QUADRI 1936 et al

#1894

Dear all,

constructing the 16 n-angle points P(1/4) for a triangle ABC, there is a special point X (see attached file):

Let Ia, Ib, Ic be the excenters of ABC,
... let Ja, Jb, Jc be the incenters of Ia,B,C and A,Ib,C and A,B,Ic,
... the circumcircles of Ja,B,C and A,Jb,C and A,B,Jc have a common point X.

Properties:
... X is a point of the JaJbJc-circumcircle,
... <ABC = 4<AXC mod Pi, <BCA = 4<BXA mod Pi, <CAB = 4<CXB mod Pi,

Let Ma, Mb, Mc be the circumcenters of Ja,B,C and A,Jb,C and A,B,Jc,
... X is the MaMbMc-isogonal conjugate of the ABC-circumcenter,

... X is the intersection of parallels to MiMj through Jk,
... let M be the MaMbMc-circumcenter and H the JaJbJc-orthocenter (= IaIbIc-incenter),
... the ABC-incenter I and the JaJbJc-orthocenter H lie diametral on the circumcircle of MaMbMc,
... the triples Ia, Ja, Ma and Ib, Jb, Mb and Ic, Jc, Mc are collinear with the orthocenter H of JaJbJc,
... the ABC-isogonal conjugate of  X is a point X* on the line I.M.H,
... the Simson line of X wrt JaJbJc is a parallel to I.M.H in half the distance,
... the inversion of X in the MaMbMc-circumcircle is the inversion of X* in the ABC-circumcircle.

The point X is not in ETC!

Best regards Eckart Schmidt

 

PS:

... let Ja´, Jb´, Jc´ be the incenters of I,B,C and A,I,C and A,B,I,

... the intersection of AJa´, BJb´, CJc´ is the Hofstadter point H(1/4).

 

---------------------------

#1895

 

[Eckart Schmidt]:
Let Ia, Ib, Ic be the excenters of ABC,
... let Ja, Jb, Jc be the incenters of Ia,B,C and A,Ib,C and A,B,Ic,
... the circumcircles of Ja,B,C and A,Jb,C and A,B,Jc have a common point X.

 

Dear Eckart,

And by the "main cyclologic theorem" the circumcircles of AJbJc, BJcJa, CJaJb are concurrent (at a point on the circumcircle of ABC)

Antreas P. Hatzipolakis

---------------------------

#1896

[Eckart Schmidt]:
Let Ia, Ib, Ic be the excenters of ABC,
... let Ja, Jb, Jc be the incenters of Ia,B,C and A,Ib,C and A,B,Ic,
... the circumcircles of Ja,B,C and A,Jb,C and A,B,Jc have a common point X.

 

[APH]:

Dear Eckart,

And by the "main cyclologic theorem" the circumcircles of AJbJc, BJcJa, CJaJb are concurrent (at a point on the circumcircle of ABC)

 

 

The same is true if we replace the incenters Ja, Jb, Jc with circumcenters Oa, Ob, Oc or orthocenters Ha, Hb, Hc of IaBC, IbCA, IcAB, resp.

Antreas P. Hatzipolakis

---------------------------

#1897

 

Dear Antreas,

 

thanks for your informations.

The circumcircles of AJbJc, BJcJa, CJaJb intersect in X(3659),

the points Oa, Ob, Oc lie on the circumcircle of ABC,

the circumcircles of ABHc, BCHa, CAHb intersect in X(80).

But I haven´t found the considered special n-angle point X in ETC.

This point has a further interesting property:

X is the Hofstadter point H(-3) of the triangle MaMbMc.

 

Best regards Eckart Schmidt

---------------------------

#1933

[Eckart Schmidt] :

Dear all,

Let Ia, Ib, Ic be the excenters of ABC,
... let Ja, Jb, Jc be the incenters of Ia,B,C and A,Ib,C and A,B,Ic,
... the circumcircles of Ja,B,C and A,Jb,C and A,B,Jc have a common point X.

 

[...]

 

Dear Eckart

The point is now in ETC

X(10215) =  SCHMIDT-LOZADA CYCLOLOGIC CENTER
http://faculty.evansville.edu/ck6/encyclopedia/ETCPart6.html#X10215

Antreas P. Hatzipolakis

 

---------------------------

#1934

 

Dear Antreas,

 

thanks for information, never thought, that this could be possible.

 

There are further cyclologic centers, X(10215) is the 2nd in a row of n-angle points, beginning with the incenter X(1). I shall describe it in the next days.

 

Best regards Eckart Schmidt

 

---------------------------

#1936

 

Dear Antreas,

 

This is not the announced message, but another possibility, to get cyclologic centers analog to the construction of X(10215) (see attached file):

 

Consider an ETC-point X(n) ,

 

... its anticevian triangle X(n)a,b,c,

 

... the points Y(n)a,b,c, which are the X(n) of ABY(n)c, BCY(n)a, CAY(n)b,

 

... the cyclologic center Z(n) = (Y(n)aY(n)bY(n)c, ABC).

 

This center doesn´t exist for all X(n):

 

Z(1) = X(10215), Z(2) = X(671), Z(3) not in ETC, Z(4) = X(10152), Z(5) and Z(6) doesn´t exist ...

 

But there are Z(n) for the Hofstadter points, Cabri-proved for X(3), X(4), X(35), X(79), X(186), X(265), X(5961), X(5962), X(5963), X(5964). But the corresponding Z(n) are not in ETC.

 

Question: Are there beside the Hofstadter points and X(2) other points with a center Z(n)?

 

Best regards Eckart Schmidt

 

PS: Perhaps Z(3) for ETC. First barycentric coordinate:

 

 ---------------------------

#1949

 

Dear Antreas, Bernard, Chris,

 

there was no reaction on #1936, but two of the described points are now in ETC: X(10230 and X(10231).

 

Attached another concept for new ETC-points.

 

Best regards Eckart Schmidt

 

---------------------------

#1951

 

Dear Eckart,

 

Let's denote QAP_acev(k,n) the QA-P(k) point of the quadrangle {X(n),vertices-of-the-anticevian-of-X(n)}

 

Then:

QAP_acev(3,4) = 

= sec(A)^2/((5*cos(2*A)+7)*cos(B-C)-cos(A)*cos(2*(B-C))-10*cos(A)-cos(3*A)) :: (trilinears)

= Trilinear pole of the line {393,6587}

= reflection of X(i) in X(j) for these (i,j): (107,6523), (3346,122)

= On lines: {20,107}, {122,3346}, {133,1249}, {2777,3183}

= [ -2.158256348287418, -2.68134960274278, 6.493101752246650 ]

 

QAP_acev(3,5) = 

= 1/((4*cos(2*A)+2*cos(4*A)+3)*cos(B-C)+2*cos(A)*cos(2*(B-C))-7*cos(A)-3*cos(3*A)-cos(5*A)) : : (trilinears)

= antigonal conjugate of X(6662)

= reflection of X(i) in X(j) for these (i,j): (476,6663), (6662,3258)

= On lines: {476,6663}, {3258,6662}, {9214,10201}

= [ -0.053793793378817, -0.08040278306128, 3.721155851740104 ]

 

QAP_acev(4,4) =

= ((2*cos(2*A)+3)*cos(B-C)-6*cos(A)-cos(3*A))*sec(A)^2 : : (trilinears)

= antigonal conjugate of X(2071)

= polar circle-inverse-of-X(185)

= midpoint of X(i),X(j) for these {i,j}: {4,6761}

= reflection of X(i) in X(j) for these (i,j): (1304,403), (6760,5)   

= on cubic K025 and these lines: 

  {4,51}, {5,6760}, {20,6526}, {30,107}, {316,6528}, {403,1300}, {1503,1559}, {1596,1629}, {3146,6523}, {3543,6525}, {3839,10002}, {5523,6529}, {6530,10151}

= [ -2.719974554883476, -3.27985730468963, 7.166707795100341 ]

 

QAP_acev(4,5) =

= (6*cos(2*A)+2*cos(4*A)+7)*cos(B-C)+(-6*cos(A)-2*cos(3*A))*cos(2*(B-C))+(2*cos(2*A)+2)*cos(3*(B-C))-cos(5*A)-7*cos(A)-cos(3*A) : : (trilinears)

= (2*SW^2-(17*R^2-2*SA)*SW+(39*R^2-7*SA)*R^2)*S^2-(18*R^4-11*R^2*SW+2*SW^2)*(SB+SC)*SA : : (barycentrics)

= polar circle-inverse-of-X(6746)

= On lines: {4,94}, {110,2888}, {9143,10201}

= [ -0.110950800319549, -0.09306259735188, 3.756300495606244 ]

 

Regards,

 

César Lozada

 

---------------------------

#1952

 

Dear Eckart, dear Cesar,

 

Cesar, very nice analyses!

 

I think also QAP_acev(k,n) for

k = 5,10 (evt. 11,12,13),16 and

n = 2,3,4,5

can be very interesting.

 

And what about QLP_acev(k,n) for

k = 1 (Miquel Point to start with), 12, 13, 26 and

n = 2 (Euler line), 3 (Brocard Axis)

 

Where QLP_acev(k,n) is the point defined by:

For an ETC-central-line L(n)

·         and its antiLINEcevian triangle L(n)_a L(n)_b L(n)_c  (Let antiLINEcevian triangle = ABC-circumscribed triangle that is perspective with ABC having perspectrix L(n))

·         consider the quadrilateral  L(n) L(n)_a L(n)_b L(n)_c

·         and take the QL-points as triangle centers.

 

Best regards,

 

Chris van Tienhoven

 

---------------------------

#1953

 

Dear  César Lozada, dear Chris, dear Bernard,

 

I am glad, you are interested in the concept for new ETC-points, and I am fascinated of the possibilities in Cesars calculations! I had only calculated the 1st barycentric coordinate for

 QAP_acev(4,4)  =

 

There are some evident results, for the reference triangle is the diagonal triangle of X(n)X(n)aX(n)bX(n)c:

QAP_avec(10,n) = X(2), QAP_avec(11,n) = X(3), QAP_avec(12,n) = X(4), QAP_avec(13,n) = X(5) for arbitrary  n.

 

In addition to Chris´ generalisations this concept for new ETC-points can also be modified wrt the dual QL of X(n)X(n)aX(n)bX(n)c, taking QL-points for triangle centers. Here some examples wrt wellknown X(n):

 ...  X(1), QL-P1 --> X(115),

...  X(1), QL-P2 --> X(3),

...  X(1), QL-P5 --> X(6140),

...  X(1), QL-P7 --> X(187),

...  X(1), QL-P8 --> X(2),

...  X(1), QL-P12 --> X(351),

...  X(1), QL-P13 --> X(6),

...  X(3), QL-P13 --> X(577),

...  X(4), QL-P2 --> X(389),

...  X(4), QL-P1 --> X(115),

...  X(4), QL-P2 --> X(389),

...  X(4), QL-P13 --> X(393),

...  X(6), QL-P13 --> X(32),

... and the evident cases for X(2), X(3), X(4), X(5) and QL-P8, QL-P9, QL-P10, QL-P11.

 

Best regards Eckart Schmidt

 

PS: This modification was also mentioned in a private mail of Bernard.

 

 ---------------------------

#1954

 

Dear Cesar Lozada, dear Chris, dear Bernard,

 

there is a further modification of my concept, to get new ETC-points:

Take for a reference triangle ABC

... the quarangle QA with one vertex X(n) and ABC as Miquel triangle QA-Tr2.

(For X(1) this gives the anticevian case.)

Then QA-Px  can be interpreted as triangle center, for example:

 

... X(3), QA-P3 --> X(1263),

... X(3), QA-P4 --> X(1157),

... X(4), QA-P3 --> X(8439),

... X(4), QA-P4 --> X(3484).

 

Best regards Eckart Schmidt

 

 ---------------------------

#1955

 

Dear Eckart,

 

As Bob Dylan said : The times, they are changing !

I remember an old time, when I was interested in properties for QA vertices isotomic wrt DT (the result swaps QL-P1 and QL-P25), your short comment was : That's not QA or QL geometry, that's triangle geometry !

Anyhow, a short comment to these results.

It appears that QL-P1 is X115 for the incenter X1 and the orthocenter X4 ; in fact, it is also X115 for X2.

For I1, QL-P1 = X115, QL-P25 is X125 and QL-P17 is X110.

For X75 = isotomic of X1, QL-P17 = X99 ...

Conversely, the locus of the points for which QL-P1 is X115 is the Stammler quartic Q066 in Bernard Gibert (isogonal of the Stammler rectangular hyperbola), which passes in fact through X1, X2 and X4 and some other points. This property was mentionned by Angel.

More generally, for a given point on the Euler circle as focus, the parabola QL-Co1 is given (inscribed in the medial triangle).

The locus of the QA vertices corresponding to a QL formed by 4 tangents to the parabola and having the same DT is a curve with similar form ; the isotomic of these vertices give a 2nd QA and the dual QL is formed by the tangents to QL-Co3.

 

Best regards

Bernard Keizer

 

---------------------------

#1956

 

Dear Eckart,

It doesn't matter if it is QA/QL- or triangle geometry, the important thing is that it leads to a wide range of reflexions !

Only a detail more on your table in message 1949 : you say for X(1) that QA-P2, P3 and P4 do not exist.

It would be more precise to say for P2 and P3 undetermined on the circumcircircle and for P4 undtermined on the infinity line.

Any point on the infinity line leads to an isogonal circular cubic, for example K001 = pK(X6,X30) or K021 = pK(X6,X512).

Best regards

Bernard

PS For X7, QL-P1 = X11

I'm awaiting impatiently for a bigger table with plenty of ETC and EQF points ...

 

Bernard Keizer

 

---------------------------

#1957

 

Dear Bernard,

 

thanks for additional examples wrt # 1953.

 

Wrt X(2) we have to respect, that the dual QL degenerates.

 

With great interest I have constructed the Stammler quartic as locus of the points for which QL-P1 is X(115),

thanks for information.

 

It took a long time, to understand and do your last construction:

 

... QL1 reference quadrilateral,

 

... QA1 dual of QL1,

 

... QA2 isotomic conjugate of QA1 wrt the common DT,

 

... QL2 dual of QA2,

 

... with the result,

 

... ... that QL-P1 of  QL2 is QL-P25 of  QL1

 

... ... and QL1 and QL2 have the same contacts with their inscribed parabolas.

 

So I have to cancel my comment - excuse!

 

But I don´t see the reference to the results in the middle of the message.

 

 Best regards Eckart Schmidt

 

 

---------------------------

#1958

 

Dear Eckart,

 

Angel's message was 1448, my construction was explained in 1449 and 1451.

 

Best regards

Bernard Keizer

 

---------------------------

#1959

 

Dear Eckart,

I was desperate as I couldn't find my initial messages !

But it's now done : they were in messages 1315 and 1316.

The property was also explained in EQF at QL-Co3.

QL-P1 is necessary one of the 2 intersections between the Euler circle (circummedial circle) and the cevian triangles of the vertices of QA. That's the way I found X115 for X1, X11 for X7. And of course, for X2, the point is undetermined on the Euler circle ...

This time, I hope I was complete.

 

Best regards

Bernard Keizel