- Antreas wrote in message 1 of Jan 1 :

? Theorem

The lines AF_a, BF_b, CF_c are concurrent where the nine-point circle

touches the excircles at F_a, F_b, F_c.

This is true. The intersection of the three lines is barycenter of A, B, C

with coefficients :

(b + c)^2 (a + b - c) (a + c - b)

(c + a)^2 (b + c - a) (b + a - c)

(a + b)^2 (c + a - b) (c + b - a)

where a, b, c are the legths of the sides BC, CA, AB.

May be there is a special name for this point. Why not the Hatzipolakis

point or the first Hatzipolakis point - many other ones will perhaps

follow -?

Proof by brute force :

A barycentric equation for the nine-point circle is

(b^2 + c^2 - a^2) u^2 + (c^2 + a^2 - b^2) v^2 + (a^2 + b^2 - c^2) w^2 - 2

c^2 u v - 2 a^2 v w - 2 b^2 w u = 0

and for the A-excircle

(a + b + c)^2 u^2 + (a + b - c)^2 v^2 + (a + c - b)^2 w^2 + 2 (a + b + c) (a

+ b - c) u v - 2 (a + b - c) (a + c - b) v w + 2 (a + b +c) (a + c - b) w u

= 0

Hence F_a is barycenter of A, B, C with coefficients

- (b - c)^2 (a + b + c)

(a + c)^2 (a + b - c)

(a + b)^2 (a + c - b)

By symetry, we have, for F_b

(b + c)^2 (b + a - c)

- (c - a)^2 (b + c + a)

(b + a)^2 (b + c - a)

and, for F_c,

(c + b)^2 (c + a - b)

(c + a)^2 (c + b - a)

- (a - b)^2 (c + a + b)

and the result with an easy computation.

Friendly from France.

Jean-Pierre Ehrmann - Dear Jean-Pierre

You wrote:

>Antreas wrote in message 1 of Jan 1 :

This Theorem is well-known. The (?) means that I don't know who first

>? Theorem

>The lines AF_a, BF_b, CF_c are concurrent where the nine-point circle

>touches the excircles at F_a, F_b, F_c.

discovered it.

See:

http://cedar.evansville.edu/~ck6/tcenters/class/feuer.html

>This is true. The intersection of the three lines is barycenter of A, B, C

:-))

>with coefficients :

>(b + c)^2 (a + b - c) (a + c - b)

>(c + a)^2 (b + c - a) (b + a - c)

>(a + b)^2 (c + a - b) (c + b - a)

>where a, b, c are the legths of the sides BC, CA, AB.

>May be there is a special name for this point. Why not the Hatzipolakis

>point or the first Hatzipolakis point - many other ones will perhaps

>follow -?

>Proof by brute force :

Greetings from Athens

>A barycentric equation for the nine-point circle is

>(b^2 + c^2 - a^2) u^2 + (c^2 + a^2 - b^2) v^2 + (a^2 + b^2 - c^2) w^2 - 2

>c^2 u v - 2 a^2 v w - 2 b^2 w u = 0

>and for the A-excircle

>(a + b + c)^2 u^2 + (a + b - c)^2 v^2 + (a + c - b)^2 w^2 + 2 (a + b + c) (a

>+ b - c) u v - 2 (a + b - c) (a + c - b) v w + 2 (a + b +c) (a + c - b) w u

>= 0

>Hence F_a is barycenter of A, B, C with coefficients

>- (b - c)^2 (a + b + c)

>(a + c)^2 (a + b - c)

>(a + b)^2 (a + c - b)

>By symetry, we have, for F_b

>(b + c)^2 (b + a - c)

>- (c - a)^2 (b + c + a)

>(b + a)^2 (b + c - a)

>and, for F_c,

>(c + b)^2 (c + a - b)

>(c + a)^2 (c + b - a)

>- (a - b)^2 (c + a + b)

>and the result with an easy computation.

>Friendly from France.

>Jean-Pierre Ehrmann

Antreas - On Sat, 8 Jan 2000, Jean-Pierre.EHRMANN wrote:

> Antreas wrote in message 1 of Jan 1 :

Antreas - this is in the Greek book on triangle geometry that you

> ? Theorem

> The lines AF_a, BF_b, CF_c are concurrent where the nine-point circle

> touches the excircles at F_a, F_b, F_c.

kindly sent me. It is in fact part of the "desmic" theory. There are

four points Fa,Fb,Fc,Fo where the NPC touches the four incircles,

and four perspectors of ABC with the triangles formed by 3 of them,

namely

Fo' of Fa,Fb,Fc

Fa' of Fo,Fc,Fb

Fb' of Fc,Fo,Fa

Fc' of Fb,Fa,Fo

Then symmetrically, Fo,Fa,Fb,Fc are the perspectors of ABC with

the triangles formed by taking triples from Fo',Fa',Fb',Fc', and these

two quartets form a desmic triple with A,B,C,D, where D (the "desmon")

is where the lines FoFo',FaFa',FbFb',FcFc' meet.

John Conway - on 1/7/00 8:16 PM Jean-Pierre.EHRMANN wrote

>The lines AF_a, BF_b, CF_c are concurrent where the nine-point circle

I call this point the Feuerbach mate but perhaps it deserves a better

>touches the excircles at F_a, F_b, F_c.

>This is true. The intersection of the three lines is barycenter of A, B, C

>with coefficients :

>(b + c)^2 (a + b - c) (a + c - b)

>(c + a)^2 (b + c - a) (b + a - c)

>(a + b)^2 (c + a - b) (c + b - a)

name. It is the second center of similarity of the incircle and the 9pt

circle. This point, along with Fo, is on the line connecting the incenter

to the 9pt circle.

The Feuerbach points are the first centers of similarity between the 9pt

circles and the 4 in/excircles.

Since it is weak and the above formulas depend on the sides, this point

has 3 extra versions, which are themselves the second centers of

similarity of the 9PC and the excircles.

I denote these 4 points Fmo, Fma, Fmb, and Fmc.

The points Fx, Fmx, and ABCN form a desmic triple of quadrangles in the

following way

o a b c

I N A B C

II Fo Fa Fb Fc

III Fmo Fma Fmb Fmc

N, the 9 pt center is the strong point that hold them together (the

"desmon," in Conway's terminology). The four incenter Ix are the harmonic

"halo" accompanying these points.

Steve - I am going to resurrect a discussion which ended two years ago.

Hereby, I refer to #61, #93, #94, #110, #125.

The problem is to prove that the lines AFa, BFb, CFc are concurrent,

where Fa, Fb, Fc are the points of touch of the 9pt circle with the

excircles. I am going to show that all three lines pass through the

point F', the internal center of similtude of the 9pt circle with the

incircle.

In fact, note that the Feuerbach point F is the EXTERNAL CENTER OF

SIMILTUDE of 9pt and incircle, and Fa, Fb, Fc are the INTERNAL

CENTERS OF SIMILTUDE of 9pt and respective excircles. Now,

* F' is the internal center of similtude of 9pt and incircle;

* Fa is the internal center of similtude of 9pt and a-excircle;

* A is the external center of similtude of incircle and a-excircle

(really easy to prove).

By Monge's theorem, the point A, F' and Fa are collinear, i. e. AFa

passes through F'. Similarly, BFb and CFc pass through F', qed.

The point F' is Clark Kimberling's X(12); I call it HARMONICAL

FEUERBACH POINT. Similar points can be defined for the 9pt circle and

the excircles.

Sincerely,

Darij Grinberg