## Re: [EMHL] X(1986) generalized

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• 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,
Message 1 of 16 , Jun 5, 2006
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:
>
>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.

We have

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.
>
>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* ?

APH
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• Dear Antreas & Tuan, [APH] ... For point P = (x:y:z), the point of concurrence, Q, is a^2 ((c^4 - a^2 c^2) x y - (a^4 - a^2 b^2 - a^2 c^2) y z + (b^4 - a^2
Message 2 of 16 , Jun 5, 2006
Dear Antreas & Tuan,

[APH]
> Let ABC be a triangle, P = (x:y:z) a point and A'B'C' the pedal
> triangle of P.
>
> Denote
> Ab, Ac : the reflections of A in B',C'
> Bc, Ba : the reflections of B in C',A'
> Ca, Cb : the reflections of C in A',B'
>
> If P = H, then the Nine Point Circles of AAbAc, BBcBa, CCaCb
> concur (at X(1986))
>
> Now, which points else have that property?
>
> Obviously for P = O (trivial). How about P = I ?
>
> If the locus is the entire plane, then which are the coordinates
> of the point of concurrence?

For point P = (x:y:z), the point of concurrence, Q, is

a^2 ((c^4 - a^2 c^2) x y - (a^4 - a^2 b^2 - a^2 c^2) y z +
(b^4 - a^2 b^2) z x - b^2 c^2 x^2)*
(a^2 b^2 c^2 x - b^2 z (S^2 - SB^2) - c^2 y (S^2 - SC^2))::

For points ETC that map to other points in ETC we have ..
X(4) -> X(1986)
X(20) -> Anticomp X(974)
X(23) -> Comp X(323)
X(36) -> X(1737)
X(186) -> X(403)
X(187) -> X(230)
X(1691) -> X(1692)

P = I, Q is
a (b c - 2 SA) (a^3 b c + b (SB^2 - S^2) + c (SC^2 - S^2)) ::
Intersection of lines
{1,104},{7,80},{11,113},{36,214},{46,100},{119,912}, ...

P = G, Q is the intersection of lines
{25,110},{51,542},{113,403}, ...

P = N, Q is the intersection of lines
{51,265},{52,110},{113,403},{389,546},{399,568}, ...

P = K, Q is the intersection of lines
{6,13,14,...},{99,193},{114,230},{187,524}, ...

For P on the Euler line, Q is on line {113,403,1986}
For P on the Brocard axis, Q is on line {114,230,1692}
For P on the OI line, Q is on line {119,912,1737}

[QTB]
>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.

The point at infinity is
a^2 (y + z) SA - x (S^2 + SB SC) ::
the intersection of line[X(3) P] with infinity, i.e. line[AOa] is
parallel to line [X(3)P]
The triangle OaObOc is ABC shifted by vector [X(3), P], so the
radius of the circles (Oa), (Ob), (Oc) is the distance |X(3) P|.

Suppose P is on the circumcircle.
The 3 other pairs of intersections of AAbAc, BBcBa, CCaCb are
colinear (the Simson line of P) and bisect PQ perpendicularly.
As P moves around the Circumcircle, the envelope of PQ traces out
the Steiner Deltoid of the Anticomplementary triangle.

The triangle formed by the centers of the NP circles of AAbAc,
BBcBa, CCaCb is similar to ABC. It is perspective to ABC for P on
the Neuberg cubic.

Best regards,
Peter.
• Dear Antreas, ... Should have mentioned it, Yes .. Barycentric. Best regards, Peter.
Message 3 of 16 , Jun 5, 2006
Dear Antreas,

>For point P = (x:y:z), the point of concurrence, Q, is
>
>a^2 ((c^4 - a^2 c^2) x y - (a^4 - a^2 b^2 - a^2 c^2) y z +
>(b^4 - a^2 b^2) z x - b^2 c^2 x^2)*
>(a^2 b^2 c^2 x - b^2 z (S^2 - SB^2) - c^2 y (S^2 - SC^2))::

>THANKS!!

>They are barycentric coordinates. Right?

Should have mentioned it, Yes .. Barycentric.

Best regards,
Peter.
• Dear Tuan, ... AAbAc, BBcBa, CCaCb are always concurrent at one common point E. ... with A B C at P1 ... through midpoints of each of them. We denote these
Message 4 of 16 , Jun 5, 2006
Dear Tuan,

> It is very interesting generalization! I see some thing may be
> 1. With any point P on the plane, the nine point circles of
AAbAc, BBcBa, CCaCb are always concurrent at one common point E.
> 2. Three lines AbAc, BcBa, CaCb bound one triangle homothetic
with A'B'C' at P1
> 3. Three perpedicular lines from P to AbAc, BcBa, CaCb pass
through midpoints of each of them. We denote these midpoints as Ma,
Mb, Mc respectively.
> 4. Denote center of rectangular circumhyperbola of MaMbMc
passing through P as Q so we have:
> - P, Q, E are collinear
> - Q is midpoint of PE.

1. as per #13194
2. P1 = a^2 (c^2 x + z SB) (b^2 x + y SC) ::
This point is the homothetic center of the pedal triangle of P and
the antipedal triangle of the isogonal of P.
3. Ma, Mb ,Mc are on the circles (Na),(Nb),(Nc) respectively. (Na)
being the NP circle of AAbAc.
4. Rectangular circumhyperbola of MaMbMc passing through P, also
passes thru A'B'C'

Best regards,
Peter.
• Dear Antreas, The construction problem is interesting itself but please check to see when applying to AAbAc: if B , C are midpoints of HAb, HAc then B , C
Message 5 of 16 , Jun 5, 2006
Dear Antreas,
The construction problem is interesting itself but please check to see when
applying to AAbAc: if B', C' are midpoints of HAb, HAc then B', C' could be not on AC, AB or Ab, Ac could be not on AB, AC?
Best regards,
Bui Quang Tuan

"Antreas P. Hatzipolakis" <xpolakis@...> wrote:
Dear Tuan

So far we have seen the 3+3 points the 9 Point Circle is
passing through.

How about the rest three ones?

But first a triangle construction problem:

Let ABC be a triangle, H its orthocenter,
and EaEbEc its Euler triangle (ie Ea,Eb,Ec = midpoints
of AH, BH, CH, resp.)

Problem:
To construct triangle ABC if are given the points A, Eb, Ec.

Now back in the Nine point Circles concurrence.

Let ABC be a triangle, and A'B'C' a triangle
inscribed in ABC.

Construct triangle AAbAc so that B',C' be the points
Eb,Ec of AAbAc (as defined above)

Similarly construct triangles BBcBa, CCbCa.

Are the Nine Point Circles of AAbAc, BBcBa, CCaCb concurrent?

Antreas

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• Dear Peter ... THANKS!! They are barycentric coordinates. Right? APH --
Message 6 of 16 , Jun 5, 2006
Dear Peter

>For point P = (x:y:z), the point of concurrence, Q, is
>
>a^2 ((c^4 - a^2 c^2) x y - (a^4 - a^2 b^2 - a^2 c^2) y z +
>(b^4 - a^2 b^2) z x - b^2 c^2 x^2)*
>(a^2 b^2 c^2 x - b^2 z (S^2 - SB^2) - c^2 y (S^2 - SC^2))::

THANKS!!

They are barycentric coordinates. Right?

APH

--
• Dear Tuan So far we have seen the 3+3 points the 9 Point Circle is passing through. How about the rest three ones? But first a triangle construction problem:
Message 7 of 16 , Jun 5, 2006
Dear Tuan

So far we have seen the 3+3 points the 9 Point Circle is
passing through.

How about the rest three ones?

But first a triangle construction problem:

Let ABC be a triangle, H its orthocenter,
and EaEbEc its Euler triangle (ie Ea,Eb,Ec = midpoints
of AH, BH, CH, resp.)

Problem:
To construct triangle ABC if are given the points A, Eb, Ec.

Now back in the Nine point Circles concurrence.

Let ABC be a triangle, and A'B'C' a triangle
inscribed in ABC.

Construct triangle AAbAc so that B',C' be the points
Eb,Ec of AAbAc (as defined above)

Similarly construct triangles BBcBa, CCbCa.

Are the Nine Point Circles of AAbAc, BBcBa, CCaCb concurrent?

Antreas

[Bui Quang Tuan]

> 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.
>

[APH]

>>I think that, more generally, 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.
>
>We have
>
>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.
>>
>>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* ?
>
>APH

--

--
• ... could be not on AC, AB or Ab, Ac could be not on AB, AC? Dear Tuan Yes. The points Ab,Ac etc could not be on the sidelines of ABC. We are trying the Nine
Message 8 of 16 , Jun 5, 2006
--- In Hyacinthos@yahoogroups.com, Quang Tuan Bui <bqtuan1962@...> wrote:
>
> Dear Antreas,
> The construction problem is interesting itself but please check to
> see when
> applying to AAbAc: if B', C' are midpoints of HAb, HAc then B', C'
could be not on AC, AB or Ab, Ac could be not on AB, AC?

Dear Tuan

Yes. The points Ab,Ac etc could not be on the sidelines of ABC.

We are trying the Nine Point Circles be passing
through A',B',C', regardless the positions of Ab, Ac etc.

Antreas

PS Your H above is NOT the H of ABC. It is the H of AAbAc.
• Dear Antreas, [APH] ... 1. As per 13194 2. Q* = (a^2 (Y Z - X^2) - X (b^2 - c^2) (Y - Z))* (a^2 Y^2 Z^2 - b^2 Z^2 X^2 - c^2 X^2 Y^2 - 2 SA X Y Z (X - Y - Z))::
Message 9 of 16 , Jun 5, 2006
Dear Antreas,

[APH]
>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.

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

>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* ?

1. As per 13194
2. Q* = (a^2 (Y Z - X^2) - X (b^2 - c^2) (Y - Z))*
(a^2 Y^2 Z^2 - b^2 Z^2 X^2 - c^2 X^2 Y^2 - 2 SA X Y Z (X - Y - Z))::
(barys)

For Q on the circumcircle, Q* = Q.

Best regards
Peter.
• Dear Antreas and all, sorry I didn t followed all your messages about this problem but I think that you wrote as a generalization: [APH] ... If A is the
Message 10 of 16 , Jun 6, 2006
Dear Antreas and all, sorry
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.

If A" is the reflection of A wrt B'C' and similarly define B", C"
the Nine Point circle of AAbAc passes through the points B', C', A"
and hence is the reflection Ka' of the circumcircle Ka of triangle AB'C'.
Since the triangle ABC is a Schaal triangle wrt A'B'C' as defined by
Darij Grinberg in 7414 and again by Floor in 10386
the circles Ka, Kb, Kc are concurrent at a point P
the Schaal perspector of ABC relative to A'B'C'.
Since the triangle A"B"C" is also a Schaal
triangle wrt A'B'C' the Nine Point circles in our problem or the
circles Ka', Kb', Kc' are concurrent at a point P'
the Schaal perspector of A"B"C" relative to A'B'C'.
From this configuration I call this point P' the Schaal reflection of P
relative to triangle A'B'C'.

Let AoBoCo be the pedal triangle of P relative to ABC.
It is known that the triangles A'B'C' and AoBoCo are similar.
Let Po be the Schaal reflection of P relative to AoBoCo.
If the triangle A'B'C' is moving and remains similar to itself
(The construction of these triangles is given by rotation
of the lines PAo, PBo, PCo about P of an arbitrary angle w)
then the Schaal perspector of ABC relative to A'B'C' is
constantly the point P, and the locus of P' the Scaal reflection of P
relative to
A'B'C' is the perpendicular line to PPo at Po.

Best regards
Nikos
• Dear Nikos [ND] ... There was another generalization with Ab, Ac etc being feet of altitudes, instead of traces of medians. Let s try now to generalizing it
Message 11 of 16 , Jun 7, 2006
Dear Nikos

[ND]
>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.

There was another generalization with Ab, Ac etc
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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