
Dear Alexander,

I prefer figures rather than tables :-)

A'B'C' = Morley Triangle of ABC.

A
/\
/ \
/ \
/ \
/ C' B' \
/ A' \
/ \
B--------------C

Denote:

Ia, Ib, Ic : Incenters of AB'C', BC'A', CA'B', resp.

I'a, I'b, I'c : Incenters of A'BC, B'CA, C'AB, resp.

Oa, Ob, Oc : Circumcenters of AB'C', BC'A', CA'B', resp.

O'a, O'b, O'c : Circumcenters of A'BC, B'CA, C'AB, resp.

Ha, Hb, Hc : Orthocenters of AB'C', BC'A', CA'B', resp.

H'a, H'b, H'c : Orthocenters of A'BC, B'CA, C'AB, resp.

Perspectivities and Not:

1. ABC, IaIbIc: Perspective
[Perspector: I of ABC]

2. ABC, HaHbHc : Not perspective

3. ABC, OaObOc : Not perspective

4. ABC, I'aI'bI'c : Perspective
[Baryc. Perspector: (1/(cotA - cot(A/6)) ::) ]

5. ABC, H'aH'bH'c : Not-Perspective

6. ABC, O'aO'bO'c : Not-Perspective

-----------

1'. A'B'C', IaIbIc: Perspective
[Baryc. Perspector wrt A'B'C': ((cot60 + cot(30 + (A/6)) ::)
wrt ABC ?]

2'. A'B'C', HaHbHc : Perspective
[Baryc. Perspector wrt A'B'C': (1/(cot60 + cot(30 - (A/3)) ::)
wrt ABC ?]

3'. A'B'C', OaObOc : Perspective
[ Perspector : the center of A'B'C' = X_356 ]

4'. A'B'C', I'aI'bI'c : Not-Perspective

5'. A'B'C', H'aH'bH'c : Not-Perspective

6'. A'B'C', O'aO'bO'c : Perspective
[ Perspector : the center of A'B'C' = X_356 ]

-----------

a. ABC, A'B'C' : Perspective
[ Perspector X_357 ]

b. IaIbIc, I'aI'bI'c : ???
[I have computed the normals of the points with respect to A'B'C'
but I don't know if the det. of the matrix of the lines IaI'a, etc,
is zero or not]

c. OaObOc, O'aO'bO'c : Perspective
[ Perspector : the center of A'B'C' = X_356 ]

d. HaHbHc, H'aH'bH'c : ???

For the above Not-perspectivities there are simple proofs
(not just geometry programs evidences)

Good Night!

Antreas
• ... I guess tables are too square. But I was wrong in another respect. While triangle A B C is of course a specific isogonal case, we may consider it in its
Message 6 of 10 , Mar 8, 2002
Dear Antreas:

>
> I prefer figures rather than tables :-)

I guess tables are too square.

But I was wrong in another respect. While
triangle A"B"C" is of course a specific
isogonal case, we may consider it in its
own right together with A'B'C'. Then
the table/list may be expanded.

> 4. ABC, I'aI'bI'c : Perspective
> [Baryc. Perspector: (1/(cotA - cot(A/6)) ::) ]

It's also curious that the fact itself can be
proved in projective geometry, since the isogonal
conjugation is a perspectivity of the pencil of
lines through the vertex.

> For the above Not-perspectivities there are simple proofs
> (not just geometry programs evidences)

I am very curious to get an example when a result in Morley's
configuration does not generalize to the isogonal case.

Talking of which, in some sense Morley's configuration
generalizes that of Napoleon and is generalizable itself.

For a given triangle A'B'C', form triangles A'B'C, B'C'A and
C'A'B. How? For example,

1. Napoleon: All angles are 60 degrees.
2. Kiepert: Similar isosceles triangles.
3. Isogonal conjugation: the angles of the add-on triangles
that share a vertex are equal.

In all three cases the lines AA', BB' and CC' are concurrent.
Now, the same appears to be true if

4. (Angles) B'A'C = AC'B', A'C'B = CB'A', C'A'B = AB'C'

What is this case? What framework does it belong to?

> Good Night!
>

Good morning to you.

Alexander
• Dear Alexander, ... A f w C --------------B q / q / / / / / w / f A B
Message 7 of 10 , Mar 9, 2002
Dear Alexander,

[AB]:
> For a given triangle A'B'C', form triangles A'B'C, B'C'A and
> C'A'B. How? For example,
>
> 1. Napoleon: All angles are 60 degrees.
> 2. Kiepert: Similar isosceles triangles.
> 3. Isogonal conjugation: the angles of the add-on triangles
> that share a vertex are equal.
>
> In all three cases the lines AA', BB' and CC' are concurrent.
> Now, the same appears to be true if
>
> 4. (Angles) B'A'C = AC'B': = f, A'C'B = CB'A' := q, C'A'B = AB'C' := w
>
> What is this case? What framework does it belong to?
>

A

f w
C'--------------B'
q \ / q
\ /
\ /
\ /
\ /
\ /
w \ / f
A'
B C

ABC, A'B'C' are perspective <==>

cotB' + cotw cotC' + cotq cotA' + cotf
------------ * ------------- * ------------- = 1
cotC' + cotf cotA' + cotw cotB' + cotq

In general, not true.

Special Cases: f = q = w or A' = B' = C' [= 60]

Antreas
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