- 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

Nikos Dergiades

-----������ ������-----

���: 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

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> > 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?

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

[AB]:> For a given triangle A'B'C', form triangles A'B'C, B'C'A and

A

> 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?

>

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