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

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#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

 

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#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

 

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#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

 

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#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

 

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#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

 

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#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

 

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#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

 

 

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#1958

 

Dear Eckart,

 

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

 

Best regards

Bernard Keizer

 

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#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

 

 

ADGEOM 3553

Let ABC be a triangle and L a line passing through X(850). Let P1, P2 denote the brocardians of a point P on L, and P1', P2' its isotomic conjugates. The envelope of line P1'P2' is a parabola as P moves on L. The locus of its focus is the Parry circle as L moves around X(850). As L moves around X(850) the directrix of the parabola passes through X(5996), on the line through centroid parallel to line of the 1st and 2nd Brocard points.

A more general approach in: http://amontes.webs.ull.es/otrashtm/HGT2016.htm#HG161116

Angel Montesdeoca

ADGEOM 2040

Dear friends,

Consider the following generalizations of the Simson and Steiner lines:

Let P be a point not on the circumcircle of ABC.  Let A'B'C' be the pedal triangle of P and A"B"C" the reflection triangle of P.  Define the quasi-Simson line of P as the orthic axis of A'B'C', and define the quasi-Steiner line of P as the orthic axis of A"B"C".  As P approaches a point Q on the circumcircle, the quasi-Simson/Steiner lines of P approach the Simson/Steiner lines of Q.

Any interesting results using these lines?

Best regards,
Randy Hutson

ADGEOM 1550 * ADGEOM 1555 * ADGEOM 1556

#1550

 

Dear geometers,

Let ABC be a triangle with three Feuerbach points Fa,Fb,Fc. Let Fa' be isogonal conjugate of Fa with respect to triangle AFbFc. Smilarly we have point Fb',Fc'. Then FaFa',FbFb',FcFc' are concurrent.

Did we know about this point ?

Best regards,
Tran Quang Hung.

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#1555

Dear Tran Quang Hung and Francisco Javier

 

 

[TQH]: 

 

Let ABC be a triangle with three Feuerbach points Fa,Fb,Fc. Let Fa' be isogonal conjugate of Fa with respect to triangle AFbFc. Similarly we have point Fb',Fc'. Then FaFa',FbFb',FcFc' are concurrent.

 

 

****  My calculations give that the point of intersection of the lines FaFa',FbFb',FcFc' has barycentric coordinates:

 

 

a^11(b - c)^2+ a^10 (b + c) (b^2 - 6 b c + c^2)+ a^9( -5 b^4 - 5 b^3 c -12 b^2 c^2 - 5 b c^3 - 5 c^4)- a^8(b + c) (5 b^4 - 6 b^3 c + 6 b^2 c^2 - 6 b c^3 + 5 c^4) a^7 (10 b^6 + 22 b^5 c + 25 b^4 c^2 + 14 b^3 c^3 + 25 b^2 c^4 + 22 b c^5 + 10 c^6)+ a^6 (b + c) (10 b^6 + 18 b^5 c + 27 b^4 c^2 + 6 b^3 c^3 + 27 b^2 c^4 + 18 b c^5 + 10 c^6)+ a^5( -10 b^8 - 16 b^7 c + 22 b^6 c^2 + 57 b^5 c^3 + 54 b^4 c^4 + 57 b^3 c^5 + 22 b^2 c^6 - 16 b c^7 - 10 c^8)- 2a^4 (b + c) (5 b^8 + 15 b^7 c + 5 b^6 c^2 - 19 b^5 c^3 - 22 b^4 c^4 - 19 b^3 c^5 + 5 b^2 c^6 + 15 b c^7 + 5 c^8)+ a^3(b - c)^2 (b + c)^2 (5 b^6 - 4 b^5 c - 42 b^4 c^2 - 59 b^3 c^3 - 42 b^2 c^4 - 4 b c^5 + 5 c^6)+ a^2 (b - c)^2 (b + c)^3 (5 b^6 + 12 b^5 c - 8 b^4 c^2 - 26 b^3 c^3 - 8 b^2 c^4 + 12 b c^5 + 5 c^6)- a(b - c)^4 (b + c)^6 (b^2 - 7 b c + c^2) -(b - c)^6 (b + c)^7 : :

 

with (6-9-13)-search number -0.0381537327593557454260317

 

 

This point is not on the line X(1)X(21).

 

 

Best regards, 

 

Angel Montesdeoca

 

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#1556

 Dear Tran Quang Hung, Francisco Javier, and Angel,

I get the same search value as Angel, on lines 5,191 30,5948 442,5947 (at least).

Best regards,
Randy Hutson

ADGEOM 1427 * ADGEOM 1430

 

#1427

Dear Members!

Let ABC be a triangle. Point P is on it's circumcircle. Tangents at P to incircle of ABC intersect incircle and circumcircle in four points - let X_{P} be the intersection of it's diagonals. Then X_{P} (where P lies on circumcircle) lies on circle which contains Gergonne point of ABC and it's center in 6,9,13 is 1.084338...

Best regards, 

Dominik Burek 

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

#1430

Dear Dominik, nice circle indeed!

I find that your circle has the point X1159 as center, and radius r (R - 2 r)/(4 R + r).

Best regards,

Francisco Javier.

ADGEOM 2697 * ADGEOM 2698

#2697

 

[Tran Quang Hung]:

 

Dear geometers,

Let ABC be a triangle.

Perpendicular bisector of BC cuts CA,AB at Ab,Ac. Let da be perpendicular bisector of segment AbAc. Simiarlaly, we have the lines db,dc.

Then, triangle bound by lines da,db,dc has nine point circle which is tangent to circumcircle of triangle ABC.

Please, the tangent point is new ?

Best regards,

Tran Quang Hung.

 

#2698

[Paul Yiu]:

Dear QH,

 

[TQH]:  Let ABC be a triangle.

Perpendicular bisector of BC cuts CA,AB at Ab,Ac. Let da be perpendicular bisector of segment AbAc. Simiarlaly, we have the lines db,dc.Then, triangle bound by lines da,db,dc has nine point circle which is tangent to circumcircle of triangle ABC.

 

*** Euler reflection point X(110).

 

Best regards

Sincerely

Paul Yiu

 

HYACINTHOS 28911

[Tran Quang Hung]:

Let ABC be a triangle.with NPC center N.

A'B'C' is the circumcevian triangle of N.

N1, N2, N3, N4, N5, N6 are the NPC centers of the triangles AC'B', B'AC, CB'A', A'CB, BA'C', C'BA, respectively.

Let H1, H2, H3 be the orthocenters of the triangles N6N1N2, N2N3N4, N4N5N6, respectively.

Then the NPC center of the triangle H1H2H3 lies on the Euler line of ABC.  

[Peter Moses]:

 

Hi Antreas,

X(10285) on lines {2,3}, {54,1263}, {1154,20327}, {11671,16766}, {12026,14051}, {14140,24573}, {19552,21230}, {24385,25044}

Best regards,
Peter Moses.

div>

HYACINTHOS 28285

 

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

 

Which is the locus of P such that the reflections of PNa, PNb, PNc in AI, BI, CI, resp are concurrent?

The Euler line of NaNbNc?

 

[César Lozada]:

Locus: {Euler line of NaNbNc=Line IN of ABC} ∪ {circumcircle of NaNbNc=circle (N, NX(10)) of ABC }

 

IF P ∈ {Euler line of NaNbNc=Line IN of ABC} then the point of concurrence Q(P) ∈ X(1)X(3)-of-ABC = Line X(4)X(94) of NaNbNc.

For P such that IP/IN=t,

Q(P) = a*( (2*a^3-3*(b+c)*a^2-2*(b^2-3*b*c+c^2)*a+3*(b^2-c^2)*(b-c))*t+2*(-a+b+c)*(a-b+c)*(a+b-c)) : :  (barys)

 

and IQ/IO = (t/2)/(1-t)

 

ETC-pairs (P,Q(P)): (1,1), (5,517), (11,65), (12,3057), (80,11009), (355,1482), (495,9957), (496,942), (1484,6583), (1837,2099), (5219,1697), (5252,2098), (5443,35), (5587,7982), (5881,16200), (5886,3), (5901,1385), (7741,5903), (7951,5697), (7958,7957), (7988,7991), (7989,11531), (8227,40), (9578,7962), (9581,3340), (9624,3576), (10283,15178), (10886,12435), (10944,5048), (10948,5570), (10950,11011), (11373,999), (11374,3295), (11375,55), (11376,56), (15888,5919), (15950,2646), (16173,5563), (17718,3303), (17720,5710), (18357,11278), (19907,11567)

 

If P lies on the circumcircle of NaNbNc (through ETC’s centers of ABC 10, 502, 946, 11798, 13604, 15529), then Q(P) moves on the circle with radius=(R-2*r)/2 and center Oq given below. This circle passes through ETC’s 946, 3244

 

ETC pairs (P,Q(P)) = (10, 3244), (946,946)

 

Oq = midpoint of X(1) and X(1482)

= a*(2*a^3-3*(b+c)*a^2-2*(b^2-3*b*c+c^2)*a+3*(b^2-c^2)*(b-c)) : : (barys)

= 3*X(1)-X(3), 5*X(1)-X(40), 11*X(1)-3*X(165), 7*X(1)-3*X(3576), 4*X(1)-X(3579), 3*X(1)+X(7982), 13*X(1)-5*X(7987), 9*X(1)-X(7991), 5*X(1)+X(8148), 5*X(1)-3*X(10246), X(1)-3*X(10247), 5*X(1)+3*X(11224), 2*X(1)+X(11278), 7*X(1)+X(11531), 7*X(1)-X(12702), 5*X(1)-2*X(13624), 3*X(1)-2*X(15178), 3*X(1)+5*X(16189)

= on lines: {1, 3}, {4, 1392}, {5, 519}, {8, 3090}, {10, 3628}, {19, 23073}, {20, 3655}, {30, 4301}, {42, 19546}, {72, 1173}, {79, 14217}, {140, 551}, {145, 355}, {381, 5881}, {388, 10525}, {392, 5047}, {497, 10526}, {515, 1483}, {516, 13607}, {518, 576}, {546, 946}, {547, 4669}, {548, 5493}, {573, 3723}, {575, 1386}, {631, 3654}, {632, 1125}, {758, 11260}, {912, 15083}, {936, 11530}, {944, 3146}, {956, 3951}, {960, 18233}, {962, 3529}, {1000, 5703}, {1012, 11520}, {1056, 4323}, {1058, 4345}, {1210, 1387}, {1320, 1389}, {1457, 5399}, {1656, 3679}, {1657, 9589}, {1699, 18525}, {1766, 16884}, {1829, 10594}, {1837, 7743}, {1870, 1872}, {1953, 22356}, {2102, 15157}, {2103, 15156}, {2771, 7984}, {2800, 3881}, {2802, 19907}, {3058, 7491}, {3083, 21549}, {3084, 21546}, {3242, 11477}, {3243, 18761}, {3419, 6984}, {3434, 10597}, {3436, 10596}, {3485, 6982}, {3488, 5812}, {3518, 11363}, {3523, 3653}, {3525, 3616}, {3544, 20050}, {3555, 5887}, {3560, 12513}, {3577, 18491}, {3584, 5559}, {3585, 7972}, {3621, 5818}, {3622, 5657}, {3625, 10175}, {3632, 5079}, {3633, 5072}, {3636, 6684}, {3680, 6918}, {3751, 11482}, {3753, 17531}, {3811, 10912}, {3817, 12811}, {3827, 15581}, {3857, 19925}, {3874, 14988}, {3877, 16865}, {3880, 13374}, {3892, 5884}, {3893, 11524}, {3913, 6911}, {3915, 5398}, {3940, 4853}, {3957, 21740}, {3962, 5288}, {3991, 4919}, {4004, 5253}, {4297, 12103}, {4342, 15172}, {4511, 6946}, {4658, 15952}, {4663, 22330}, {4677, 5055}, {4691, 10172}, {4701, 10171}, {4745, 15699}, {4848, 15325}, {4864, 15310}, {4870, 6980}, {4930, 6913}, {5044, 5289}, {5054, 9588}, {5068, 20049}, {5070, 19875}, {5076, 5691}, {5198, 11396}, {5250, 19526}, {5258, 7489}, {5396, 19646}, {5440, 14923}, {5497, 10700}, {5609, 11699}, {5722, 5761}, {5727, 9669}, {5731, 17538}, {5763, 15935}, {5840, 12735}, {5853, 20330}, {5854, 10915}, {5855, 10916}, {6419, 7969}, {6420, 7968}, {6427, 18991}, {6428, 18992}, {6447, 9583}, {6519, 9616}, {6797, 12740}, {6833, 11240}, {6834, 11239}, {6842, 15888}, {6860, 12649}, {6863, 10056}, {6865, 15933}, {6914, 8666}, {6924, 8715}, {6958, 10072}, {6978, 11373}, {6988, 7320}, {7377, 17389}, {7419, 18180}, {7680, 10943}, {7681, 10942}, {7978, 15054}, {7983, 23235}, {9619, 22332}, {9708, 15829}, {10039, 15950}, {10165, 12108}, {10446, 17393}, {10573, 11376}, {10696, 19904}, {10895, 11928}, {10896, 11929}, {10944, 12047}, {11041, 14986}, {11272, 22475}, {11375, 12647}, {11496, 12559}, {11551, 11826}, {11552, 12119}, {11705, 20415}, {11706, 20416}, {11707, 21401}, {11708, 21402}, {11717, 13497}, {11724, 20399}, {11725, 20398}, {11728, 20401}, {11735, 20397}, {12053, 18527}, {12104, 22937}, {12331, 12653}, {12610, 17390}, {12773, 13253}, {12778, 15034}, {13211, 15027}, {13743, 16126}, {14563, 21625}, {16173, 19914}, {16239, 19883}, {16842, 19860}, {16862, 19861}, {17018, 19647}, {17438, 22357}, {17572, 17614}

= midpoint of X(i) and X(j) for these {i,j}: {1, 1482}, {145, 355}, {944, 12699}, {962, 18481}, {1320, 6265}, {1657, 9589}, {3555, 5887}, {3633, 12645}, {3811, 10912}, {11496, 12559}, {12331, 12653}, {12773, 13253}, {13743, 16126}

= reflection of X(i) in X(j) for these (i,j): (5, 13464), (8, 9956), (10, 5901)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 7982, 3), (1, 8148, 13624), (1, 11224, 40), (1, 11278, 3579), (1, 11531, 3576), (40, 11224, 8148), (65, 11567, 1385), (1482, 8148, 11224), (1482, 10246, 8148), (1482, 10247, 1), (3576, 11531, 12702), (7373, 10306, 10269), (7982, 16189, 1482), (7982, 16200, 16189), (8148, 10246, 40), (10680, 12000, 55)

= [ -0.8557186829126229, -0.4120126991801520, 4.3208511272995400 ]

 

Some others Q(P):

Q( X(119) ) = midpoint of X(4) and X(3885)

= a*((b+c)*a^5-(b^2+6*b*c+c^2)*a^4-2*(b+c)*(b^2-5*b*c+c^2)*a^3+2*(b^4+c^4+2*(b^2-4*b*c+c^2)*b*c)*a^2+(b^2-c^2)*(b-c)*(b^2-8*b*c+c^2)*a-(b^2-c^2)^2*(b-c)^2) : : (barys)

= 4*X(1)-3*X(10202), 5*X(1)-4*X(13373), 7*X(1)-5*X(15016), 3*X(65)-4*X(6583), 3*X(392)-2*X(5690), 2*X(942)-3*X(10247), 4*X(942)-3*X(10273), 3*X(944)-X(9961), 8*X(3628)-7*X(4002), 3*X(3655)-2*X(9943), 3*X(3656)-2*X(7686), 5*X(3698)-6*X(11230), 3*X(3753)-4*X(5901), 3*X(3877)-X(12245), 5*X(3890)-3*X(5657)

= on lines: {1, 3}, {4, 3885}, {5, 6735}, {8, 6893}, {72, 5844}, {119, 946}, {145, 912}, {355, 3880}, {392, 5690}, {496, 17622}, {519, 5887}, {944, 9961}, {952, 12672}, {962, 12115}, {971, 18526}, {1000, 5555}, {1210, 15558}, {1320, 12775}, {1519, 10942}, {1699, 11929}, {1872, 1877}, {2136, 5720}, {2800, 3244}, {2950, 12773}, {3555, 14988}, {3585, 12749}, {3625, 20117}, {3628, 4002}, {3633, 5693}, {3655, 9943}, {3656, 7686}, {3698, 11230}, {3753, 5901}, {3869, 6930}, {3877, 5084}, {3878, 5795}, {3881, 15528}, {3884, 11362}, {3890, 5657}, {3898, 6684}, {4301, 12608}, {5053, 21853}, {5252, 10525}, {5439, 10283}, {5552, 5603}, {5587, 11928}, {5734, 6970}, {5761, 6848}, {5777, 12625}, {5836, 5886}, {6256, 12699}, {6827, 9785}, {6882, 12053}, {6923, 12700}, {6958, 11373}, {6971, 7743}, {7330, 12629}, {7491, 10624}, {7680, 13463}, {7967, 13369}, {7970, 13189}, {7978, 13217}, {7983, 12189}, {7984, 12381}, {9856, 18525}, {10526, 12701}, {10595, 17567}, {10698, 13278}, {10705, 13118}, {10738, 12751}, {10866, 18527}, {10912, 11496}, {12650, 12686}, {12705, 18519}, {13099, 13313}

= midpoint of X(i) and X(j) for these {i,j}: {4, 3885}, {3633, 5693}

= reflection of X(i) in X(j) for these (i,j): (1071, 1483), (3625, 20117), (7491, 10624)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 40, 10269), (1, 2077, 1385), (1, 3359, 16203), (1, 5903, 18838), (1, 11010, 14803), (1, 12703, 11248), (942, 9957, 20789), (946, 10915, 119), (1482, 10679, 1), (2098, 11509, 1), (2099, 10965, 1), (3746, 11014, 1385), (10596, 12245, 5554), (10942, 22791, 1519), (12702, 16203, 3359)

= [ -2.2411636793303000, -1.5560108497491040, 5.7522859991169750 ]

 

Q( X(13604) ) = X(35)X(60) ∩ X(758)X(3057)

= ((b-c)^2*a^6+(b+c)*b*c*a^5-(3*b^2+4*b*c+3*c^2)*(b-c)^2*a^4-(b+c)*(3*b^2-4*b*c+3*c^2)*b*c*a^3+(3*b^4+3*c^4+(7*b^2+9*b*c+7*c^2)*b*c)*(b-c)^2*a^2+(b+c)*(2*b^4+2*c^4-(2*b^2-b*c+2*c^2)*b*c)*b*c*a-(b^2-c^2)^2*(b^4+c^4+(b^2+b*c+c^2)*b*c))*a^2 : : (barys)

= on lines: {35, 60}, {758, 3057}, {946, 6003}, {5563, 6584}, {8674, 13604}, {11009, 23153}

= [ -1.7062356821329990, 0.7822171051486402, 3.8866229547120880 ]

 

César Lozada

 

Note: I made some attempts for taking NaNbNc as the reference triangle. In fact, ABC can be obtained from NaNbNc by drawing two centers and a pair of lines in this latter. Unfortunately, the resulting algebraic expressions for A,B,C are not nice to work with.

HYACINTHOS 25511

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

Denote:

A'B'C' = the midway triangle of O
(ie A', B', C' = midpoints of OA, OB, OC, resp.)

A"B"C" = the tangential triangle of A'B'C'.
(ie the antipedal triangle of O wrt triangle A'B'C')

Ma, Mb, Mc = the midpoints of APa, BPb, CPc, resp

R1 = the radical axis of the circles (Mb, MbB'), (Mc, McC')
R2 = the radical axis of the circles (Mc, McC'), (Ma, MaA')
R3 = the radical axis of the circles (Ma, MaA'), (Mb, MaB')
[the radical center of the circles is the point they concur at = the NPC center of ABC]

Ra, Rb, Rc = the reflections of R1, R2, R3 in OA", OB", OC", resp.
[OA", OB", OC' are the perpendicular bisectors of ABC and A'B'C']

A*B*C* = the triangle bounded by Ra, Rb, Rc.

1. Which is the locus of P such that the triangles A"B"C" and A*B*C* are parallelogic?

H lies on the locus. The parallelogic center (A"B"C", A*B*C*) lies on the Euler line of ABC.

O lies on the locus. Ra, Rb, Rc are concurrent on the Euler line at the reflection of N in O.
(R1, R2, R3 are parallels to OA",OB",OC", resp.)

2. Which is the locus of P such that Ra, Rb, Rc are concurrent?
O lies on the locus.

 

[Angel Montesdeoca]:

**** The locus of P such that the triangles A"B"C" and A*B*C* are parallelogic is the Euler line and the circumcircle.

== Parallelogic center (A"B"C", A*B*C*)
 
-- P lies on Euler line:
If P = O + t H then the parallelogic center (A"B"C", A*B*C*) is  Q = (OH^2 t + R^2) O + t R^2  H (lies on Euler line).

In particular, Q=O if t=0,  ie. P=O
                     Q=H if t=-R^2/OH^2,  ie. P = OH^2 O - R^2 H =  X(11250) =
         (a^2 (a^8-2 a^6 (b^2+c^2)+6 a^4 b^2 c^2+a^2 (2 b^6-3 b^4 c^2-3 b^2 c^4+2 c^6)-(b^2-c^2)^2 (b^4+3 b^2 c^2+c^4)):...:...).
                     Q=N if  t=R^2/(R^2-OH^2),  ie. P = (R^2-OH^2)  O +R^2 H = X(3520).
                    
Other pairs {P,Q}:  {20,1658}, {30,186}, {376,7502}, {378,6644}, {548,7512}, {550,7488}, {3516,6642},  {3522,7525}, {7464,7575}, {7502,10298}, {8703,6636}, {10226,3520}, {11410,9818}, {11413,26}

-- If P lies on circumcircle  then the parallelogic center (A"B"C", A*B*C*) is  the homothetic of P under h(X(186), (a^2 b^2 c^2)/(8SA SB SC)) (lies on circumcircle of A"B"C", of center X(1658) and passes through X(5961)).


== Parallelogic center (A*B*C*, A"B"C")
 
-- P lies on Euler line then the parallelogic center (A*B*C*,A"B"C") lies on the line X(550)X(1511)
Pairs {P,parallelogic center (A*B*C*,A"B"C")} :  {3,550},  {20,10282}, {30,1511},  {? , 2777}, {? ,2883}

-- If P lies on circumcircle  then the parallelogic center  (A*B*C*,A"B"C") lies on the conic

(a^6 b^4 c^2+a^4 b^6 c^2-5 a^2 b^8 c^2+3 b^10 c^2+a^6 b^2 c^4-6 a^4 b^4 c^4+5 a^2 b^6 c^4-12 b^8 c^4+a^4 b^2 c^6+5 a^2 b^4 c^6+18 b^6 c^6-5 a^2 b^2 c^8-12 b^4 c^8+3 b^2 c^10) x^2 + (-2 a^12+7 a^10 b^2-8 a^8 b^4+2 a^6 b^6+2 a^4 b^8-a^2 b^10+7 a^10 c^2-18 a^8 b^2 c^2+22 a^6 b^4 c^2-18 a^4 b^6 c^2+7 a^2 b^8 c^2-8 a^8 c^4+22 a^6 b^2 c^4+8 a^4 b^4 c^4-6 a^2 b^6 c^4+2 a^6 c^6-18 a^4 b^2 c^6-6 a^2 b^4 c^6+2 a^4 c^8+7 a^2 b^2 c^8-a^2 c^10) y z + ... = 0,

of center X(10282) = 5 X(3) - X(64)
(see Hyacinthos 24665 ) and passes through X(3) and X(6759).
Pairs {P,parallelogic center (A*B*C*,A"B"C")} : {74,6759}, {110,3} 



Angel Montesdeoca

 

HYACINTHOS 25344

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, A'B'C' the pedal triangle of H and P a point.

Denote:

A",B",C" = the reflections of P in BC, CA, AB, resp.

A*, B*, C* = the reflections of A", B", C" in B'C', C'A', A'B', resp.

Ab, Ac = the orthogonal projections of A* on AC, AB, resp.

Bc, Ba = the orthogonal projections of B* on BA, BC, resp.

Ca, Cb = the orthogonal projections of C* on CB, CA, resp.

Oa, Ob, Oc = the circumcenters of A*AbAc, B*BcBa, C*CaCb, resp.

For P = O:

 

1. ABC, OaObOc are homothetic

2. A"B"C", OaObOc are homothetic

Which points are the homothetic centers (on the Euler line)?

3. ABC, OaObOc are orthologic.

The orthologic center (ABC, OaObOc) is the H.

Which point is the other one (OaObOc, ABC) [on the Euler line]?

Which is the locus of P such that ABC, OaObOc are
a. perspective
b. orthologic?

[Peter Moses]:

Hi Antreas,

 

1) X(26).

 

2) a^10-a^8 b^2-2 a^6 b^4+2 a^4 b^6+a^2 b^8-b^10-a^8 c^2-4 a^4 b^4 c^2+2 a^2 b^6 c^2+3 b^8 c^2-2 a^6 c^4-4 a^4 b^2 c^4-6 a^2 b^4 c^4-2 b^6 c^4+2 a^4 c^6+2 a^2 b^2 c^6-2 b^4 c^6+a^2 c^8+3 b^2 c^8-c^10::
on lines {{2,3},{68,143},...} Searches {-0.88466424430114870039903323 9615,-1.7518504005848276963283 5858389,5.26179056429748229306 831177117}.

Reflection of X(7514) in X(5).

 

3) 2 a^10-3 a^8 b^2-2 a^6 b^4+4 a^4 b^6-b^10-3 a^8 c^2-2 a^4 b^4 c^2+2 a^2 b^6 c^2+3 b^8 c^2-2 a^6 c^4-2 a^4 b^2 c^4-4 a^2 b^4 c^4-2 b^6 c^4+4 a^4 c^6+2 a^2 b^2 c^6-2 b^4 c^6+3 b^2 c^8-c^10::
on lines {{2,3},{51,11750},...}
Searches {-3.18469148443498600227301307 083,-4.04581011078426256326240 328836,7.911467551420400571041 41033483}.

Midpoint of X(i) and X(j) for these {i,j}: {{382,6240},{3575,7553}}.

Reflection of X(i) in X(j) for these {i,j}: {{5,6756},{1885,3853},{6146,14 3}}.

 

a) a circular cubic through {3,403,1658}.

b) line through{3,49,155,184,185,283,3 94,1092,1147,1181,1204,1216,14 37,1790,1800,1801,1819,3167,32 92,3796,3917,5406,5407,5408,54 09,5447,5562,7689,8913,9703,97 04,9720,9908,10132,10133,10605 ,10670,10674,10984}

& infinity.

 

Best regards,

Peter Moses.

HYACINTHOS 25342

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C', A"B"C" the cevian triangles of H,G, resp.

Denote:

A* = the reflections of A" in B'C'

B* = the reflections of B" in C'A'

C* = the reflections of C" in A'B'

Ab, Ac = the orthogonal projections of A* on AC, AB, resp.

Bc, Ba = the orthogonal projections of B* on BA, BC, resp.

Ca, Cb = the orthogonal projections of C* on CB, CA, resp.

 Oa, Ob, Oc = the circumcenters of A*AbAc, B*BcBa, C*CaCb, resp.

1. A'B'C' are orthologic.

The orthologic center (OaObOc, A'B'C') is the midpoint of ON = X140

2. ABC, OaObOc are orthologic.

 

[Peter Moses]:

 

Hi Antreas,

 

1).

{OaObOc,A’B’C’} orthologic at X(140).

{A’B’C’, OaObOc} orthologic at:
a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^4-2 a^2 b^2+b^4-2 a^2 c^2-3 b^2 c^2+c^4) (a^4 b^2-2 a^2 b^4+b^6+a^4 c^2-6 a^2 b^2 c^2-b^4 c^2-2 a^2 c^4-b^2 c^4+c^6)::
on lines {{4,2889},{6,1173},{23,9827},{ 185,7576},{428,6152},{1598, 2904},{1986,6756},{2914,5609}, {5895,11455},{7999,10516}}.

Searches {-2. 51951022758613447812683246361, -3. 79072232298120963846265690728, 7. 42786157978035553503565640613} .

5 X[1173]-7 X[9781].

X(5557) of the orthic triangle.

on the Feuerbach of the orthic triangle.

 

2)

{ABC,OaObOc} orthologic at X(1173).

{OaObOc,ABC} orthologic at X(5446).

 

Best regards,

Peter Moses.

 

HYACINTHOS 25327

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

A", B", C" = the orthogonal projections of A, B, C on PA', PB', PC', resp.

Ab, Ac = the orthogonal projections of A" on BO, CO, resp.
Bc, Ba = the orthogonal projections of B" on CO, AO, resp.
Ca, Cb = the orthogonal projections of C" on AO, BO, resp.

Na, Nb, Nc = the NPC centers of A"AbAc, B"BcBa, C"CaCb, resp.

Which is the locus of P such that ABC, NaNbNc are orthologic?

The Euler line?

And which are the loci of the orthologic centers as P moves on the Euler line?

 

[César Lozada]:

 

> The Euler line?

Yes.

 

Za = ABC->NaNbNc: The locus is the circum-conic CyclicSum[ a*(SB^2-SC^2)*SA*(SA^2+5*S^2)* v*w ]=0 (trilinears) = isogonal conjugate of the line {4, 2889, 6101, 11591}. O is the only ETC center on it.

 

Zn = NaNbNc->ABC:  The locus is the line {3, 143, 1173, 3060, 3567, 5422, 5495, 5946, 9777, 10263}.  

If OP=t*OH, then OZn=t*OX(143).

 

Zn(O) = O ; Zn(H)=X(143)

  

Za(O) = isogonal conjugate of X(11591)

= 1/((b^2+c^2)*a^2-(b^2-c^2)^2)/ (a^4-2*(b^2+c^2)*a^2+3*b^2*c^ 2+c^4+b^4) : : trilinears

= on lines: {30,54}, {95,3260}, {1990,8882}

= isogonal conjugate of X(11591)

= trilinear pole of the line {1637,2623}

= [ -5.392942000760921, 11.54709120356563, -1.864425427902168 ]

 

Za(H) = isogonal conjugate of X(6101)

= 1/(a*((b^2+c^2)*a^6-(3*b^4+4* b^2*c^2+3*c^4)*a^4+(b^2+c^2)*( 3*b^4-b^2*c^2+3*c^4)*a^2-(b^6- c^6)*(b^2-c^2))) : : (trilinears)

= on lines: {5,1614}, {53,10312}, {54,3613}, {311,1078}

= [ -5.946442603069915, -8.88166434592441, 12.534020999733550 ]

 

César Lozada

HYACINTHOS 25325

 

[Tran Quang Hung]:

 

Dear Mr Antreas Hatzipolakis, I see a similar problem as following

Let ABC be a triangle, with incenter I.

Denote:

A', B', C' = the reflections of I in midpoints of BC, CA, AB, resp.

A", B", C" = the circumcenters of IBC, ICA, IAB, resp.

The NPCs of AA'A", BB'B", CC'C" are coaxial.

Which are your intersection ?

[A', B', C' = the refletions of A, B, C, resp. in Spieker point X10]

[Angel Montesdeoca, Hyacinthos 25322]

(a^2 (b+c)+a (b^2-6 b c+c^2)-b^3+2 b^2 c+2 b c^2-c^3:...:...),
with (6-9-13)-search numbers (4.37905905513746,-2. 67770277764476,3. 47335453329039).
Is the reflection of X(i) in X(j) for these {i,j}: {10,121}, {106,1125}.