- Dear Antreas,

Yes, I see and try to study the general case. Now we have six cases:

1. General case: any INSCRIBED triangle A'B'C' (circle concurrency only)

Ab, Ac, Bc, Ba, Ca, Cb are defined so A', B', C' as:

a. Medial traces

b. Altitude foots

Of Nine Point Circles

2. PEDAL triangle A'B'C' (P and P*)

Ab, Ac, Bc, Ba, Ca, Cb are defined so A', B', C' as:

a. Medial traces

b. Altitude foots

Of Nine Point Circles

3. CEVIAN triangle A'B'C' (Q and Q*)

Ab, Ac, Bc, Ba, Ca, Cb are defined so A', B', C' as:

a. Medial traces

b. Altitude foots

Of Nine Point Circles

If I understand you well?

I think we should discover what is essential mechanism of circle concurrency in general case when A'B'C' is any INSCRIBED triangle. There are a lot of works should do if there is not a general method to resolve all these cases.

In case of PEDAL triangle and MEDIAL traces (2a) there are some things interesting:

We denote three circumcircles of ABaCa, BCbAb, CAcBc by (Oa), (Ob), (Oc) respectively so these three circles are congruent. The triangle of theire centers OaObOc are congruent with ABC and homothetic with ABC at infinity.

Best regards,

Bui Quang Tuan

"Antreas P. Hatzipolakis" <xpolakis@...> wrote: [Antreas]

>I think that, more generally, we have:

We have

>

>Let ABC be a triangle and A'B'C' a triangle INSCRIBED in ABC.

>

>Denote

>Ab, Ac : the reflections of A in B',C', resp.

>Bc, Ba : the reflections of B in C',A', resp.

>Ca, Cb : the reflections of C in A',B', resp.

B', C' are midpoints of AAb, AAc

C', A' are midpoints of BBc, BBa

A', B' are midpoints of CCa, CCb

(The Nine Point Circle of AAbAc passes through B',C'

and similarly the others)

Now, we can define Ab, Ac etc such that

B', C' etc be feet of altitudes instead traces of medians

That is:

The perpendicular to AB at C' intersects AC at Ab

The perpendicular to AC at B' intersects AB at Ac

The perpendicular to BC at A' intersects BA at Bc

The perpendicular to BA at C' intersects BC at Ba

The perpendicular to CA at B' intersects CB at Ca

The perpendicular to CB at A' intersects CA at Cb

We have again that:

>The Nine Point Circles of AAbAc, BBcBa, CCaCb are concurrent.

APH

>

>Now, we have two interesting mappings of the triangle plane

>into itself,namely:

>

>1. Let A'B'C' be the PEDAL triangle of a point P = (x:y:z),

>and P* the point of concurrence.

>

>2. Let A'B'C' be the CEVIAN triangle of a point Q = (X:Y:Z),

>and Q* the point of concurrence.

>

>PROBLEM: Which are the coordinates of the points P*,Q* ?

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

[ND]>I didn't followed all your messages about this problem

There was another generalization with Ab, Ac etc

>but I think that you wrote as a generalization:

>

>[APH]

>>A'B'C' is an arbitrary inscribed triangle in ABC.

>>Ab, Ac : the reflections of A in B',C', resp.

>>Bc, Ba : the reflections of B in C',A', resp.

>>Ca, Cb : the reflections of C in A',B', resp.

>>The Nine Point Circles of AAbAc, BBcBa, CCaCb are concurrent.

being feet of altitudes, instead of traces of medians.

Let's try now to generalizing it further

Firstly a problem:

Let AB'C' be a triangle.

To construct triangle ABC such that: the points B,C be

on AC',AB', resp. and the intersection of BB',CC' be the

point P = (x:y:z) with given homog. coordinates.

In other words:

Let ABC be a triangle and A'B'C' the cevian triangle of a given

point P = (x:y:z)

To construct ABC if are given the points A, B', C'

For P = G,H, it's easy.

For P = I: P is the intersection of the int. angle bisector

of (B'AC') and the arc (B'C', 90 + (A/2))

Now the general problem:

Let ABC be a triangle and A'B'C' a triangle

inscribed in ABC.

Construct triangle AAbAc with Ab,Ac on AC,AB resp. such that

B'Ac /\ C'Ab := Pa = (x:y:z) with respect AAbAc

Similarly construct triangles BBcBa, CCaCb.

For which P's the Nine Point Circles (or the Circumcircles)

of AAbAc, BBcBa, CCaCb are concurrent ?

The two generalizations above (for NPCs), are for P = G,H.

Antreas

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