ADGEOM 4383 * ADGEOM 4386 * ADGEOM 4389 * ADGEOM 4392 * ADGEOM 4393

 

#4383

Dear Geometers!

 

Let ABC be a triangle, the lines through X13 perpendicular to BC meets AX14 at A'; the line through X14 perspendicular to BC meets AX13 at A''. Define B', C', B'', C'' cyclically.

Then A'B'C' and A''B''C'' are two new equilateral triangle.

Which is centroids of these triangle  in ETC?

 

Best regards

Sincerely
Dao Thanh Oai

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#4386

The centroids of the triangles  A'B'C' and A''B''C'' are X(15) and X(16), respec.

Angel Montesdeoca

 

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#4389

Dear Mr Dao Thanh Oai,

 

Your triangle A'B'C' is homothetic to the first Napoleon triangle and the homothetic center lies on line OX(14).

 

Best regards,

Tran Quang Hung.

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#4392

Dear Mr Hung and Mr Oai,

Homothetic centers are:

 

P1 = (A’B’C’, outer-Napoleon) = X(2)X(14) ∩ X(3)X(13)

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

= 7*S^2+sqrt(3)*(SB+SC)*S-3*SB*SC : : (barys)

= on lines: {2, 14}, {3, 13}, {4, 5352}, {5, 5238}, {6, 5054}, {16, 396}, {18, 3526}, {30, 10645}, {61, 140}, {62, 631}, {203, 5432}, {298, 11132}, {299, 618}, {303, 3643}, {381, 11480}, {397, 3530}, {398, 632}, {547, 5321}, {624, 11299}, {629, 633}, {630, 11290}, {1606, 3131}, {3104, 7786}, {3111, 14182}, {3201, 5012}, {3364, 5420}, {3365, 5418}, {3390, 15765}, {3523, 5237}, {3524, 10646}, {3533, 10187}, {3534, 12816}, {3582, 10638}, {3584, 7051}, {3851, 10188}, {4045, 11298}, {5070, 5339}, {5318, 8703}, {5335, 15692}, {5350, 12103}, {5433, 7005}, {5444, 7052}, {5463, 5569}, {5470, 6772}, {5474, 9735}, {5642, 10658}, {6669, 11303}, {6694, 11308}, {6774, 9117}, {6778, 12042}, {7619, 9761}, {7761, 11297}, {9116, 9885}, {10124, 11543}, {10182, 11244}, {11134, 13339}, {11268, 15330}, {11481, 15693}, {11485, 15694}, {11486, 15701}, {11489, 15709}, {11542, 12100}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2, 617, 623), (2, 13083, 5464), (396, 549, 16), (397, 3530, 5351), (619, 5981, 5464), (619, 6671, 2)

= [ 2.9030202383909050, 2.5290443728959210, 0.5499321137221704 ]

 

P2 = (A”B”C”, inner-Napoleon) = X(2)X(13) ∩ X(3)X(14)

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

= 7*S^2-sqrt(3)*(SB+SC)*S-3*SB*SC : : (barys)

= on lines: {2, 13}, {3, 14}, {4, 5351}, {5, 5237}, {6, 5054}, {15, 395}, {17, 3526}, {30, 10646}, {61, 631}, {62, 140}, {202, 5432}, {298, 619}, {299, 11133}, {302, 3642}, {381, 11481}, {397, 632}, {398, 3530}, {547, 5318}, {623, 11300}, {629, 11289}, {630, 634}, {1250, 3582}, {1605, 3132}, {2306, 5442}, {3105, 7786}, {3111, 14178}, {3200, 5012}, {3364, 15765}, {3389, 5420}, {3390, 5418}, {3523, 5238}, {3524, 10645}, {3533, 10188}, {3534, 12817}, {3851, 10187}, {4045, 11297}, {5070, 5340}, {5321, 8703}, {5334, 15692}, {5349, 12103}, {5433, 7006}, {5464, 5569}, {5469, 6775}, {5473, 9736}, {5642, 10657}, {6670, 11304}, {6695, 11307}, {6771, 9115}, {7619, 9763}, {7761, 11298}, {10124, 11542}, {10182, 11243}, {11137, 13339}, {11267, 15330}, {11480, 15693}, {11485, 15701}, {11486, 15694}, {11488, 15709}, {11543, 12100}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2, 616, 624), (2, 13084, 5463), (395, 549, 15), (398, 3530, 5352), (618, 5980, 5463), (618, 6672, 2)

= [ 8.2105967007085710, 5.4179799383431200, -3.8997508757340440 ]

 

·         {P1,P2} are harmonic-conjugates w/r to {X(6), X(5054)}

 

·         The center of inverse-similitude of A’B’C’ and A”B”C” is X(98)

 

César Lozada

 

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#4393

 

Thank you very much Mr Lozada,

 

I see a general problem for problem of Mr Dao as following

 

Let ABC be a triangle. P is any point.

 

(H) is rectangular hyperbola passes through A,B,C,P.

 

Q is a point on (H).

 

Perpendicular lines from P to BC,CA,AB meet QB,QB,QC at A',B',C', reps.

 

Then we get equal angles <B'A'C'=180-<BPC.

 

When P=F1 and Q is any point on Kiepert hyperbola then we get A'B'C' is equilateral triangle, is this triangle homothetic to the first Napoleon triangle and the homothetic center is on line X(2)X(14) ?

 

Best regards,

Tran Quang Hung.

ADGEOM 833 - ADGEOM 852

 

#833

Dear friends:

Let P be a point in the plane of ABC, A’B’C’ the cevian triangle of ABC and Ia, Ib, Ic its excenters.

The lines (IaC’) cut (IaB’) cut (BC) at Ab and Ac, respectively. Build Ba, Bc, Ca, Cb cyclically.

Then:

1) The points Ab, Ac, Ba, Bc, Ca, Cb lie on a conic.

2) If P=u:v:w (trilinears) then the center of the conic is:

    Z(P) = 2*(b+c)*u+(c+2*b)*v+(2*c+b)*w-a*v*w/u : :

3) If P lies on circumcircle of ABC then Z(P)=X(37)

4) Z(P) lies on line (X(37)P*), where P*=isogonal conjugate of P

 

ETC pairs [P,Z(P)] = [X(I),X(J)] for these (I,J)=

(2,1001), (44,88), (57,1), (81,37), (88,1), (513,100), (649,190), (650,651),       

(652,653), (654,655), (656,162), (657,658), (659,660), (661,662), (672,673),       

(770,771), (798,799), (822,823), (896,897), (1155,1156), (1491,1492), (1635,3257), 

(1755,1821), (2173,2349), (2578,2580), (2579,2581), (3218,1807), (3768,4607),      

(4790,4606), (4893,4604)

 

Non-ETC:

Z( X(1) ) = 5*b + 5*c-a : :

    On lines (1,6), (2,1266), (8,4029), (10,4873), (63,89), (142,4346), (145,3707), (346,5257), 

 (527,5308), (966,3626), (968,3158), (1696,5217), (2321,3617),         

 (2345,3634), (3161,5550), (3241,4700), (3621,3686), (3625,4034), (3677,3989),      

 (3679,3943), (3729,4687), (3875,4704), (3929,5287), (4363,4755), (4384,4664),      

 (4419,4887), (4648,4896), (4677,4727)

 

Z( X(6) ) = (b+c)*a+b^2+b^2+(b*c)/2  : :

   On lines (1,89), (2,37), (8,4868), (38,4430), (42,4661), (43,3989), (63,1449), (145,3931),  

 (982,1962), (984,3240), (1150,4360), (1743,3219), (1999,5372),         

 (3187,5361), (3247,3306), (3616,3743), (3936,4389), (4361,5235), (4646,4678),      

 (4651,4734), (4709,4970)

 

Regards

César Lozada

 

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#842

Dear  César,

This may be the same conic discussed in Hyacinthos #20547.  Your point Z(X(1)) matches a point I have in my notes as 'Hyacinthos #20547 [https://aphyacinthos.blogspot.com/2019/10/hyacinthos-20547.html] conic center when P = X(1)', but the new Yahoo groups format makes it difficult to find a message based on message #, so I cannot confirm.

 

Best regards,

Randy Hutson

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#844

Thank you, Francisco Javier.  These two conics are similar: yours uses the cevian and anticevian triangles of a point P, while César's uses the cevian triangle of P and the anticevian triangle of X(1).

Reviewing my notes from back then, I had found a (unique?) point P such that the conic (
Francisco Javier's) is a circle.  It is a non-ETC center, P^ (search=0.834038702291580), as is the center, O^ (search=2.227553680841584).  P^ and O^ are collinear with X(3534), but I could find nothing else of interest about these points, except that P^ is also the perspector of the circle.  (Does this hold in general?)

 

I wonder which is the (unique?) point P such that César's conic is a circle.

 

Best regards,

Randy Hutson

 

--Francisco Javier wrote:

Dear Randy, according to my record, this is Hyacinthos #20547:

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

From: "Francisco Javier" <garciacapitan@...>
Date: Mon Dec 19, 2011 10:47 am
Subject: The anticevian intersections conic     garciacapitan
  
Dear friends,

As a little Christmas present, in

http://garciacapitan.blogspot.com/2011/12/anticevian-intersection-conic.html

you can read a little research on a conic through six intersection points.

I gladly wish to express my gratitude to Jean-Pierre Ehrmann, Bernard Gibert and Sung Hyun Lim for several helpful comments concerning this problem.

Have a Happy Christmas and New Year 2012.

Francisco Javier García Capitán.

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

 

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#848

Another interesting property of P^: it is the trilinear pole of its polar wrt circle (O^).  Similar to K wrt circumcircle.

Coordinates?

Best regards,
Randy Hutson

 

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#852

[Randy Hutson]:

>> I wonder which is the (unique?) point P such that César's conic is a circle.

[César Lozada]:

The conic is a circle for P with ETC-search 1.41380272606812.., 3.85478662709369.., 0.319441712657307.. (at least)

The center Z of the circle for this P has ETC-search 5.302845236905069.., 5.39706500845055.., -2.543232171745264..

No relations were found between P, Z and other ETC centers.

Best regards

César Lozada

 

 

ADGEOM 2197 * ADGEOM 2198 * ADGEOM 2199

#2197

 

Dear Dr. Garcia Capitan, Dr. Randy Hutson and all geometer,

 

Let ABC be a triangle, let a circle through A_c,A_b,C_b,C_a,B_a,B_c,A_c

 such that the center is Incenter (X1) and radius =2r/sqrt{3). Let A_1,B_1,C_1 be 

intersections of three circle (IB_aB_c), (IC_aC_b), (IA_bA_c) then

A_1B_1C_1 be an equilateral triangle. 

 

1-Which is the center of A1B1C1 in Kimberling center?

 

2-ABC and A_1B_1C_1 are perpective which is the perpector in Kimberling center?

 

Thank to You very much,

 

Best regards

Sincerely

Dao Thanh Oai

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

 

#2198

 

Dear Mr. Dao:

 

Maybe I didn´t understand well, but A1B1C1 is not equilateral.

I supposed:

   {Ab,Ac} = {I,2*r/sqrt(3)} /\ (BC)

   A1 =Circle{ I,Ba,Bc}/\ Circle{ I,Ca,Cb}-{I}

 

Perspector ABC, A1B1C1 is I.

 

Trilinears of A1 are:

A1 = [((b+c)*a^2+3*a*b*c-(b^2-c^2)*(b-c))/(a*(c^2+b*c+b^2-a^2)), 1, 1]

and cyclically for B1, C1

 

César Lozada

 

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

 

#2199

 

Dear Dao,

 

If I am correct the perspector is the point with barycentrics

S/(s-a)+32a/Sqrt[3] : . . .

Search number  1.58801253144064532

 

Best regards

Nikos Dergiades

 

 

 

 

HYACINTHOS 27565

[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 B'C', C'A', A'B', resp.

(Oa), (Ob), (Oc) = the circles with diameters AA", BB", CC", resp.

Ra = the radical axis of (Ob), (Oc)
Rb = the radical axis of (Oc), (Oa)
Rc = the radical axis of (Oa), (Ob)

R1, R2, R3 = the reflections of Ra, Rb, Rc in BC, CA, AB, resp.

A*B*C* = the triangle bounded by R1, R2, R3

Which is the locus of P such that 

 

1. R1, R2, R3 are concurrent?
O lies on the locus

2. ABC, A*B*C* are parallelogic?
O lies on the locus

3. A'B'C', A*B*C* are parallelogic?
O, H lie on the locus

 

 

 

[César Lozada]:

 

Loci are monster problems!

 

1)      For P=O, {R1, R2, R3} concur at:

Q1 = isogonal conjugate of X(14457)

=  a*(a^8-2*(b^2+c^2)*a^6+6*b^2* c^2*a^4+2*(b^2+c^2)*(b^4-4*b^2 *c^2+c^4)*a^2-(b^4+c^4)*(b^2-c ^2)^2)

= 2*X(3)+X(10594), X(20)+2*X(1906)

= as a point on the Euler line, this center has Shinagawa coefficients (E-2*F, 2*F)

= on lines: {2, 3}, {6, 2929}, {35, 4319}, {36, 4320}, {51, 13346}, {54, 9932}, {64, 15305}, {74, 15058}, {110, 974}, {155, 5890}, {182, 6467}, {184, 9729}, {185, 9306}, {323, 12160}, {371, 9694}, {372, 9682}, {389, 1092}, {394, 5889}, {568, 16266}, {569, 5892}, {578, 5422}, {895, 14528}, {962, 11365}, {1105, 14249}, {1147, 7592}, {1192, 11444}, {1204, 5651}, {1498, 15072}, {1578, 11417}, {1579, 11418}, {1587, 8276}, {1588, 8277}, {1602, 1603}, {1619, 12324}, {1899, 14516}, {1941, 3168}, {1994, 11432}, {2079, 5013}, {2207, 15355}, {2931, 15035}, {2979, 17834}, {2986, 13599}, {3053, 9608}, {3100, 11399}, {3357, 15030}, {3425, 13335}, {3796, 17821}, {4293, 10037}, {4294, 10046}, {4296, 11398}, {4297, 8185}, {5012, 11449}, {5085, 8549}, {5206, 9699}, {5218, 10831}, {5432, 9659}, {5433, 9672}, {5462, 13352}, {5562, 11438}, {5621, 6698}, {5640, 10982}, {5731, 9798}, {5891, 7689}, {5943, 11424}, {6090, 12164}, {6193, 12421}, {6194, 9917}, {6449, 9695}, {6684, 15177}, {6746, 11255}, {6800, 10282}, {7288, 10832}, {7691, 7998}, {7987, 9590}, {8567, 10117}, {9545, 11402}, {9591, 16192}, {9658, 15326}, {9673, 15338}, {9700, 15515}, {9706, 17809}, {9778, 9911}, {9826, 15463}, {9938, 14852}, {10539, 11456}, {10546, 12279}, {10601, 11425}, {10605, 12111}, {10606, 15062}, {10643, 11476}, {10644, 11475}, {10961, 11474}, {10963, 11473}, {11064, 12233}, {11430, 11695}, {11439, 13445}, {11440, 15056}, {11457, 12134}, {11459, 12163}, {12022, 12118}, {12302, 14644}, {14986, 16541}, {15024, 15033}

= isogonal conjugate of X(14457)

= X(1906) of Ara triangle

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (3, 4, 11413), (3, 1593, 2071), (3, 5899, 15696), (3, 7387, 376), (3, 9715, 6636), (3, 12083, 548), (4, 631, 3546), (4, 3546, 858), (4, 6642, 1995), (25, 16542, 6995), (1593, 5020, 3091), (3146, 13595, 1598), (3530, 7502, 3), (5004, 5005, 7386), (5056, 7527, 11479), (15760, 16238, 7505)

= [ 3.9507895877465620, 3.0708473821603260, -0.3087481308557978 ]

 

2)      For P=O, {R1, R2, R3} concur and triangle A*B*C* degenerates to a point. For P=H A*B*C* and ABC are not parallelogic

3)      For P=O, {R1, R2, R3} concur and triangle A*B*C* degenerates to a point. For P=H, parallelogic centers (A’->A*, A*->A’) are (1593, 1181)

 

César Lozada

HYACINTHOS 27556

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, IaIbIc the antipedal tringle of I and Fe, Fa, Fb, Fc  the Feuerbach point and the three ex-Feuerbach points.

The circumcircles of FeAIa, FeBIb, FeCIc are coaxial.
Let F0 be the second, other than Fe, intersection.

 

The circumcircles of FaAIa, FaBIb, FaCIc are coaxial.
Let F1 be the second, other than Fa, intersection.

The circumcircles of FbAIa, FbBIb, FbCIc are coaxial.
Let F2 be the second, other than Fb, intersection.

The circumcircles of FcAIa, FcBIb, FcCIc are coaxial.
Let F3 be the second, other than Fc, intersection.

The points F0, F1, F2, F3 are concyclic.

1. Point F0 ?

2. Center of the circle?

 

 

[Peter Moses]:

Hi Antreas,

1). X(5531).

2).
Radius of circle = 2 R^2 / (R - r).

Center of circle = 

a (a^6-2 a^5 b-a^4 b^2+4 a^3 b^3-a^2 b^4-2 a b^5+b^6-2 a^5 c+2 a^3 b^2 c-2 a^2 b^3 c+2 b^5 c-a^4 c^2+2 a^3 b c^2-2 a^2 b^2 c^2+2 a b^3 c^2-b^4 c^2+4 a^3 c^3-2 a^2 b c^3+2 a b^2 c^3-4 b^3 c^3-a^2 c^4-b^2 c^4-2 a c^5+2 b c^5+c^6):: 

 

2 X[1837] - 3 X[5587], 4 X[496] - 5 X[8227], 3 X[5660] - X[12750].
(3 R + r) X[1] - 4 R X[5].
(2 R + r) X[3] - 3 R X[210].

 

= on the cubic K364 and these lines: {{1,5},{3,210},{4,2900},{8,626 1},{9,10902},{10,6889},{20,442 0},{33,2910},{35,7330},{40,64} ,{46,912},{55,1898},{63,6796}, {78,515},{84,1259},{100,1158}, {101,7079},{224,5552},{254,182 6},{354,6918},{404,12757},{411 ,3681},{474,12675},{480,971},{ 518,3149},{581,612},{610,2917} ,{936,958},{944,997},{946,3870 },{962,3935},{990,4878},{1071, 1376},{1260,12664},{1709,11248 },{1745,8270},{1750,5812},{176 6,3949},{1858,11501},{1867,319 1},{2948,12890},{3085,10393},{ 3158,12705},{3174,11372},{3185 ,15623},{3295,9848},{3306,1200 5},{3338,6911},{3359,15071},{ 3434,12608},{3475,6864},{3586, 10953},{3689,10306},{3694, 5776},{3711,5584},{3873,6915}, {3880,6765},{3913,12672},{ 3940,14110},{4294,5811},{4413, 9940},{4557,15622},{4853, 11014},{4855,5450},{5119,5887} ,{5178,6932},{5220,5732},{ 5248,15064},{5268,5788},{5440, 12114},{5657,12520},{5691, 10526},{5709,5904},{5779,15837 },{5791,8580},{5794,15843},{ 5818,6877},{5927,11496},{6245, 6745},{6256,10522},{6259, 11826},{6282,11827},{6684, 10884},{6826,13407},{6834, 10916},{7702,12831},{8583, 16842},{10175,10585},{10830, 15177},{11012,12687},{11491, 12514},{12115,17647},{12332, 17661},{12407,13214},{12775, 13548},{16139,16143}}.

 

= reflection of X(i) in X(j) for these {i,j}: {{40, 5687}, {46, 11499}, {1898, 5777}, {6264, 12740}, {10073, 119}, {10085, 3}, {12704, 3149}}.

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 5531, 5534), (12, 355, 5587), (72, 11500, 40), (100, 12528, 1158), (200, 1490, 40), (355, 10942, 10827), (3689, 12688, 10306), (3913, 12672, 12703), (5534, 5720, 1), (5881, 6326, 1).

 

Best regards,
Peter Moses.

 

HYACINTHOS 27555

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, IaIbIc the antipedal tringle of I and Fe, Fa, Fb, Fc  the Feuerbach point and the three ex-Feuerbach points.

The circumcircles of FeAIa, FeBIb, FeCIc are coaxial.
Let F0 be the second, other than Fe, intersection.

 

The circumcircles of FaAIa, FaBIb, FaCIc are coaxial.
Let F1 be the second, other than Fa, intersection.

The circumcircles of FbAIa, FbBIb, FbCIc are coaxial.
Let F2 be the second, other than Fb, intersection.

The circumcircles of FcAIa, FcBIb, FcCIc are coaxial.
Let F3 be the second, other than Fc, intersection.

The points F0, F1, F2, F3 are concyclic.

Center of the circle?

[Angel Montesdeoca]:

The points F0=X(5531), F1, F2, F3 lies on circle of center:

W = (r+3R) X(1) - 4R X(5) =  X(1)X(5) /\ X(40)X(64)

with first barycentric coordinate:
 
 a (-a^6+2 a^5 (b+c)+a^4 (b^2+c^2)-2 a^3 (2 b^3+b^2 c+b c^2+2 c^3)+a^2 (b+c)^2 (b^2+c^2)+2 a (b^5-b^3 c^2-b^2 c^3+c^5)-(b-c)^2 (b+c)^4)

 
W is the reflection of X(i) in X(j), for these {i, j}: {40,5687}, {46,11499}, {1898,5777}, {6264,12740}, {10073,119}, {10085,3}, {12704,3149}.

W lies on lines X(i)X(j) for these {i, j}:  {1,5}, {3,210}, {4,2900}, {8,6261}, {9,10902}, {10,6889}, {20,4420}, {33,2910}, {35,7330}, {40,64}, {46,912}, {55,1898}, {63,6796}, {78,515}, {84,1259}, {100,1158}, {101,7079}, {224,5552}, {254,1826}, {354,6918}, {404,12757}, {411,3681}, {474,12675}, {480,971}, {518,3149}, {581,612}, {610,2917}, {936,958}, {944,997}, {946,3870}, {962,3935}, {990,4878}, {1071,1376}, {1260,12664}, {1709,11248}, {1745,8270}, {1750,5812}, {1766,3949}, {1858,11501}, {1867,3191}, {2948,12890}, {3085,10393}, {3158,12705}, {3174,11372}, {3185,15623}, {3295,9848}, {3306,12005}, {3338,6911}, {3359,15071}, {3434,12608}, {3475,6864}, {3586,10953}, {3689,10306}, {3694,5776}, {3711,5584}, {3873,6915}, {3880,6765}, {3913,12672}, {3940,14110}, {4294,5811}, {4413,9940}, {4557,15622}, {4853,11014}, {4855,5450}, {5119,5887}, {5178,6932}, {5220,5732}, {5248,15064}, {5268,5788}, {5440,12114}, {5657,12520}, {5691,10526}, {5709,5904}, {5779,15837}, {5791,8580}, {5794,15843}, {5818,6877}, {5927,11496}, {6245,6745}, {6256,10522}, {6259,11826}, {6282,11827}, {6684,10884}, {6826,13407}, {6834,10916}, {7702,12831}, {8583,16842}, {10175,10585}, {10830,15177}, {11012,12687}, {11491,12514}, {12407,13214}, {12775,13548}, {16139,16143}.

 (6 - 9 - 13) - search numbers  of W: (-4.20253732826351, -8.71491257242104, 11.6136981066283).
 
 Angel Montesdeoca

HYACINTHOS 27554

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, IaIbIc the antipedal tringle of I and Fe, Fa, Fb, Fc  the Feuerbach point and the three ex-Feuerbach points.

The circumcircles of FeAIa, FeBIb, FeCIc are coaxial.
Let F0 be the second, other than Fe, intersection.

 

The circumcircles of FaAIa, FaBIb, FaCIc are coaxial.
Let F1 be the second, other than Fa, intersection.

The circumcircles of FbAIa, FbBIb, FbCIc are coaxial.
Let F2 be the second, other than Fb, intersection.

The circumcircles of FcAIa, FcBIb, FcCIc are coaxial.
Let F3 be the second, other than Fc, intersection.

The points F0, F1, F2, F3 are concyclic.

[Randy Hutson]:

Hi Antreas,

The construction you describe is the Gibert-Burek-Moses concurrent circles image, described at X(5524).  F0 = X(5531) = GIBERT-BUREK-MOSES CONCURRENT CIRCLES IMAGE OF X(11), and F1, F2, F3 are the Gibert-Burek-Moses concurrent circles images of Fa, Fb, Fc, resp.  The circle is the Gibert-Burek-Moses concurrent circles image of the nine-point circle, and its center is not yet in ETC.  It has the following properties:

Trilinears: a^6 - 2 a^5 (b + c) - a^4 (b^2 + c^2) +  2 a^3 (b + c) (2 b^2 - b c + 2 c^2) - a^2 (b + c)^2 (b^2 + c^2) - 2 a (b - c)^2 (b + c) (b^2 + b c + c^2) + (b - c)^2 (b + c)^4 : :
Lies on lines {1,5}, {3,210}, {8,6261}, {9,10902}, {10,6889}, {20,4420}, {33,2910}, {35,7330}, {40,64}, {46,912}, {55,1898}, {63,6796}, {78,515}, {84,1259}, {100,1158}, {101,7079} et al
inner-Johnson-to-outer-Johnson similarity image of X(1)
X(1837)-of-outer-Johnson-trian gle
(6,9,13) values = (4.202537328263515, 15.289361876597825, 8.714912572421045)

Best regards,
Randy Hutson.

 

HYACINTHOS 27552

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P a point.

Denote:

A' = the 2nd intersection of the circle (P, PH) and AH
B' = the 2nd intersection of the circle (P, PH) and BH
C' = the 2nd intersection of the circle (P, PH) and CH

[or, to avoid circles:
Pa, Pb, Pc = the orthogonal projections of P on AH, BH, CH, resp.
A', B', C' = the reflections of H in Pa, Pb, Pc, resp.]

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

La = the Euler line or AAbAc.
Similarly Lb, Lc.

Which is the locus of P such that La, Lb, Lc are concurrent?
The Euler line + something else ?
And which is the locus of the point of concurrence?

[César Lozada]:

 

 

Locus = {Linf} ∪ {Euler line}

 

If OP/OH=t then the point of intersection Q(P) lies on the line X(125)X(403), through ETC’s 125, 403, 6000, 12292, 13399, and

X(125)*Q/X(125)X(403) = (OH/R)^2*(t-1)

 

ETC pairs (P,Q): (4,125), (382,12292), (3153,13399)

 

Some others Q(P):

 

Q(X(2)) = X(74)X(184) ∩ X(125)X(403)

= (SB+SC)*((-6*R^2+5*SA+SW)*S^2+ (216*R^4-9*R^2*(3*SA+7*SW)+5*S A^2-5*SB*SC+4*SW^2)*SA) : : (barys)

= 5*X(125)-2*X(12292), 2*X(185)+X(10990), 5*X(185)-2*X(13148), 4*X(185)-X(14448), 4*X(974)-X(13202), 5*X(10990)+4*X(13148), 2*X(10990)+X(14448), X(11381)-4*X(16270), 8*X(13148)-5*X(14448), 2*X(13491)+X(16003)

= on lines: {51, 974}, {74, 184}, {125, 403}, {185, 1205}, {542, 15072}, {549, 5642}, {1154, 16111}, {2777, 5890}, {2935, 11402}, {3819, 12825}, {7687, 11455}, {7729, 9971}, {11381, 16270}, {13491, 16003}, {14855, 16163}, {15054, 15462}

= reflection of X(51) in X(974)

= {X(185), X(10990)}-Harmonic conjugate of X(14448)

= [ 15.0847616889502900, 15.8297465474735500, -14.2805885227820600 ]

 

Q(X(3)) = X(3)X(74) ∩ X(125)X(403)

= (SB+SC)*((-5*R^2+3*SA+SW)*S^2+ (108*R^4-R^2*(15*SA+32*SW)+3*S A^2-3*SB*SC+2*SW^2)*SA) : : (barys)

= 3*X(4)-4*X(11746), 3*X(74)-X(12281), X(110)-3*X(15072), 3*X(185)-X(13417), 3*X(974)-2*X(11746), 2*X(1112)-3*X(5890), 2*X(1539)-3*X(16222), 3*X(1986)-2*X(13417), 3*X(5890)-X(10721), 3*X(6241)+X(12281), 4*X(6723)-3*X(15030), 3*X(11459)-4*X(13416), X(12111)-3*X(15055), 2*X(12358)-3*X(15055), 4*X(13630)-3*X(16222)

= on lines: {3, 74}, {4, 974}, {24, 9934}, {125, 403}, {184, 13293}, {185, 1986}, {186, 15647}, {265, 11457}, {378, 13198}, {382, 12236}, {389, 13202}, {1112, 5890}, {1181, 2935}, {1204, 13289}, {1539, 13630}, {2071, 3047}, {2693, 10419}, {2781, 6776}, {3028, 9627}, {3520, 11598}, {5504, 11413}, {5656, 7505}, {6699, 12162}, {6723, 15030}, {7592, 15472}, {7687, 11381}, {7728, 12824}, {7731, 13148}, {9786, 17812}, {9826, 10574}, {10018, 14862}, {10117, 10605}, {10264, 15761}, {10293, 10938}, {10575, 17702}, {10619, 10628}, {10706, 16072}, {10733, 12279}, {11455, 12099}, {11806, 12295}, {11807, 13382}, {12133, 12290}, {12140, 16659}, {12283, 12284}, {13754, 16111}, {15059, 15305}

= midpoint of X(i) and X(j) for these {i,j}: {10293, 10938}, {10733, 12279}

= reflection of X(i) in X(j) for these (i,j): (4, 974), (382, 12236), (1539, 13630), (7728, 14708), (7731, 13148), (11455, 12099), (11807, 13382), (12133, 16270)

= 2nd Droz-Farny circle-inverse-of X(74)

= X(974) of anti-Euler triangle

= X(12825) of ABC-X3 reflections triangle

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (1181, 2935, 15463), (1539, 13630, 16222), (5890, 10721, 1112), (6241, 15072, 11456), (7728, 14708, 12824), (10620, 13171, 74), (12111, 15055, 12358), (12133, 16270, 14644), (12290, 14644, 12133)

= [ 20.8820066825843300, 22.2047506268805900, -21.3697044209718700 ]

 

Q(X(5)) = X(54)X(74) ∩ X(125)X(403)
= (SB+SC)*((-R^2+2*SA)*S^2+(108* R^4-R^2*(12*SA+31*SW)+2*SA^2-2 *SB*SC+2*SW^2)*SA) : : (barys)

= 3*X(51)-X(10721), 3*X(74)+X(7722), 3*X(125)-X(12292), X(146)-5*X(10574), X(146)-3*X(16223), 3*X(185)-X(7722), 3*X(376)+X(12284), 3*X(389)-2*X(1112), 5*X(389)-2*X(16105), 3*X(974)-X(1112), 4*X(974)-X(11807), 5*X(974)-X(16105), 4*X(1112)-3*X(11807), 5*X(1112)-3*X(16105), 2*X(5972)-3*X(16836), 5*X(10574)-3*X(16223), 3*X(11800)-4*X(13358), 3*X(11806)-2*X(13358), 5*X(11807)-4*X(16105)

= on lines: {3, 17838}, {30, 11800}, {51, 10721}, {54, 74}, {67, 13623}, {113, 9729}, {125, 403}, {140, 5663}, {146, 10574}, {182, 5621}, {265, 10575}, {376, 12284}, {389, 974}, {511, 16111}, {578, 2935}, {1539, 5462}, {1598, 17812}, {1614, 17701}, {1986, 10990}, {2807, 11709}, {3448, 15072}, {3522, 12273}, {5449, 10264}, {5562, 15055}, {5622, 12294}, {5890, 12244}, {6102, 14677}, {7687, 13474}, {7728, 9730}, {7729, 10117}, {9786, 9919}, {10110, 13202}, {10113, 14915}, {10605, 13171}, {11381, 14644}, {11557, 13630}, {11558, 11801}, {11793, 12825}, {12041, 13754}, {12121, 14855}, {12162, 15061}, {12219, 15021}, {12236, 13598}, {15012, 16222}, {15030, 15059}

= midpoint of X(i) and X(j) for these {i,j}: {265, 10575}, {1986, 10990}, {6102, 14677}

= reflection of X(i) in X(j) for these (i,j): (113, 9729), (389, 974), (1539, 5462), (1986, 13382), (7687, 16270), (11557, 13630)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (74, 13198, 13293), (146, 10574, 16223), (5890, 12244, 13417), (13198, 13293, 11430)

= [ 12.1861391921332800, 12.6422445077700400, -10.7360305736871600 ]

 

Q(X(20)) = X(74)X(6759) ∩ X(125)X(403)

= (SB+SC)*((-14*R^2+7*SA+3*SW)*S ^2+(216*R^4-R^2*(33*SA+65*SW)+ 7*SA^2-7*SB*SC+4*SW^2)*SA) : : (barys)

= 3*X(125)-2*X(12292), 3*X(185)-2*X(1112), 5*X(974)-4*X(15465), 4*X(1112)-3*X(13202), 2*X(5972)-3*X(15072), 3*X(6241)-X(7722), 5*X(6241)-X(7731), 4*X(6241)-X(14448), 4*X(6723)-3*X(15305), 5*X(7722)-3*X(7731), 4*X(7722)-3*X(14448), 4*X(7731)-5*X(14448), 5*X(11381)-8*X(15465), 3*X(12295)-4*X(13358)

= on lines: {74, 6759}, {113, 13491}, {125, 403}, {185, 1112}, {550, 5562}, {974, 11381}, {1204, 9934}, {2777, 6241}, {2935, 12174}, {3047, 13445}, {5972, 15072}, {6723, 15305}, {7687, 12290}, {10620, 12893}, {11456, 13293}, {11598, 13367}, {12295, 13358}

= reflection of X(113) in X(13491)

= [ 38.2737416634864200, 41.3297628651016900, -42.6370521155412900 ]

 

César Lozada

HYACINTHOS 27538

[Tran Quang Hung]:

Let ABC be a triangle. and P a point.

A',B',C' are midpoints of PA,PB,PC, resp.

Q is the radical center of the circles (A',BC), (B',CA), (C',AB).

Then the line PQ passes through a fixed point on the Euler line.

Which is this point ?


[Angel Montesdeoca]:

**** The line PQ passes through a fixed point on Euler line:  

W = 9 a^4-4 (b^2-c^2)^2-5 a^2 (b^2+c^2) ::

 = 5 O - 4 H

= lies on these lines: {2,3}, {35,9656}, {36,9671}, {485,6445}, {486,6446}, {516,8148}, {999,4325}, {1154,12279}, {1159,1770}, {1384,7748}, {1482,9589}, {1539,15040}, {1975,7850}, {2777,12308}, {2794,14692}, {3068,9691}, {3070,9681}, {3295,4330}, {3411,5339}, {3412,5340}, {3527,13623}, {4296,9642}, {4299,9668}, {4301,10247}, {4302,9655}, {4309,6767}, {4316,12953}, {4317,6284}, {4324,12943}, {4746,11362}, {4816,5881}, {4902,7967}, {5024,7747}, {5093,12007}, {5840,8158}, {5925,13093}, {6101,12290}, {6241,13391}, {6243,10575}, {6361,12645}, {6407,13665}, {6408,13785}, {6417,6560}, {6418,6561}, {6455,6564}, {6456,6565}, {6472,9693}, {6496,10576}, {6497,10577}, {6519,8960}, {7737,9607}, {7756,9605}, {8717,11424}, {8976,9680}, {9606,15484}, {9654,15338}, {9669,15326}, {9690,13903}, {9692,13886}, {9703,13346}, {10113,15057}, {10263,15072}, {10574,13321}, {10627,15305}, {10706,15039}, {10733,15041}, {11439,15067}, {11455,11591}, {11485,16965}, {11486,16964}, {11820,12316}, {12002,16226}, {12121,15063}, {12162,13340}, {12902,16003}, {13348,16194}, {13598,14855}, {13881,15655}, {15046,15051}.

= the midpoint of X(3529) and X(5059).

= the reflection of X(i) in X(j), for these {i, j}: {3,1657}, {4,15704}, {381,11001}, {382,20}, {1657,3529}, {3146,550}, {3534,15683}, {3830,15681}, {5073,3}, {6243,10575}, {12290,6101}, {12645,6361}, {13093,5925}, {15640,549}, {15681,15685}, {15682,15686}, {15684,3534}.

 (6 - 9 - 13) - search numbers  of W:  (56.6182921146879, 55.5994117075985, -60.9827553685936)
 
 Angel Montesdeoca

HYACINTHOS 27235

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of  O.

Denote:

Ab, Ac = the reflections of A' in BO, CO, resp.
Bc, Ba = the reflections of B' in CO, AO, resp.
Ca, Cb = the reflections of C' in AO, BO, resp.

P, Pa, Pb, Pc = same points on the Euler lines of ABC, AAbAc, BBcBa, CCaCb, resp. such that PO/PH = t (number) etc.

ABC, PaPbPc are orthologic.
The orthologic center (PaPbPc, ABC) is the P.
Which is the locus of the other one (ABC, PaPbPc) as P moves on the Euler line?


[César Lozada]:

 

When P is an OH-absolutely positioned center on the Euler line(*),   triangles ABC and PaPbPc are orthologic and the orthologic center A->Pa = Qa(P) lies on the circumconic:

 

q = ∑ [(SB-SC)*(SA^2+SB*SC)*(SA-8*R^ 2+SW)*y*z ] =0 (barys)

This conic passes through ETC’s 264, 1975, 10621, 10622 and has perspector:

 

q_pers = X(297)X(525) ∩ X(2451)X(17215)

= (b^2-c^2)*(a^4-(b^2+c^2)*a^2+ 2*b^2*c^2)*((b^2+c^2)*a^4-2*( b^2-c^2)^2*a^2+(b^4-c^4)*(b^2- c^2)) : : (barys)

= on lines: {297, 525}, {2451, 17215}

= [ 5.4619738886654760, -0.7348553712368464, 1.6284994826104380 ]

 

The center of q is not related to any ETC center X(i) for 1<=i<=17772

 

IF OP/OH=t, then

Qa(P) = f(SB,SC,SA)*f(SC,SA,SB) : : (barys)

where

f(SA,SB,SC)=(SB+SC)* ((35*R^2+4*SA-8*SW)*S^2+(3*R^ 2-4*SW)*SA^2)*t-(S^2+SA^2)*R^2

 

ETC pairs (P,Qa): (2,264), (3,1975), (4,10621), (5,10622)

 

No other interesting Qa(P) found.

 

César Lozada

 

 

 

(*) Centers P for which OP/OH=t=real constant not depending on the shape of the triangle.

HYACINTHOS 27522

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle:

Denote:

Ra = the radical axis of (N) and (A, AN)
(ie the radical axis of the circles: NPC and the circle centered at A with radius AN)
Rb = the radical axis of (N) and (B, BN)
Rc = the radical axis of (N) and (C, CN)

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

1. ABC, A*B*C* are perspective.

2. The NPC center of A*B*C* lies on the Euler line.

[Peter Moses]:


Hi Antreas,

1. X(16837).

2. 

4 a^12 b^4-20 a^10 b^6+40 a^8 b^8-40 a^6 b^10+20 a^4 b^12-4 a^2 b^14+6 a^12 b^2 c^2-25 a^10 b^4 c^2+22 a^8 b^6 c^2+28 a^6 b^8 c^2-56 a^4 b^10 c^2+29 a^2 b^12 c^2-4 b^14 c^2+4 a^12 c^4-25 a^10 b^2 c^4+22 a^8 b^4 c^4+9 a^6 b^6 c^4+29 a^4 b^8 c^4-63 a^2 b^10 c^4+24 b^12 c^4-20 a^10 c^6+22 a^8 b^2 c^6+9 a^6 b^4 c^6+14 a^4 b^6 c^6+38 a^2 b^8 c^6-60 b^10 c^6+40 a^8 c^8+28 a^6 b^2 c^8+29 a^4 b^4 c^8+38 a^2 b^6 c^8+80 b^8 c^8-40 a^6 c^10-56 a^4 b^2 c^10-63 a^2 b^4 c^10-60 b^6 c^10+20 a^4 c^12+29 a^2 b^2 c^12+24 b^4 c^12-4 a^2 c^14-4 b^2 c^14:: 

 

2 X[10289]+X[15557].

 

= on the line {2,3}.

Best regards,
Peter Moses.

 

HYACINTHOS 27520

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the pedal triangle of O.

Denote:

A1, B1, C1 = the orthogonal projections of A, B, C on OA', OB', OC', resp.

A2, B2, C2 = the orthogonal projections of A', B', C' on OA, OB, OC, resp

Na, Nb, Nc = the NPC centers of A'A1A2, B'B1B2, C'C1C2, resp.

[...]

2. ABC, NaNbNc are orthologic

 

[Peter Moses]:

 

 

Hi Antreas,

2) (ABC, NaNbNc):  It is now X(17711)
.
Isogonal Conjugate of X(17711):

a^2 (a^8-2 a^6 b^2+2 a^2 b^6-b^8-2 a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2+3 b^6 c^2+a^2 b^2 c^4-4 b^4 c^4+2 a^2 c^6+3 b^2 c^6-c^8)::

X[3] - 3 X[26], 2 X[3] - 3 X[1658], X[1658] + 2 X[7387], X[3] + 3 X[7387], X[3] - 9 X[9909], X[1658] - 6 X[9909], X[26] - 3 X[9909], X[7387] + 3 X[9909], 3 X[159] - X[9925], 5 X[632] - 6 X[10020], 11 X[3525] - 12 X[10125], 5 X[632] - 9 X[10154], 2 X[10020] - 3 X[10154], 7 X[3090] - 9 X[10201], 7 X[3090] - 6 X[10224], 3 X[10201] - 2 X[10224], 7 X[3] - 6 X[10226], 7 X[1658] - 4 X[10226], 7 X[26] - 2 X[10226], 7 X[7387] + 2 X[10226], 5 X[1658] - 14 X[10244], 5 X[26] - 7 X[10244], 15 X[9909] - 7 X[10244], 5 X[7387] + 7 X[10244], 11 X[1658] - 18 X[10245], 11 X[26] - 9 X[10245], 11 X[9909] - 3 X[10245], 11 X[7387] + 9 X[10245], 12 X[5498] - 13 X[10303], 8 X[10226] - 7 X[11250], 4 X[3] - 3 X[11250], 4 X[26] - X[11250], 12 X[9909] - X[11250], 4 X[7387] + X[11250], 7 X[3090] - 8 X[12010], 9 X[10201] - 8 X[12010], 3 X[10224] - 4 X[12010], 10 X[10226] - 7 X[12084], 5 X[11250] - 4 X[12084], 5 X[3] - 3 X[12084], 5 X[1658] - 2 X[12084], 5 X[26] - X[12084], 15 X[9909] - X[12084], 7 X[10244] - X[12084], 5 X[7387] + X[12084], 7 X[12084] - 5 X[12085], 7 X[11250] - 4 X[12085], 7 X[3] - 3 X[12085], 7 X[1658] - 2 X[12085], 7 X[26] - X[12085], 7 X[7387] + X[12085], 3 X[12085] - 14 X[12107], 3 X[12084] - 10 X[12107], 3 X[11250] - 8 X[12107], 3 X[10226] - 7 X[12107], 3 X[1658] - 4 X[12107], 3 X[26] - 2 X[12107], 9 X[9909] - 2 X[12107], 3 X[7387] + 2 X[12107], 4 X[3628] - 3 X[13371], 2 X[3628] - 3 X[13383], 5 X[3091] - 6 X[13406], 5 X[11250] - 12 X[14070], 15 X[10245] - 11 X[14070], 5 X[3] - 9 X[14070], 10 X[12107] - 9 X[14070], 5 X[1658] - 6 X[14070], 5 X[26] - 3 X[14070], 7 X[10244] - 3 X[14070], X[12084] - 3 X[14070], 5 X[9909] - X[14070], 5 X[7387] + 3 X[14070], 7 X[3090] - 3 X[14790], 8 X[12010] - 3 X[14790], 3 X[10201] - X[14790], 3 X[6759] - X[15083], 5 X[631] - 6 X[15330], 5 X[12085] - 14 X[15331], 5 X[11250] - 8 X[15331], 5 X[10226] - 7 X[15331], 5 X[3] - 6 X[15331], 5 X[1658] - 4 X[15331], 5 X[12107] - 3 X[15331], 5 X[26] - 2 X[15331], 15 X[9909] - 2 X[15331], 7 X[10244] - 2 X[15331], 3 X[14070] - 2 X[15331], 5 X[7387] + 2 X[15331], 2 X[546] - 3 X[15761], 7 X[3] - 15 X[16195], 14 X[12107] - 15 X[16195], 7 X[1658] - 10 X[16195], 7 X[26] - 5 X[16195], 2 X[10226] - 5 X[16195], X[12085] - 5 X[16195], 7 X[7387] + 5 X[16195], 3 X[154] - X[16266].

= lies on these lines:{{2,3},{61,11267},{62,11268},{154,16266},{156,511},{159,9925},{182,10095},{184,1493},{355,9591},{495,9658},{496,9673},{576,11536},{577,14577},{1147,13391},{1154,6759},{1173,5012},{1495,10625},{1511,16105},{1614,6243},{2916,14561},{3060,11423},{3303,9645},{3564,15581},{3581,6241},{3746,8144},{3796,13451},{5449,15579},{5690,8185},{5844,9798},{5944,13352},{5946,10984},{5965,15580},{6101,10539},{6225,9919},{6419,11265},{6420,11266},{7745,9700},{8546,11255},{8981,9683},{9306,10627},{9626,12699},{9781,13353},{9833,10117},{9920,11206},{9972,9973},{10540,11412},{10610,11424},{10833,15172},{11003,13472},{11411,12310},{11440,16835},{12161,14449},{13336,15026},{13339,15024}}

= isogonal conjugate of  X(17711)

= midpoint of X(26) and X(7387).

= reflection of X(i) in X(j) for these {i,j}: {{3, 12107}, {1658, 26}, {11250, 1658}, {12084, 15331}, {12085, 10226}, {13371, 13383}, {14790, 10224}}.

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3, 26, 12107), (3, 382, 14865), (3, 1995, 632), (3, 5072, 7550), (3, 5076, 7527), (3, 7517, 10594), (3, 7530, 546), (3, 7545, 3090), (3, 10594, 5), (3, 11403, 7526), (3, 12082, 15704), (3, 12107, 1658), (3, 13861, 3628), (4, 2937, 7502), (5, 22, 7525), (22, 7517, 5), (22, 10594, 3), (23, 12088, 3), (24, 12083, 550), (26, 12084, 14070), (186, 12087, 1657), (546, 7555, 3), (1598, 7514, 3850), (1614, 15107, 6243), (2937, 5899, 4), (3090, 7492, 3), (3090, 10201, 12010), (3146, 7556, 3), (6644, 11414, 548), (7387, 9909, 26), (7387, 10244, 12084), (7488, 14865, 3), (7506, 10323, 549), (7516, 7529, 547), (7575, 15704, 3), (9714, 11414, 6644), (9715, 11403, 3), (9781, 15080, 13353), (10201, 14790, 10224), (10224, 12010, 3090), (10226, 16195, 1658), (10244, 14070, 26), (12084, 14070, 15331), (14070, 15331, 1658).


Best regards,
Peter Moses.

 

HYACINTHOS 27517

[Antreas P. Hatzipolakis]:

 
Let ABC be a triangle and A'B'C' the pedal triangle of O.

Denote:

A1, B1, C1 = the orthogonal projections of A, B, C on OA', OB', OC', resp.

A2, B2, C2 = the orthogonal projections of A', B', C' on OA, OB, OC, resp

Na, Nb, Nc = the NPC centers of A'A1A2, B'B1B2, C'C1C2, resp.

1. A1A2, B1B2, C1C2 are concurrent.

2. ABC, NaNbNc are orthologic

3. A'B'C', NaNbNc are orthologic.

4. The centroid of NaNbNc lies on the Euler line.

 

[Peter Moses]:

 

 

Hi Antreas,

1)  

a^2 (a^4 b^2-b^6+a^4 c^2-2 a^2 b^2 c^2-c^6):: 

 

9 X[373] - 5 X[1843], 5 X[6] - 3 X[3060], 9 X[373] - 10 X[3589], 5 X[141] - 6 X[3819], 2 X[3631] - 3 X[3917], 3 X[51] - 4 X[6329], 3 X[5085] - X[6403], 3 X[597] - 2 X[9969], 5 X[3618] - 3 X[9971], 5 X[9822] - 6 X[10219], 5 X[3763] - 3 X[11188], 3 X[3819] - 5 X[11574], 5 X[182] - 4 X[12006], 5 X[9967] - X[12162], 3 X[3060] + 5 X[12220], 3 X[599] - X[12272], 9 X[3819] - 5 X[14913], 3 X[141] - 2 X[14913], 3 X[11574] - X[14913], X[6144] - 3 X[15531], 6 X[373] - 5 X[16776], 2 X[1843] - 3 X[16776], 4 X[3589] - 3 X[16776].

= on lines {{2,9973},{3,9972},{6,22},{51, 6329},{52,12007},{67,69},{141, 1368},{159,8780},{182,1658},{2 06,6593},{373,468},{376,974},{ 511,550},{524,3313},{597,9969} ,{599,12272},{888,13232},{1350 ,15073},{1503,9967},{3098,9976 },{3618,9971},{3630,8681},{ 3631,3917},{3763,11188},{5085, 6403},{5092,11649},{5097,10263 },{5157,8541},{5965,6101},{ 6144,15531},{7493,11746},{7502 ,12236},{7712,12824},{7804, 16983},{9822,10219},{10168, 16532},{10539,15580},{12283, 15069},{13201,16176}}.
= complement X(9973).
= midpoint of X(i) and X(j) for these {i,j}: {{6, 12220}, {1350, 15073}, {3313, 6467}, {12283, 15069}, {13201, 16176}}.
= reflection of X(i) in X(j) for these {i,j}: {{52, 12007}, {141, 11574}, {1843, 3589}, {10263, 5097}, {16983, 7804}}.
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1176, 11416, 6), (1843, 3589, 16776).

2)
(ABC, NaNbNc): 

 

(a^8-3 a^6 b^2+4 a^4 b^4-3 a^2 b^6+b^8-2 a^6 c^2-a^4 b^2 c^2-a^2 b^4 c^2-2 b^6 c^2+2 a^2 b^2 c^4+2 a^2 c^6+2 b^2 c^6-c^8) (a^8-2 a^6 b^2+2 a^2 b^6-b^8-3 a^6 c^2-a^4 b^2 c^2+2 a^2 b^4 c^2+2 b^6 c^2+4 a^4 c^4-a^2 b^2 c^4-3 a^2 c^6-2 b^2 c^6+c^8)::

on the Jerabek hyperbola and the lines {{265,13201},{2979,3519},{3521 ,15305},{6241,14861},{9833,112 70},{11457,15108},{13432,16266 }}.
.
= X(14864)-cross conjugate of X(4).

(NaNbNc,  ABC): 

 

2 a^10-4 a^8 b^2+a^6 b^4+a^4 b^6+a^2 b^8-b^10-4 a^8 c^2-8 a^6 b^2 c^2+9 a^4 b^4 c^2+3 b^8 c^2+a^6 c^4+9 a^4 b^2 c^4-2 a^2 b^4 c^4-2 b^6 c^4+a^4 c^6-2 b^4 c^6+a^2 c^8+3 b^2 c^8-c^10:: 

 

3 X[5892] - X[7553], X[1216] - 3 X[7667], 3 X[376] + X[11750], 9 X[376] - X[12278], 3 X[11750] + X[12278], 3 X[549] - X[13419], X[12225] + 3 X[14855], 3 X[10170] - X[16655], X[11819] - 3 X[16836].

on lines {{3,2918},{4,8717},{30,5462},{ 113,8718},{125,13564},{376,117 50},{382,15805},{539,15644},{5 41,10575},{542,10627},{548,669 6},{549,13419},{550,10264},{12 16,7667},{1503,5447},{1657,128 97},{5448,14791},{5892,7553},{ 6030,6143},{6699,7488},{7391, 13336},{7525,15578},{7691, 16003},{9530,15912},{10116,106 25},{10170,16655},{10539,16063 },{11585,12900},{11818,13347}, {11819,16836},{12103,13470},{ 12140,17506},{12225,14855},{ 12362,14915},{12605,14641},{ 13403,15704},{14810,14864}}.
= midpoint of X(i) and X(j) for these {i,j}: {{1657, 12897}, {10116, 10625}, {12103, 13470}, {12605, 14641}, {13403, 15704}}.

3)
(A'B'C',  NaNbNc): 

 

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

 

on the Jerabek hyperbola of the medial triangle and the lines {{51,1493},{1147,3060},{1511,1 0625},{4550,6241},{10539,11423 }}.

(NaNbNc, A'B'C'): as (NaNbNc, ABC).

4) X(10691)

Best regards,
Peter.

HYACINTHOS 27510

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle and A'B'C' the pedal triangle of N.

Denote:

Bc, Cb = the reflections of B', C' in NC', NB', resp.

A1 = the orthogonal projection of N on BcCb
A2 = BcCb /\ NA'

We have A1 = A2 := A*

Similarly B*, C*

 

ABC, A*B*C* are perspective.

Perspector?

LOCI:

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

Denote:

Bc, Cb = the reflections of B', C' in PC', PB', resp.

A1 = the orthogonal projection of P on BcCb
A2 = BcCb /\ PA' 

Similarly B1, B2 and C1, C2

Which is the locus of P such that
1. ABC, A1B1C1 are perspective?

2. ABC, A2B2C2 are perspective?

 

 

 

 [César Lozada]:

 

[...] 

 

 

> LOCI:

> Which is the locus of P such that
> 1. ABC, A1B1C1 are perspective?

> 2. ABC, A2B2C2 are perspective?

A1 = A2 for any P

 

Locus: An excentral-circum-sixtic through ETC’s 1, 5, 54.

Perspectors: Q(X(1)) = X(1) & Q(X(54))= X(16030)

 

 

[César Lozada]:

 

 

CORRECTION:

Sorry. I confused A’ with A1 in A2 = BcCb /\ PA'.

> Which is the locus of P such that
> 2. ABC, A2B2C2 are perspective?

Locus = {l=Line X(5)X(523)} {q=rectangular circum-hyperbola with center=X(5099) and perspector=X(2492)}

 

ETC centers on l: 5, 523, 6757, 8151, 10287, 10288, 10412, 14254, 14356, 14566, 14592, 14670, 15475

ETC centers on q: 4, 23, 316, 842, 1383, 8744, 10422, 10561, 10630, 13485, 13574, 14246

 

l is the tripolar of X(94) and is perpendicular to the Euler line at X(5)=N

 

q is the isogonal conjugate of the line X(3)X(67), perpendicular to the Fermat axis X(6)X(13) at X(542), having tripole:

 

ETC-pairs (P,Q=perspector): (4,4), (523,4)

 

T = isogonal conjugate of X(2492)

= (a^4-c^2*a^2-b^4+c^4)*(a^4-b^ 2*a^2+b^4-c^4)*(c^2-a^2)*(a^2- b^2) : : (barys)

= on MacBeath circumconic  and these lines: {67, 524}, {110, 525}, {249, 4576}, {287, 323}, {441, 14919}, {476, 2799}, {648, 850}, {671, 9141}, {805, 9076}, {2966, 3268}, {2987, 3580}, {3265, 4558}, {3455, 9888}, {5503, 10511}, {6390, 15900}

= isogonal conjugate of X(2492)

= isotomic conjugate of X(9979)

= trilinear pole of the line {3, 67}

= [ -5.4729673199657220, 2.8243990691198060, 4.2112961971163850 ]

 

One perspector:

Q(X(23)) = X(4)X(83) ∩ X(9138)X(9979)

= ((SB+SC)*(3*SA^2-3*R^2*(3*SA- 2*SW)-3*SB*SC-SW^2)-(4*SA+SB+ SC-12*R^2)*S^2)*(S^2+SA^2-4* SB*SC) : : (barys)

= on lines: {4, 83}, {9138, 9979}

= [ -2.2113879199533490, -2.8271461470550880, 6.6185600852317540 ]

 

César Lozada

HYACINTHOS 27509

[Antreas P. Hatzipolakis]:

 

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

Denote:

Ab, Ac = the orthogonal projections of A' on AB, AC, resp.
A2, A3 = the orthogonal projections of A' on BP, CP, resp.

Ma, M1 = the midpoints of AbAc, A2A2, resp.

Similarly Mb, M2 and Mc, M3

Mi, Mii, Miii = the midpoints of MaM1, MbM2. McM3.

 

 

Which is the locus of P such that:

1. ABC, MiMiiMiii are orthologic?
O, H lie on the locus

 

2. A'B'C', MiMiiMiii are orthologic?
O, H lie on the locus

 

3. The perpendiculars to MaM1, MbM2, McM3 at Mi, Mii, Miii, resp. (perpendicular bisectors) are concurrent?
O, H lie on the locus.

For P = O,  the  MaM1, MbM2. McM3. are parallels to Euler line
(their perpendicular bisectors concur at the infinity point of the direction of the perpendicular to Euler line)
For P = H, the perpendicular bisectors concur at the NPC center of A'B'C'

[César Lozada]:

 

1)      Locus = Linf ∪ {q7: circum-septic through ETC’s 3,4,5,1113,1114}

 

q7 =∑ [ y*z*(2*(b^2-c^2)*(-a^2+b^2+c^ 2)^2*b^2*c^2*x^5+(((3*b^2-c^2) *a^6-(5*b^4+3*b^2*c^2-3*c^4)* a^4+(b^6+7*b^4*c^2-3*c^6)*a^2+ (b^2-c^2)*(b^6-c^6-b^2*c^2*(2* b^2+c^2)))*c^2*y+((b^2-3*c^2)* a^6-(3*b^4-3*b^2*c^2-5*c^4)*a^ 4+(3*b^6-7*b^2*c^4-c^6)*a^2-( b^2-c^2)*(b^6-c^6+b^2*c^2*(b^ 2+2*c^2)))*b^2*z)*x^4+(2*a^8- 5*(b^2+c^2)*a^6+3*(b^4+4*b^2* c^2+c^4)*a^4+(b^2+c^2)*(b^4-9* b^2*c^2+c^4)*a^2-(b^6-c^6)*(b^ 2-c^2))*(b^2-c^2)*z*y*x^3+a^2* (a^6-(b^2+c^2)*a^4-((b^2-c^2)^ 2-4*b^2*c^2)*a^2+(b^4-c^4)*(b^ 2-c^2))*(b^2-c^2)*z^2*y^2*x-2* a^4*c^2*(a^2-b^2+c^2)*(a^2-c^ 2)*z*y^4+a^4*(a^6-(2*b^2-c^2)* a^4+(b^4+5*b^2*c^2-4*c^4)*a^2- 2*(b^2-c^2)*(2*b^2+c^2)*c^2)* z^2*y^3-a^4*(a^6+(b^2-2*c^2)* a^4-(4*b^4-5*b^2*c^2-c^4)*a^2+ 2*(b^2-c^2)*(b^2+2*c^2)*b^2)* z^3*y^2+2*a^4*b^2*(a^2+b^2-c^ 2)*(a^2-b^2)*z^4*y)] = 0

 

Some orthologic centers Q1=A->Mii, Q2=Mii->A

 

Q1(X(3)) = X(5)X(2883) ∩ X(53)X(1906)

= (3*S^2-16*R^2*SB+SB^2-2*SC*SA) *(3*S^2-16*R^2*SC+SC^2-2*SA* SB) : : (barys)

on lines: {5, 2883}, {53, 1906}, {3832, 17500}

= [ -5.6953736993560960, -7.1890583417472480, 11.2463388874352900 ]

 

Z1 = Q2(X(3)) = X(2)X(11381) ∩ X(3)X(6)

= (SB+SC)*(SA^2-16*R^2*SA-SB*SC) : : (barycentrics)

= 9*X(2)-X(11381), 21*X(2)-5*X(11439), 15*X(2)+X(12279), 7*X(3)+X(52), 3*X(3)+X(389), 13*X(3)+3*X(568), 15*X(3)+X(6243), 5*X(3)+3*X(9730), 9*X(3)-X(10625), 19*X(3)-3*X(13340), 3*X(3)-X(13348), 2*X(3)+X(15012), 5*X(3)-X(15644), 5*X(3)+X(16625), X(3)+3*X(16836), 3*X(52)-7*X(389), 15*X(52)-7*X(6243), X(52)-7*X(9729), 9*X(52)+7*X(10625), 3*X(52)+7*X(13348), 2*X(52)-7*X(15012), 5*X(52)+7*X(15644), 5*X(52)-7*X(16625), 13*X(389)-9*X(568), 5*X(389)-X(6243), X(389)-3*X(9729), 5*X(389)-9*X(9730), 3*X(389)+X(10625), 2*X(389)-3*X(15012), 5*X(389)+3*X(15644), 5*X(389)-3*X(16625), X(389)-9*X(16836), 3*X(568)-13*X(9729), 5*X(568)-13*X(9730), 6*X(568)-13*X(15012), 15*X(568)-13*X(16625), X(568)-13*X(16836), 7*X(11381)-15*X(11439), 5*X(11381)+3*X(12279)

= on lines: {2, 11381}, {3, 6}, {4, 6688}, {5, 10219}, {20, 5943}, {30, 11695}, {51, 3522}, {64, 16419}, {140, 6000}, {185, 3523}, {373, 3146}, {376, 9781}, {382, 13570}, {517, 6744}, {548, 5462}, {549, 5876}, {550, 5892}, {631, 5907}, {1204, 7485}, {1216, 13382}, {1656, 13474}, {2808, 5044}, {3090, 12045}, {3357, 7393}, {3524, 5562}, {3525, 15030}, {3526, 10575}, {3528, 15045}, {3530, 13754}, {3533, 12290}, {3628, 14915}, {3818, 6803}, {3917, 10574}, {4550, 13154}, {5054, 12162}, {5059, 11451}, {5070, 16194}, {5073, 14845}, {5446, 8703}, {5447, 12100}, {5650, 12111}, {5663, 12108}, {5889, 15692}, {5890, 10299}, {5891, 15720}, {6101, 17504}, {6102, 15606}, {6689, 15122}, {6699, 7568}, {6865, 15488}, {7496, 11440}, {8717, 13861}, {9707, 10984}, {10128, 16656}, {10170, 13491}, {10303, 15072}, {10304, 15043}, {10691, 13568}, {11812, 14128}, {12041, 13402}, {12103, 13363}, {13446, 16386}, {14831, 15698}, {15058, 15702}, {15759, 16881}

= midpoint of X(i) and X(j) for these {i,j}: {3, 9729}, {548, 5462}, {550, 10110}, {1216, 13382}, {5447, 13630}, {6102, 15606}, {13446, 16386}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (3, 389, 13348), (3, 9730, 15644), (3, 9786, 3098), (3, 13336, 11430), (3, 13347, 5092), (3, 13353, 10564), (3, 16836, 9729), (185, 3523, 3819), (389, 6243, 16625), (389, 15644, 6243), (550, 5892, 10110), (6243, 9730, 389), (9729, 13348, 389), (9729, 16625, 9730), (9730, 15644, 16625)

= [ 5.8415841905283880, 5.1790486577048490, -2.6409465229013110 ]

 

Q1(X(4)) = X(3);  Q2(X(4)) = X(10110);

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

2)      Locus= {sidelines} ∪ {Linf} ∪ {circumcircle} ∪ {McCay cubic K003}

Orthologic centers Q1=A’->Mii, Q2=Mii->A’:

 

Q1(X(1)) =

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

= on line {1510, 4017}

= [ 15.6339469766864400, 16.7524683344941300, -15.1728660465976000 ]

 

Q2(X(1)) = midpoint of X(3874) and X(5795)

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

= 5*X(65)+3*X(3058), 3*X(65)+X(10624), 3*X(553)+X(10572), 3*X(942)-X(4298), 3*X(950)+X(1770), X(950)+3*X(5902), X(1770)-9*X(5902), 9*X(3058)-5*X(10624), X(4298)+3*X(6738), 5*X(5439)-X(6737)

= on lines: {1, 631}, {5, 12563}, {10, 15934}, {57, 4305}, {65, 3058}, {226, 10826}, {355, 5542}, {515, 942}, {516, 12433}, {517, 6744}, {519, 3812}, {553, 10572}, {938, 946}, {944, 10980}, {950, 1770}, {952, 12577}, {999, 6796}, {1159, 4301}, {1210, 11375}, {1376, 3244}, {1837, 9656}, {3241, 4917}, {3333, 4308}, {3339, 3488}, {3452, 12559}, {3487, 10175}, {3579, 15935}, {3585, 3982}, {3626, 3826}, {3634, 5719}, {3636, 6690}, {3671, 5722}, {3754, 5853}, {3874, 5795}, {4031, 4299}, {4297, 5708}, {4304, 5221}, {4311, 4860}, {5122, 15174}, {5326, 13411}, {5439, 6737}, {5587, 11036}, {5728, 5884}, {5881, 11037}, {7982, 10580}, {9578, 11518}, {9797, 11525}, {9843, 12635}, {10172, 11374}, {11019, 11373}, {11020, 15016}, {12564, 16004}

= midpoint of X(3874) and X(5795)

= {X(938), X(11529)}-Harmonic conjugate of X(946)

= [ 1.3199013978167580, 1.4453769740964900, 2.0308336239252300 ]

 

Q1(X(3)) = X(140)X(6000) ∩ X(570)X(1990)

= (4*S^2-(SB+SC)*(16*R^2+SA-SW)) *(S^2-(16*R^2+SA-2*SW)*SA) : : (barys)

= on lines: {140, 6000}, {570, 1990}

= [ 6.7917400384111270, 6.2238979633623420, -3.8028372027639160 ]

 

Q2(X(3)) = Z1

Q1(X(4)) = X(52); Q2(X(4)) = X(143)

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3)       

Sorry. Too long.

 

César Lozada