ADGEOM 4653 * ADGEOM 4654 * ADGEOM 4660

#4653

 

Dear geometers,

 

Let ABC be a triangle with centroid G and orthocenter H.

 

P=X(381) is midpoint of GH which is the center of orthocentroidal circle.

 

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

 

Then APa,BPb,CPc are concurrent at.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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#4654

 

[Tran Quang Hung]:

Let ABC be a triangle with centroid G and orthocenter H.

P=X(381) is midpoint of GH which is the center of orthocentroidal
circle.

Let Pa,Pb,Pc be the X(381) of the triangles PBC, PCA, PAB respectively.

Then APa,BPb,CPc are concurrent.

Which is this point?

*** APa,BPb,CPc are concurrent at

W = 1/((4 a^4-5 a^2 b^2+b^4-5 a^2 c^2-2 b^2 c^2+c^4)(2 a^4-4 a^2 b^2+2
b^4-4 a^2 c^2-b^2 c^2+2 c^4)) : .... : ....

on the line X(381)X(14483), with (6 - 9 - 13) - search numbers
(0.125506022546977, 4.78594924329040, 0.269389149223193)

Angel Montesdeoca

 

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

#4660

 

 

Dear geometers,

I have seen general problem for this,

G and H be the centroid and orthocenter of ABC respectively.

Let ABC be a triangle and P is a point on Euler line of ABC which has Shinagawa coefficients P=kG+H.

Let Ga,Ha be the centroid and orthocenter of PBC respectively.

Pa is a point on Euler line of PBC which has Shinagawa coefficients Pa=kGa+Ha.
 
Define similarly the points Pb and Pc.

Then APa,BPb and CPc are concurrent. Which is this concurrent point in term of k?

Best regards,
Tran Quang Hung.

 

ADGEOM 4649 * ADGEOM 4650

#4649

Dear Geometers,

1. Let DEF be cevian triangle of Nagels point Na. Ha,Hb,Hc be orthocenters of NaEF, NaFD, NaDE respectively. Perpendiculars from Ha to BC, Hb to CA and Hc to AB concur. (ıts a construction for X(3057).

2. Let DEF be cevian triangle of Nagels point Na. Ga,Gb,Gc be centroids of NaEF, NaFD, NaDE respectively. Perpendiculars from Ga to BC, Gb to CA and Gc to AB concur. (ıts a not listed in ETC).

Best regards

Abdilkadir Altıntaş

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

#4650

[Abdilkadir Altıntas]:

2. Let DEF be cevian triangle of Nagels point Na. Ga,Gb,Gc be centroids
of NaEF, NaFD, NaDE respectively. Perpendiculars from Ga to BC, Gb to CA
and Gc to AB concur. (ıts a not listed in ETC).

**** The perpendiculars from Ga to BC, Gb to CA and Gc to AB concur at

W = (r-2 R) O - (r+4 R)Na

with first barycentric coordinate:

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

W is the reflection of X(i) in X(j), for these {i, j}: {354,10175},
{3576,3740}, {3753,5790}, {3892,10171}, {5603,10157}, {10164,3956}.

W lies on lines X(i)X(j) for these {i, j}: {3,3697}, {4,8}, {5,3555},
{10,1071}, {12,354}, {40,4662}, {84,165}, {200,1012}, {210,515},
{392,952}, {405,5534}, {495,5728}, {518,5587}, {519,15064}, {912,3753},
{936,958}, {942,5261}, {944,5044}, {956,5720}, {960,5881}, {971,5657},
{997,18236}, {1385,5260}, {1532,4847}, {1698,12675}, {1737,17625},
{2836,13214}, {3036,12665}, {3086,17624}, {3090,5045}, {3295,9844},
{3617,12528}, {3621,13600}, {3678,14110}, {3679,6001}, {3698,5884},
{3870,6913}, {3889,5056}, {3892,10171}, {3916,11499}, {3921,10167},
{3935,6912}, {3956,10164}, {3983,6684}, {4015,4297}, {4533,18525},
{4882,12705}, {5049,11374}, {5173,10590}, {5251,5531}, {5290,5902},
{5302,10902}, {5439,9956}, {5603,10157}, {5687,7330}, {5693,5836},
{5726,18412}, {5770,17612}, {5780,10246}, {5791,10786}, {5886,11240},
{5904,7686}, {5919,10950}, {6927,12125}, {6965,18527}, {7580,18528},
{7989,13374}, {8164,11018}, {8168,12703}, {8728,10202}, {9004,12587},
{9708,18446}, {9780,9940}, {10039,12711}, {10176,18250}, {10573,12709},
{10588,16193}, {11362,12688}, {12599,12692}, {12629,12635}

(6 - 9 - 13) - search numbers of W: (1.49964716737654,
-0.605533359430132, 3.36773503810838)

Angel Montesdeoca

 

ADGEOM 4589 * ADGEOM 4613

#4589

 

Dear geometers,

 

Let ABC be a triangle and P,Q are isogonal conjugate points.

 

Ab, Ac are isogonal conjugate of B, C wrt triangle APQ.

 

Define similarly, the points Bc, Ba, Ca, Cb.

 

Then six points Ab, Ac, Bc, Ba, Ca, Cb lie on a conic.

 

When is it a circle?

 

Best regards,

Tran Quang Hung.

 

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#4613

 

Dear Tran Quang Hung,

 

I think the locus of points P such that the conic is a circle is complicated.  I found several such points on the Euler line, none are in ETC.

 

I did find several interesting observations from this configuration:

 

When P,Q are the Brocard points, PU(1):

The center is the trilinear pole of the line through X(351) parallel to the trilinear polar of X(351).

On lines {3,9217}, {6,694}, {32,249}, {39,512}, {543,598}, {574,805}, {733,12074}, {2086,3229} et al.

Barycentrics: a^2 (2 a^2 - b^2 - c^2) / (a^4 - b^2 c^2) : :

(6,9,13) values: (-0.765454011614075, 7.206375474085383, -0.995078225560544).

 

The perspector of the conic lies on lines {262,6036}, {523,3629}, {576,2065} et al

Barycentrics: a^2 (a^2 b^2 + a^2 c^2 - b^4 - c^4) / (a^4 + 2 b^4 + 2 c^4 - b^2 c^2 - 2 a^2 b^2 - 2 a^2 c^2) : :

(6,9,13) values:(-0.177721420347812, -0.199750198006888, 3.860978351457369).

 

The centroid of AbAcBcBaCaCb is X(10567).

 

The inverse of X(39) in the conic is X(882).

 

 

When P,Q are X(2), X(6):

The center of the conic is X(6).  The conic passes through X(2) and its antipode, X(1992) which are vertices of the conic.

 

The perspector of the conic has barycentrics: (5 a^2 - b^2 - c^2) / (14 a^4 + 2 b^4 + 2 c^4 - 20 a^2 b^2 - 20 a^2 c^2 + 13 b^2 c^2) : :

(6,9,13) values:(-0.654815572698841, -1.061665363990563, 4.677886152069619).

 

 

When P,Q are X(3), X(4), the conic is the Yff hyperbola, with center X(381) and perspector X(13481).

 

The centroid of AbAcBcBaCaCb is X(13448).

 

 

When P,Q are X(13), X(15):

The conic is degenerate, 2 lines intersecting at X(13).

 

The centroid of AbAcBcBaCaCb lies on lines {13,15}, {1989,9140}.

 

 

When P,Q are X(14), X(16):

The conic is degenerate, 2 lines intersecting at X(14).

 

The centroid of AbAcBcBaCaCb lies on lines {14,16}, {1989,9140}.

 

 

Best regards,

Randy Hutson

ADGEOM 4604 * ADGEOM 4608

#4604

 

Dear geometers,

 

Let ABC be a triangle with circumcenter O.

 

P is on its Euler line.

 

Oa, Ob, Oc are isogonal conjugate of O wrt triangles PBC, PCA, PAB.

 

Then centroid of OaObOc lies on Euler line of ABC.

 

Which is this point in term of P?

 

Best regards,

Tran Quang Hung.

 

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

#4608

 

Dear Tran Quang Hung.

If P is on its Euler line such that OP:PH=t , then centroid Q of
OaObOc ( lies on Euler line of ABC) is:

Q = (a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) t (a^16 t^2-(b^2-c^2)^6
(b^2+c^2)^2 t^2-a^14 (b^2+c^2) t (1+3 t)+a^2 (b^2-c^2)^4 (b^2+c^2) t (-2
b^2 c^2 (-2+t)+b^4 (1+3 t)+c^4 (1+3 t))-a^6 (b^2-c^2)^2 (b^2+c^2) (b^4
(-5+t) t+c^4 (-5+t) t+b^2 c^2 (-5+7 t-4 t^2))-a^8 (b^4 c^4 (-7+14 t-12
t^2)+b^6 c^2 (3-9 t+4 t^2)+b^2 c^6 (3-9 t+4 t^2))-a^4 (b^2-c^2)^2 (2 b^8
t (2+t)+2 c^8 t (2+t)+b^6 c^2 (2+5 t-3 t^2)+b^2 c^6 (2+5 t-3 t^2)+b^4
c^4 (5-6 t+6 t^2))+a^10 (b^2+c^2) (b^4 (-5+t) t+c^4 (-5+t) t-b^2 c^2
(1-5 t+8 t^2))+a^12 (2 b^4 t (2+t)+2 c^4 t (2+t)+b^2 c^2 (1+9
t^2)))-(-a^2+b^2+c^2) (a^8 t-2 a^6 (b^2+c^2) t+a^4 b^2 c^2 (-2+3
t)-(b^2-c^2)^2 (b^2 c^2 (-1+t)+b^4 t+c^4 t)+a^2 (b^2+c^2) (b^2 c^2 (1-4
t)+2 b^4 t+2 c^4 t)) ((b^2-c^2) (-a^2 c^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2
b^4+b^2 c^2 (1-2 t))-a^8 t-(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 c^2 t+b^2
(1+t))+a^2 (b^2-c^2) (-b^2 c^2 (-2+t)+2 c^4 t+b^4 (1+t)))+(-b^2+c^2)
(-a^2 b^2-a^4 t+(b^2-c^2)^2 t) (a^4 (-2 c^4+b^2 c^2 (1-2 t))-a^8
t+(b^2-c^2)^3 (b^2+c^2) t+a^6 (2 b^2 t+c^2 (1+t))-a^2 (b^2-c^2) (-b^2
c^2 (-2+t)+2 b^4 t+c^4 (1+t)))) : ... : ...

Pairs {P=X(i),Q=X(j)} (safe error or omission) , for {i,j}: {3,1650},
{23,4}, {186,2}, {1113,10720}, {1114,10719}, {1583,11321}, {2060,14020},
{7486,16950}, {7512,14006}, {8359,17686}, {10128,8364}, {11343,16896},
{14142,17531}, {16355,14001}, {16383,11335}, {16399,14461}

*** If P=X(2)

Q = a^16-2 a^14 (b^2+c^2)-3 a^12 (b^4-4 b^2 c^2+c^4)-(b^2-c^2)^4
(b^2+c^2)^2 (2 b^4-3 b^2 c^2+2 c^4)+2 a^10 (4 b^6-5 b^4 c^2-5 b^2 c^4+4
c^6)-2 a^6 (b^2-c^2)^2 (5 b^6-2 b^4 c^2-2 b^2 c^4+5 c^6)+a^8 (b^8-15 b^6
c^2+29 b^4 c^4-15 b^2 c^6+c^8)+a^4 (b^2-c^2)^2 (3 b^8+2 b^6 c^2-15 b^4
c^4+2 b^2 c^6+3 c^8)+4 a^2 (b^2-c^2)^2 (b^10-b^8 c^2+2 b^6 c^4+2 b^4
c^6-b^2 c^8+c^10) : ... : ...

with (6 - 9 - 13) - search numbers (2.25482286270048,
1.37935466015752, 1.64503916516741).


*** If P=X(4) then Q is the reflection of X(2) in X(13448) :

Q = a^12-2 a^10 (b^2+c^2)+2 a^8 (b^4+b^2 c^2+c^4)-(b^2-c^2)^4 (2
b^4+b^2 c^2+2 c^4)-2 a^6 (b^6+c^6)+2 a^2 (b^2-c^2)^2 (2 b^6-b^4 c^2-b^2
c^4+2 c^6)-a^4 (b^8-3 b^6 c^2+3 b^4 c^4-3 b^2 c^6+c^8) : ... : ...,

on lines X(i)X(j) for these {i, j}: {2,3}, {94,9140}, {2088,14846}.
(6 - 9 - 13) - search numbers ( -54.1570303678729, -54.8836826772129,
66.6326126589962).


*** If P=X(5)

Q = (b^2-c^2)^2 (-a^2+b^2+c^2) (a^8+a^4 b^2 c^2-2 a^6
(b^2+c^2)-(b^2-c^2)^2 (b^4+c^4)+a^2 (b^2+c^2) (2 b^4-3 b^2 c^2+2 c^4))
(a^10-4 a^8 (b^2+c^2)+2 (b^2-c^2)^4 (b^2+c^2)+a^6 (4 b^4+b^2 c^2+4
c^4)-a^2 (b^2-c^2)^2 (5 b^4+4 b^2 c^2+5 c^4)+a^4 (2 b^6+3 b^4 c^2+3 b^2
c^4+2 c^6))+(a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) (a^16-4 a^14
(b^2+c^2)-(b^2-c^2)^6 (b^2+c^2)^2-4 a^10 (b^2+c^2) (b^4+b^2 c^2+c^4)+2
a^6 (b^2-c^2)^2 (b^2+c^2) (2 b^4+b^2 c^2+2 c^4)+2 a^2 (b^2-c^2)^4
(b^2+c^2) (2 b^4+b^2 c^2+2 c^4)+2 a^12 (3 b^4+5 b^2 c^2+3 c^4)+a^8 (2
b^6 c^2+5 b^4 c^4+2 b^2 c^6)-a^4 (b^2-c^2)^2 (6 b^8+4 b^6 c^2+5 b^4
c^4+4 b^2 c^6+6 c^8)) : ... : ....

with (6 - 9 - 13) - search numbers (-9.94914860821929,
-10.7924224258757, 15.7042562882303).

*** If P=X(20)

Q = (b^2-c^2)^2 (-a^2+b^2+c^2) (-a^^2 c^2+c^4)-2 a^4 (5 b^6-9 b^4
c^2-9 b^2 c^4+5 c^6))-(a^2-b^2) (a^2-c^2) (a^4-(b^2-c^2)^2) (a^16-34 a^8
b^2 c^2 (b^2-c^2)^2-a^14 (b^2+c^2)+11 a^10 (b^2-c^2)^2
(b^2+c^2)-(b^2-c^2)^6 (b^2+c^2)^2+a^12 (-6 b^4+13 b^2 c^2-6 c^4)+a^2
(b^2-c^2)^4 (b^6-9 b^4 c^2-9 b^2 c^4+c^6)-a^6 (b^2-c^2)^2 (11 b^6-8-7 a^4 b^2 c^2+2 a^6
(b^2+c^2)+(b^2-c^2)^2 (b^4+3 b^2 c^2+c^4)-2 a^2 (b^6-2 b^4 c^2-2 b^2
c^4+c^6)) (-5 a^10+8 a^8 (b^2+c^2)+2 (b^2-c^2)^4 (b^2+c^2)+4 a^6 (b^4-8
b^2 c^2+c^4)+a^2 (b^2-c^2)^2 (b^4+14 b27 b^4
c^2-27 b^2 c^4+11 c^6)+a^4 (b^2-c^2)^2 (6 b^8+5 b^6 c^2-38 b^4 c^4+5 b^2
c^6+6 c^8)) : ... : ...

with (6 - 9 - 13) - search numbers (5.41707665942407,
4.53326635151949, -1.99793991195560).

*** If P=X(30)

Q = a^12-a^10 (b^2+c^2)-5 a^6 (b^2-c^2)^2 (b^2+c^2)+a^8 (b^4-b^2
c^2+c^4)-(b^2-c^2)^4 (2 b^4+3 b^2 c^2+2 c^4)+2 a^4 (b^2-c^2)^2 (2 b^4+5
b^2 c^2+2 c^4)+2 a^2 (b^2-c^2)^2 (b^6-3 b^4 c^2-3 b^2 c^4+c^6) : ... :
....

on lines X(i)X(j) for these {i, j}: {2,3}, {125,9530}, {523,1853},
{1899,2452}, {2972,10714}, {3258,11550}}.
(6 - 9 - 13) - search numbers (7.18058761893295, 6.29212512458911,
-4.02953950539252)

Best regards,
Angel Montesdeoca

 

 

ADGEOM 4584 * ADGEOM 4585 * ADGEOM 4587

#4584

 

Dear geometers,

 

Let ABC be a triangle.

 

O and H are circumcenter and orthocenter.

 

I is incenter of ABC.

 

Ja, Jb, Jc are incenters of triangles AHO, BHO, CHO.

 

Then isogonal conjugate of I wrt triangle JaJbJc lies on line OH.

 

Which is this point?

 

Best regards,

Tran Quang Hung.

 

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

#4585

 

I find huge coordinates ( T stands for S·OH where S is twice the area of ABC.

 

Best regards, 

Francisco Javier.

 

{-a^13 + a^12 b + 3 a^11 b^2 - 3 a^10 b^3 - 2 a^9 b^4 + 2 a^8 b^5 - 

  2 a^7 b^6 + 2 a^6 b^7 + 3 a^5 b^8 - 3 a^4 b^9 - a^3 b^10 + 

  a^2 b^11 + a^12 c - 2 a^11 b c - 3 a^10 b^2 c + 2 a^9 b^3 c + 

  3 a^8 b^4 c + 4 a^7 b^5 c - a^6 b^6 c - 4 a^5 b^7 c - 2 a^3 b^9 c + 

  2 a b^11 c + 3 a^11 c^2 - 3 a^10 b c^2 - 5 a^9 b^2 c^2 + 

  4 a^8 b^3 c^2 + 4 a^7 b^4 c^2 - 2 a^6 b^5 c^2 - 4 a^5 b^6 c^2 + 

  4 a^4 b^7 c^2 + a^3 b^8 c^2 - 3 a^2 b^9 c^2 + a b^10 c^2 - 

  3 a^10 c^3 + 2 a^9 b c^3 + 4 a^8 b^2 c^3 - 10 a^7 b^3 c^3 - 

  3 a^6 b^4 c^3 + 4 a^5 b^5 c^3 + 2 a^4 b^6 c^3 + 10 a^3 b^7 c^3 - 

  6 a b^9 c^3 - 2 a^9 c^4 + 3 a^8 b c^4 + 4 a^7 b^2 c^4 - 

  3 a^6 b^3 c^4 + 2 a^5 b^4 c^4 - 3 a^4 b^5 c^4 + 3 a^2 b^7 c^4 - 

  4 a b^8 c^4 + 2 a^8 c^5 + 4 a^7 b c^5 - 2 a^6 b^2 c^5 + 

  4 a^5 b^3 c^5 - 3 a^4 b^4 c^5 - 16 a^3 b^5 c^5 - a^2 b^6 c^5 + 

  4 a b^7 c^5 - 2 a^7 c^6 - a^6 b c^6 - 4 a^5 b^2 c^6 + 

  2 a^4 b^3 c^6 - a^2 b^5 c^6 + 6 a b^6 c^6 + 2 a^6 c^7 - 

  4 a^5 b c^7 + 4 a^4 b^2 c^7 + 10 a^3 b^3 c^7 + 3 a^2 b^4 c^7 + 

  4 a b^5 c^7 + 3 a^5 c^8 + a^3 b^2 c^8 - 4 a b^4 c^8 - 3 a^4 c^9 - 

  2 a^3 b c^9 - 3 a^2 b^2 c^9 - 6 a b^3 c^9 - a^3 c^10 + a b^2 c^10 + 

  a^2 c^11 + 2 a b c^11 + 2 a^10 T - 4 a^9 b T - 4 a^8 b^2 T + 

  10 a^7 b^3 T - 6 a^5 b^5 T + 4 a^4 b^6 T - 2 a^3 b^7 T - 

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

  8 a^5 b^4 c T - 8 a^4 b^5 c T + 4 a^2 b^7 c T - 4 a b^8 c T + 

  2 b^9 c T - 4 a^8 c^2 T + 4 a^6 b^2 c^2 T - 10 a^5 b^3 c^2 T - 

  4 a^4 b^4 c^2 T + 12 a^3 b^5 c^2 T + 4 a^2 b^6 c^2 T - 

  2 a b^7 c^2 T + 10 a^7 c^3 T - 4 a^6 b c^3 T - 10 a^5 b^2 c^3 T + 

  16 a^4 b^3 c^3 T - 10 a^3 b^4 c^3 T - 4 a^2 b^5 c^3 T + 

  10 a b^6 c^3 T - 8 b^7 c^3 T + 8 a^5 b c^4 T - 4 a^4 b^2 c^4 T - 

  10 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T - 6 a b^5 c^4 T - 6 a^5 c^5 T - 

  8 a^4 b c^5 T + 12 a^3 b^2 c^5 T - 4 a^2 b^3 c^5 T - 

  6 a b^4 c^5 T + 12 b^5 c^5 T + 4 a^4 c^6 T + 4 a^2 b^2 c^6 T + 

  10 a b^3 c^6 T - 2 a^3 c^7 T + 4 a^2 b c^7 T - 2 a b^2 c^7 T - 

  8 b^3 c^7 T - 2 a^2 c^8 T - 4 a b c^8 T + 2 a c^9 T + 2 b c^9 T, 

 a^11 b^2 - a^10 b^3 - 3 a^9 b^4 + 3 a^8 b^5 + 2 a^7 b^6 - 

  2 a^6 b^7 + 2 a^5 b^8 - 2 a^4 b^9 - 3 a^3 b^10 + 3 a^2 b^11 + 

  a b^12 - b^13 + 2 a^11 b c - 2 a^9 b^3 c - 4 a^7 b^5 c - 

  a^6 b^6 c + 4 a^5 b^7 c + 3 a^4 b^8 c + 2 a^3 b^9 c - 

  3 a^2 b^10 c - 2 a b^11 c + b^12 c + a^10 b c^2 - 3 a^9 b^2 c^2 + 

  a^8 b^3 c^2 + 4 a^7 b^4 c^2 - 4 a^6 b^5 c^2 - 2 a^5 b^6 c^2 + 

  4 a^4 b^7 c^2 + 4 a^3 b^8 c^2 - 5 a^2 b^9 c^2 - 3 a b^10 c^2 + 

  3 b^11 c^2 - 6 a^9 b c^3 + 10 a^7 b^3 c^3 + 2 a^6 b^4 c^3 + 

  4 a^5 b^5 c^3 - 3 a^4 b^6 c^3 - 10 a^3 b^7 c^3 + 4 a^2 b^8 c^3 + 

  2 a b^9 c^3 - 3 b^10 c^3 - 4 a^8 b c^4 + 3 a^7 b^2 c^4 - 

  3 a^5 b^4 c^4 + 2 a^4 b^5 c^4 - 3 a^3 b^6 c^4 + 4 a^2 b^7 c^4 + 

  3 a b^8 c^4 - 2 b^9 c^4 + 4 a^7 b c^5 - a^6 b^2 c^5 - 

  16 a^5 b^3 c^5 - 3 a^4 b^4 c^5 + 4 a^3 b^5 c^5 - 2 a^2 b^6 c^5 + 

  4 a b^7 c^5 + 2 b^8 c^5 + 6 a^6 b c^6 - a^5 b^2 c^6 + 

  2 a^3 b^4 c^6 - 4 a^2 b^5 c^6 - a b^6 c^6 - 2 b^7 c^6 + 

  4 a^5 b c^7 + 3 a^4 b^2 c^7 + 10 a^3 b^3 c^7 + 4 a^2 b^4 c^7 - 

  4 a b^5 c^7 + 2 b^6 c^7 - 4 a^4 b c^8 + a^2 b^3 c^8 + 3 b^5 c^8 - 

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

   b^3 c^10 + 2 a b c^11 + b^2 c^11 + 2 a^9 b T - 2 a^8 b^2 T - 

  2 a^7 b^3 T + 4 a^6 b^4 T - 6 a^5 b^5 T + 10 a^3 b^7 T - 

  4 a^2 b^8 T - 4 a b^9 T + 2 b^10 T + 2 a^9 c T - 4 a^8 b c T + 

  4 a^7 b^2 c T - 8 a^5 b^4 c T + 8 a^4 b^5 c T - 4 a^3 b^6 c T + 

  6 a b^8 c T - 4 b^9 c T - 2 a^7 b c^2 T + 4 a^6 b^2 c^2 T + 

  12 a^5 b^3 c^2 T - 4 a^4 b^4 c^2 T - 10 a^3 b^5 c^2 T + 

  4 a^2 b^6 c^2 T - 4 b^8 c^2 T - 8 a^7 c^3 T + 10 a^6 b c^3 T - 

  4 a^5 b^2 c^3 T - 10 a^4 b^3 c^3 T + 16 a^3 b^4 c^3 T - 

  10 a^2 b^5 c^3 T - 4 a b^6 c^3 T + 10 b^7 c^3 T - 6 a^5 b c^4 T - 

  4 a^4 b^2 c^4 T - 10 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T + 

  8 a b^5 c^4 T + 12 a^5 c^5 T - 6 a^4 b c^5 T - 4 a^3 b^2 c^5 T + 

  12 a^2 b^3 c^5 T - 8 a b^4 c^5 T - 6 b^5 c^5 T + 10 a^3 b c^6 T + 

  4 a^2 b^2 c^6 T + 4 b^4 c^6 T - 8 a^3 c^7 T - 2 a^2 b c^7 T + 

  4 a b^2 c^7 T - 2 b^3 c^7 T - 4 a b c^8 T - 2 b^2 c^8 T + 

  2 a c^9 T + 2 b c^9 T, 

 2 a^11 b c + a^10 b^2 c - 6 a^9 b^3 c - 4 a^8 b^4 c + 4 a^7 b^5 c + 

  6 a^6 b^6 c + 4 a^5 b^7 c - 4 a^4 b^8 c - 6 a^3 b^9 c + 

  a^2 b^10 c + 2 a b^11 c + a^11 c^2 - 3 a^9 b^2 c^2 + 3 a^7 b^4 c^2 -

   a^6 b^5 c^2 - a^5 b^6 c^2 + 3 a^4 b^7 c^2 - 3 a^2 b^9 c^2 + 

  b^11 c^2 - a^10 c^3 - 2 a^9 b c^3 + a^8 b^2 c^3 + 10 a^7 b^3 c^3 - 

  16 a^5 b^5 c^3 + 10 a^3 b^7 c^3 + a^2 b^8 c^3 - 2 a b^9 c^3 - 

  b^10 c^3 - 3 a^9 c^4 + 4 a^7 b^2 c^4 + 2 a^6 b^3 c^4 - 

  3 a^5 b^4 c^4 - 3 a^4 b^5 c^4 + 2 a^3 b^6 c^4 + 4 a^2 b^7 c^4 - 

  3 b^9 c^4 + 3 a^8 c^5 - 4 a^7 b c^5 - 4 a^6 b^2 c^5 + 

  4 a^5 b^3 c^5 + 2 a^4 b^4 c^5 + 4 a^3 b^5 c^5 - 4 a^2 b^6 c^5 - 

  4 a b^7 c^5 + 3 b^8 c^5 + 2 a^7 c^6 - a^6 b c^6 - 2 a^5 b^2 c^6 - 

  3 a^4 b^3 c^6 - 3 a^3 b^4 c^6 - 2 a^2 b^5 c^6 - a b^6 c^6 + 

  2 b^7 c^6 - 2 a^6 c^7 + 4 a^5 b c^7 + 4 a^4 b^2 c^7 - 

  10 a^3 b^3 c^7 + 4 a^2 b^4 c^7 + 4 a b^5 c^7 - 2 b^6 c^7 + 

  2 a^5 c^8 + 3 a^4 b c^8 + 4 a^3 b^2 c^8 + 4 a^2 b^3 c^8 + 

  3 a b^4 c^8 + 2 b^5 c^8 - 2 a^4 c^9 + 2 a^3 b c^9 - 5 a^2 b^2 c^9 + 

  2 a b^3 c^9 - 2 b^4 c^9 - 3 a^3 c^10 - 3 a^2 b c^10 - 

  3 a b^2 c^10 - 3 b^3 c^10 + 3 a^2 c^11 - 2 a b c^11 + 3 b^2 c^11 + 

  a c^12 + b c^12 - c^13 + 2 a^9 b T - 8 a^7 b^3 T + 12 a^5 b^5 T - 

  8 a^3 b^7 T + 2 a b^9 T + 2 a^9 c T - 4 a^8 b c T - 2 a^7 b^2 c T + 

  10 a^6 b^3 c T - 6 a^5 b^4 c T - 6 a^4 b^5 c T + 10 a^3 b^6 c T - 

  2 a^2 b^7 c T - 4 a b^8 c T + 2 b^9 c T - 2 a^8 c^2 T + 

  4 a^7 b c^2 T + 4 a^6 b^2 c^2 T - 4 a^5 b^3 c^2 T - 

  4 a^4 b^4 c^2 T - 4 a^3 b^5 c^2 T + 4 a^2 b^6 c^2 T + 

  4 a b^7 c^2 T - 2 b^8 c^2 T - 2 a^7 c^3 T + 12 a^5 b^2 c^3 T - 

  10 a^4 b^3 c^3 T - 10 a^3 b^4 c^3 T + 12 a^2 b^5 c^3 T - 

  2 b^7 c^3 T + 4 a^6 c^4 T - 8 a^5 b c^4 T - 4 a^4 b^2 c^4 T + 

  16 a^3 b^3 c^4 T - 4 a^2 b^4 c^4 T - 8 a b^5 c^4 T + 4 b^6 c^4 T - 

  6 a^5 c^5 T + 8 a^4 b c^5 T - 10 a^3 b^2 c^5 T - 10 a^2 b^3 c^5 T + 

  8 a b^4 c^5 T - 6 b^5 c^5 T - 4 a^3 b c^6 T + 4 a^2 b^2 c^6 T - 

  4 a b^3 c^6 T + 10 a^3 c^7 T + 10 b^3 c^7 T - 4 a^2 c^8 T + 

  6 a b c^8 T - 4 b^2 c^8 T - 4 a c^9 T - 4 b c^9 T + 2 c^10 T}

 

Frabcisco Javier Garcia Capitan

 

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

#4587

Dear Mr Francisco and friends,

 

I see similar problem with excenters.

 

H,O and I are orthocenter, circumcenter and incenter of ABC.

 

If Ja, Jb, Jc are incenters of AOH, BOH and COH.

 

If Ja’, Jb’, Jc’ are A,B,C excenters of AOH, BOH, COH.

 

I1 is isogonal conjugate of I wrt triangle JaJbJc.

 

I2 is isogonal conjugate of I wrt triangle Ja’Jb’Jc’.

 

Then I1, I2 are on line OH and they are reflection in NPC center of ABC.

 

Best regards,

Tran Quang Hung.

 

ADGEOM 4554

[archimedes_26 = Kadir Altintas]:

Dear geometers

Let ABC be triangle and P any point.

Let O1,O2,O3 circumcenters of circles BCP, CAP and ABP.

Let Ha be hyperbola whose foci are O2 and O3 and passes through A.

Define Hb,Hc similarly.

These hyperbolas have always a second intersection point other than P. 

Is it possibelto find (in barycentrics) hyperbolic images of notable points ? (P=X(1), X(2).....)

What is the locus of points  if P moves on Euler line?

Thank you

With my regards....

 

[Randy Hutson]:

The hyperbolic image of P is the hyperbolic conjugate of P (discussed here sometime earlier) wrt O1O2O3.  Bernard Gibert has found that the coordinates for hyperbolic conjugates are in general very complicated, so I would expect the same for hyperbolic images.  One exception I found is when P = X(1), the hyperbolic image lies on lines {1,164}, {40,[X(108) of excentral triangle]}, {168,3973}, {361,1743}, {8688,12518} and has trilinears:

Cos[A/2] (Csc[B/2] - Csc[C/2]) - Cos[B/2] (Csc[C/2] - Csc[A/2]) - Cos[C/2] (Csc[A/2] - Csc[B/2]) - 2 Cot[B/2] (1 - Csc[C/2] Sin[A/2]) + 2 Cot[C/2] (1 - Csc[B/2] Sin[A/2]) : :

(6,9,13) values: (13.169691395932970, 9.337732849541849, -8.902238904359810)

Note that in this case, O1O2O3 is the 2nd circumperp triangle.


For P = X(2), the hyperbolic image is collinear with X(2) and [X(1) of circummedial triangle], but I could find no other lines through it.


For P = X(4), the hyperbolas are degenerate (the altitudes of ABC), and meet only at X(4).  O1O2O3 is the Johnson triangle in this case.


Best regards,
Randy Hutson

HYACINTHOS 28554

[Antreas P. Hatzipolakis]:

 

 

Variation 3:

 

Let ABC be a triangle.

 

Denote:

 

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

 

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

 

(Oa), (Ob), (Oc) = the circles (A', A'Na), (B', B'Nb), (C', C'Nc), resp.

 

Ra = the radical axis of (Ob), (Oc)

Rb = the radical axis of (Oc), (Oa)

Rc = the radical axis of (Oa), (Ob)

 

Point of concurrence of Ra, Rb, Rc (radical center of the circles) =: U.

The reflections of Ra, Rb, Rc in AI, BI, CI, resp. are concurrent at a point W such that W and U are symmetric in I.

 

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

 

 

[Ercole Suppa]

 

*** Ra, Rb, Rc are concurrent at a point U

 

U = MIDPOINT OF X(4) AND X(10284)

 

= a (a^5 b-a^4 b^2-2 a^3 b^3+2 a^2 b^4+a b^5-b^6+a^5 c-4 a^4 b c+7 a^3 b^2 c+4 a^2 b^3 c-8 a b^4 c-a^4 c^2+7 a^3 b c^2-14 a^2 b^2 c^2+7 a b^3 c^2+b^4 c^2-2 a^3 c^3+4 a^2 b c^3+7 a b^2 c^3+2 a^2 c^4-8 a b c^4+b^2 c^4+a c^5-c^6)  : : (barys)

 

= X[4]+X[10284], X[546]-X[2802], X[550]-3*X[3898], 3*X[1482]+X[5693], X[2771]-X[7984], X[2800]-X[6583], X[5694]+X[7982], X[5885]-2*X[13464], X[5887]+X[11278], 2*X[5901]-X[13145], 5*X[11522]-X[25413]

 

= lies on these lines: {4,10284}, {5,10}, {65,16173}, {392,17531}, {546,2802}, {550,3898}, {962,6951}, {1385,6909}, {1482,5693}, {2771,7984}, {2800,6583}, {3057,3585}, {3579,6940}, {5441,5919}, {5603,6972}, {5694,7982}, {5697,10895}, {5885,13464}, {5887,11278}, {5901,13145}, {6284,9957}, {10058,24928}, {10738,12751}, {11009,17638}, {11522,25413}, {15558,18990}, {18393,25414}

 

= midpoint of X(i) and X(j) for these {i,j}: {4,10284}, {3057,22793}, {5887,11278}, {10222,12672}, {18480,23340}

 

= reflection of X(i) in X(j) for these {i,j}: {5885,13464}, {13145,5901}

 

= (6-8-13) search numbers [-2.68587794894358968, -2.99242695479864457, 6.95198065781894365]

 

 

*** The reflections of Ra, Rb, Rc in AI, BI, CI, resp. are concurrent at a point W 

 

W =  MIDPOINT OF X(550) AND X(3874)

 

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

 

= 5*X[3]-X[5904], X[30]-X[6583], X[140]-X[2801], X[382]-5*X[18398], X[389]-2*X[15229], X[515]-X[5885], X[517]-X[550], X[518]-X[14810], 5*X[632]-3*X[15064], X[912]-X[12038], X[952]-X[13145], X[971]-X[9955], 2*X[3530]-X[3678], 3*X[3576]-X[5694], X[3579]-3*X[10167], 3*X[3656]+X[9961], X[6001]-X[15178], X[6102]+X[23156], 3*X[7967]-X[10284], 3*X[10202]+X[12680], 3*X[10246]+X[15071], 3*X[11220]+X[12699], 3*X[11231]-X[14872], 5*X[15016]-X[18525]

 

= lies on these lines: {3,5904}, {21,104}, {30,6583}, {35,17660}, {65,4325}, {72,5303}, {79,354}, {140,2801}, {382,18398}, {389,15229}, {515,5885}, {517,550}, {518,14810}, {632,15064}, {912,12038}, {942,7354}, {946,12267}, {952,13145}, {971,9955}, {1858,5126}, {3057,11571}, {3530,3678}, {3576,5694}, {3579,10167}, {3583,13751}, {3656,9961}, {5045,10391}, {5083,15171}, {5536,16117}, {5563,17637}, {6001,15178}, {6102,23156}, {6940,12738}, {7967,10284}, {8582,8728}, {10202,12680}, {10225,11491}, {10246,15071}, {10268,24645}, {11220,12699}, {11231,14872}, {15016,18525}, {15931,22937}, {16132,22765}

 

= midpoint of X(i) and X(j) for these {i,j}: {550,3874}, {6102,23156}, {12675,13369}, {12680,18480}

 

= reflection of X(i) in X(j) for these {i,j}: {389,15229}, {3678,3530}, {6583,12005}, {9955,13373}, {9956,9940}

 

= {X(i),X(j)}-harmonic conjugate of X(k) for these {i,j,k}: {10202,12680,18480}

 

= (6-8-13) search numbers [6.06649496785765599, 6.37304397371271088, -3.57136363890487734]

 

 

Best regards

Ercole Suppa

HYACINTHOS 27809

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

A', B', C' = the midpoints of AN, BN, CN, resp.

(Na), (Nb), (Nc) = the NPCs of NBC, NCA, NAB, resp.

The line B'C' intersects again (Nb), (Nc) at B1, C1, resp.
The line C'A' intersects again (Nc), (Na) at C2, A2, resp.
The line A'B' intersects again (Na), (Nb) at A3, B3, resp.

N1, N2, N3 = the NPC centers of A'B1C1, B'C2A2, C'A3B3, resp.

The circumcenter of N1N2N3 lies on the Euler line of ABC


[Peter Moses]:

Hi Antreas,

2 a^16-13 a^14 b^2+27 a^12 b^4-9 a^10 b^6-45 a^8 b^8+73 a^6 b^10-47 a^4 b^12+13 a^2 b^14-b^16-13 a^14 c^2+30 a^12 b^2 c^2-a^10 b^4 c^2-16 a^8 b^6 c^2-67 a^6 b^8 c^2+130 a^4 b^10 c^2-79 a^2 b^12 c^2+16 b^14 c^2+27 a^12 c^4-a^10 b^2 c^4-28 a^8 b^4 c^4-15 a^6 b^6 c^4-66 a^4 b^8 c^4+159 a^2 b^10 c^4-76 b^12 c^4-9 a^10 c^6-16 a^8 b^2 c^6-15 a^6 b^4 c^6-34 a^4 b^6 c^6-93 a^2 b^8 c^6+176 b^10 c^6-45 a^8 c^8-67 a^6 b^2 c^8-66 a^4 b^4 c^8-93 a^2 b^6 c^8-230 b^8 c^8+73 a^6 c^10+130 a^4 b^2 c^10+159 a^2 b^4 c^10+176 b^6 c^10-47 a^4 c^12-79 a^2 b^2 c^12-76 b^4 c^12+13 a^2 c^14+16 b^2 c^14-c^16::

= lies on these lines: {2, 3}, {3574, 14051}.
= midpoint of X(546) and X(5501).
= reflection of X(i) in X(j) for these {i, j}: {140, 13469}, {3530, 12056}, {10289, 5}, {15333, 10109}, {15334, 3628}, {15335, 3850}, {15336, 3530}.

Best regards,
Peter Moses.

HYACINTHOS 27803

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

A', B', C' = the midpoints of AN, BN, CN, resp.

(Na), (Nb), (Nc) = the NPCs of NBC, NCA, NAB, resp.

The line B'C' intersects again (Nb), (Nc) at B1, C1, resp.
The line C'A' intersects again (Nc), (Na) at C2, A2, resp.
The line A'B' intersects again (Na), (Nb) at A3, B3, resp.

La, Lb, Lc = the Euler lines of A'B1C1, B'C2A2, C'A3B3, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

ABC, A*B*C* are parallelogic.

Parallelogic centers?


[César Lozada]

 

 

ABC->A*B*C* = X(11584)

 

A*B*C*->ABC = (Nasty )

= (4*(29*R^2-2*SW)*S^4+(503*R^6- 2*R^4*(48*SA+185*SW)-4*R^2*( 17*SA^2-30*SW*SA-14*SW^2)+8* SA*SW*(-6*SW+5*SA))*S^2+(32*R^ 6-R^4*(19*SA+54*SW)+2*R^2*SW*( -6*SW+29*SA)-8*SW^2*(-2*SW+3* SA))*R^2*SA)*(4*S^2-(3*SA+SW)* (SB+SC)) : : (barys)

= reflection of X(137) in the line X(5501)X(13856)

= [ -2.0844932304853440, -4.6924677145285710, 7.8513697752666250 ]

 

César Lozada

HYACINTHOS 27802

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

A', B', C' = the midpoints of AN, BN, CN, resp.

(Na), (Nb), (Nc) = the NPCs of NBC, NCA, NAB, resp.

The line B'C' intersects again (Nb), (Nc) at B1, C1, resp.
The line C'A' intersects again (Nc), (Na) at C2, A2, resp.
The line A'B' intersects again (Na), (Nb) at A3, B3, resp.

La, Lb, Lc = the Euler lines of A'B1C1, B'C2A2, C'A3B3, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

ABC, A*B*C* are parallelogic.

Parallelogic centers?


[Peter Moses]:

Hi Antreas,

(ABC,  A*B*C*):  X(11584).

(A*B*C*,  ABC):

(b-c)^2 (b+c)^2 (-3 a^26+30 a^24 b^2-132 a^22 b^4+330 a^20 b^6-495 a^18 b^8+396 a^16 b^10-396 a^12 b^14+495 a^10 b^16-330 a^8 b^18+132 a^6 b^20-30 a^4 b^22+3 a^2 b^24+30 a^24 c^2-216 a^22 b^2 c^2+642 a^20 b^4 c^2-977 a^18 b^6 c^2+733 a^16 b^8 c^2-204 a^14 b^10 c^2+224 a^12 b^12 c^2-758 a^10 b^14 c^2+972 a^8 b^16 c^2-628 a^6 b^18 c^2+214 a^4 b^20 c^2-33 a^2 b^22 c^2+b^24 c^2-132 a^22 c^4+642 a^20 b^2 c^4-1174 a^18 b^4 c^4+947 a^16 b^6 c^4-333 a^14 b^8 c^4+128 a^12 b^10 c^4+227 a^10 b^12 c^4-1026 a^8 b^14 c^4+1253 a^6 b^16 c^4-682 a^4 b^18 c^4+159 a^2 b^20 c^4-9 b^22 c^4+330 a^20 c^6-977 a^18 b^2 c^6+947 a^16 b^4 c^6-374 a^14 b^6 c^6+120 a^12 b^8 c^6+39 a^10 b^10 c^6+315 a^8 b^12 c^6-1215 a^6 b^14 c^6+1225 a^4 b^16 c^6-445 a^2 b^18 c^6+35 b^20 c^6-495 a^18 c^8+733 a^16 b^2 c^8-333 a^14 b^4 c^8+120 a^12 b^6 c^8+27 a^10 b^8 c^8+51 a^8 b^10 c^8+420 a^6 b^12 c^8-1273 a^4 b^14 c^8+825 a^2 b^16 c^8-75 b^18 c^8+396 a^16 c^10-204 a^14 b^2 c^10+128 a^12 b^4 c^10+39 a^10 b^6 c^10+51 a^8 b^8 c^10+76 a^6 b^10 c^10+546 a^4 b^12 c^10-1122 a^2 b^14 c^10+90 b^16 c^10+224 a^12 b^2 c^12+227 a^10 b^4 c^12+315 a^8 b^6 c^12+420 a^6 b^8 c^12+546 a^4 b^10 c^12+1226 a^2 b^12 c^12-42 b^14 c^12-396 a^12 c^14-758 a^10 b^2 c^14-1026 a^8 b^4 c^14-1215 a^6 b^6 c^14-1273 a^4 b^8 c^14-1122 a^2 b^10 c^14-42 b^12 c^14+495 a^10 c^16+972 a^8 b^2 c^16+1253 a^6 b^4 c^16+1225 a^4 b^6 c^16+825 a^2 b^8 c^16+90 b^10 c^16-330 a^8 c^18-628 a^6 b^2 c^18-682 a^4 b^4 c^18-445 a^2 b^6 c^18-75 b^8 c^18+132 a^6 c^20+214 a^4 b^2 c^20+159 a^2 b^4 c^20+35 b^6 c^20-30 a^4 c^22-33 a^2 b^2 c^22-9 b^4 c^22+3 a^2 c^24+b^2 c^24)::

Best regards,
Peter Moses.

 

HYACINTHOS 27781

[Alexandr Skutin]:

Let ABC be a triangle.

Denote:

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

N1, N2, N3 = the isogonal conjugates of I wrt triangles NaBC, NbCA,
NcAB, resp.

1. N1, N2,N3 are collinear.

Which is this line ?

2. ABC, N1N2N3 are circumcyclologic.
ie the circumcircles of AN2N3, BN3N1, CN1N2 and ABC are concurrent
the circumcircles of N1BC, N2CA,N3AB and [degenerated] N1N2N3 [ = line
N1N2N3] are concurrent.

Cyclologic centers ?

 

[Angel Montesdeoca]:

**** 1. N1, N2,N3 are collinear.
Which is this line ?

a^6 (-b+c)+(b-c)^3 (b+c)^2 (b^2+b c+c^2)+a^4 (3 b^3-b^2 c+b c^2-3
c^3)+a^3 (-b^3 c+b c^3)+a b c (b^4+b^3 c-b c^3-c^4)-3 a^2 (b^5-c^5) x +
.... = 0.

{5, 79, 1749, 3336, 3467, 3652, 11246}

**** 2. ABC, N1N2N3 are circumcyclologic.

The cyclologic center (on circumcircle of ABC) of ABC with respect to
N1N2N3 is:

U = a^2/((b-c)(a^8 (b-c)^2+a^9 (b+c)+(b-c)^4 (b+c)^6+a (b-c)^2 (b+c)^5
(b^2+b c+c^2)-2 a^6 (b-c)^2 (2 b^2+3 b c+2 c^2)-a^7 (4 b^3+5 b^2 c+5 b
c^2+4 c^3)-a^2 (b^2-c^2)^2 (4 b^4+6 b^3 c+5 b^2 c^2+6 b c^3+4 c^4)+a^4
(b-c)^2 (6 b^4+16 b^3 c+21 b^2 c^2+16 b c^3+6 c^4)+a^5 (6 b^5+11 b^4 c+6
b^3 c^2+6 b^2 c^3+11 b c^4+6 c^5)-a^3 (4 b^7+11 b^6 c+6 b^5 c^2-6 b^4
c^3-6 b^3 c^4+6 b^2 c^5+11 b c^6+4 c^7))) : .... : ...

(6 - 9 - 13) - search numbers of U: (0.0573063005303804,
-0.0831477543240013, 3.67177925004082).

The cyclologic center of N1N2N3 with respect to ABC is:

V = a^13+2 a^12 (b+c)-(b-c)^6 (b+c)^7+a^9 b c (-5 b^2+4 b c-5
c^2)+a^11 (-3 b^2+2 b c-3 c^2)-a (b-c)^4 (b+c)^6 (2 b^2-b c+2 c^2)-a^10
(9 b^3+5 b^2 c+5 b c^2+9 c^3)+a^2 (b-c)^4 (b+c)^3 (3 b^4+4 b^3 c+3 b^2
c^2+4 b c^3+3 c^4)+5 a^8 (3 b^5+b^4 c+2 b^3 c^2+2 b^2 c^3+b c^4+3
c^5)+a^3 (b^2-c^2)^2 (9 b^6+8 b^5 c+2 b^4 c^2+b^3 c^3+2 b^2 c^4+8 b
c^5+9 c^6)+a^7 (10 b^6+6 b^5 c-3 b^4 c^2+10 b^3 c^3-3 b^2 c^4+6 b c^5+10
c^6)+2 a^4 b c (b^7+b^6 c-2 b^5 c^2-2 b^2 c^5+b c^6+c^7)-a^6 (10 b^7+4
b^6 c+3 b^5 c^2+b^4 c^3+b^3 c^4+3 b^2 c^5+4 b c^6+10 c^7)+a^5 (-15 b^8-8
b^7 c+12 b^6 c^2+b^5 c^3-7 b^4 c^4+b^3 c^5+12 b^2 c^6-8 b c^7-15 c^8) :
.... : ...

V lies on lines X(i)X(j) for these {i, j}: {5,79}, {30,5685}, {265,14452}.

(6 - 9 - 13) - search numbers of V: (0.820540785269206,
-0.513508968663262, 3.61745955931931).

Angel Montesdeoca

HYACINTHOS 27780

[Alexandr Skutin]:

Let ABC be a triangle.

Denote:

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

N1, N2, N3 = the isogonal conjugates of I wrt triangles NaBC, NbCA, NcAB, resp.

1. N1, N2,N3 are collinear.

Which is this line ?

2. ABC, N1N2N3 are circumcyclologic.

ie the circumcircles of AN2N3, BN3N1, CN1N2 and ABC are concurrent
the circumcircles of N1BC, N2CA,N3AB and [degenerated] N1N2N3 [ = line N1N2N3] are concurrent.

Cyclologic centers ?

 

[Peter Moses]:


1) line through X{5,79,1749,3336,3467,3652,11246}.

2) 

 

(ABC, N1N2N3):

a^2 (a-b) (a-c) (a^10+2 a^9 b-3 a^8 b^2-8 a^7 b^3+2 a^6 b^4+12 a^5 b^5+2 a^4 b^6-8 a^3 b^7-3 a^2 b^8+2 a b^9+b^10+a^9 c+4 a^8 b c+5 a^7 b^2 c-a^6 b^3 c-9 a^5 b^4 c-9 a^4 b^5 c-a^3 b^6 c+5 a^2 b^7 c+4 a b^8 c+b^9 c-4 a^8 c^2-6 a^7 b c^2+3 a^6 b^2 c^2+6 a^5 b^3 c^2+2 a^4 b^4 c^2+6 a^3 b^5 c^2+3 a^2 b^6 c^2-6 a b^7 c^2-4 b^8 c^2-4 a^7 c^3-11 a^6 b c^3-6 a^5 b^2 c^3+6 a^4 b^3 c^3+6 a^3 b^4 c^3-6 a^2 b^5 c^3-11 a b^6 c^3-4 b^7 c^3+6 a^6 c^4+4 a^5 b c^4-5 a^4 b^2 c^4-10 a^3 b^3 c^4-5 a^2 b^4 c^4+4 a b^5 c^4+6 b^6 c^4+6 a^5 c^5+11 a^4 b c^5+6 a^3 b^2 c^5+6 a^2 b^3 c^5+11 a b^4 c^5+6 b^5 c^5-4 a^4 c^6+2 a^3 b c^6+4 a^2 b^2 c^6+2 a b^3 c^6-4 b^4 c^6-4 a^3 c^7-5 a^2 b c^7-5 a b^2 c^7-4 b^3 c^7+a^2 c^8-2 a b c^8+b^2 c^8+a c^9+b c^9) (a^10+a^9 b-4 a^8 b^2-4 a^7 b^3+6 a^6 b^4+6 a^5 b^5-4 a^4 b^6-4 a^3 b^7+a^2 b^8+a b^9+2 a^9 c+4 a^8 b c-6 a^7 b^2 c-11 a^6 b^3 c+4 a^5 b^4 c+11 a^4 b^5 c+2 a^3 b^6 c-5 a^2 b^7 c-2 a b^8 c+b^9 c-3 a^8 c^2+5 a^7 b c^2+3 a^6 b^2 c^2-6 a^5 b^3 c^2-5 a^4 b^4 c^2+6 a^3 b^5 c^2+4 a^2 b^6 c^2-5 a b^7 c^2+b^8 c^2-8 a^7 c^3-a^6 b c^3+6 a^5 b^2 c^3+6 a^4 b^3 c^3-10 a^3 b^4 c^3+6 a^2 b^5 c^3+2 a b^6 c^3-4 b^7 c^3+2 a^6 c^4-9 a^5 b c^4+2 a^4 b^2 c^4+6 a^3 b^3 c^4-5 a^2 b^4 c^4+11 a b^5 c^4-4 b^6 c^4+12 a^5 c^5-9 a^4 b c^5+6 a^3 b^2 c^5-6 a^2 b^3 c^5+4 a b^4 c^5+6 b^5 c^5+2 a^4 c^6-a^3 b c^6+3 a^2 b^2 c^6-11 a b^3 c^6+6 b^4 c^6-8 a^3 c^7+5 a^2 b c^7-6 a b^2 c^7-4 b^3 c^7-3 a^2 c^8+4 a b c^8-4 b^2 c^8+2 a c^9+b c^9+c^10)::
= lies on the circumcircle.

(N1N2N3, ABC):

(a^2+a b+b^2-c^2) (a^2-b^2+a c+c^2) (a^3-a^2 b-a b^2+b^3+a^2 c-a b c+b^2 c-a c^2-b c^2-c^3) (a^3+a^2 b-a b^2-b^3+a^2 c-a b c+b^2 c-a c^2+b c^2-c^3) (a^3+a^2 b-a b^2-b^3-a^2 c-a b c-b^2 c-a c^2+b c^2+c^3):: 

 

= lies on the cubic K060 and these lines: {5,79},{30,5685},{265,14452}

.
= antigonal image of X(484).
= symgonal image of X(11813).
= X(i)-isoconjugate of X(j) for these (i,j): {3336, 7343}, {6149, 14452}.
= cevapoint of X(3467) and X(5685).
= barycentric quotient X(i)/X(j) for these {i,j}: {1989, 14452}, {11076, 3336}.

 

Best regards,
Peter Moses.

HYACINTHOS 27777

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

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

N1, N2, N3 = the isogonal conjugates of N wrt triangles A'BC, B'CA, C'AB, resp.

ABC, N1N2N3 are cyclologic.

Cyclologic centers?

[Peter Moses]:

Hi Antreas,
 
(ABC, N1N2N3):

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

on lines {{30,54},{93,186},{476,1141},{477,933},{1173,16106},{5899,16035},{15646,19176}}.
crosssum of X(1154) and X(14128).
barycentric product X(95) {{6,382},{50,112},{93,393},...}.

and

(N1N2N3, ABC):

X(54).

Best regards,
Peter Moses.

HYACINTHOS 27769

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

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

N1, N2, N3 = the isogonal conjugates of Na,Nb, Nc wrt triangles IBC, ICA, IAB, resp.

(ie N1,N2,N3 are the X(54)'s of IBC, ICA, IAB, resp)

ABC and the (degenerated) triangle N1N2N3 are circumcyclologic

ie the circumcircles of AN2N3, BN3N1, CN1N2, ABC are concurrent
the circumcircles of N1BC, N2CA, N3AB, N1N2N3 [= line OI] are concurrent.

 

Cyclologic centers?

Note: We can continue but the cyclologic centers probably are not interesting

Denote:
Oa, Ob, Oc = the circumcenters of AN2N3, BN3N1, CN1N2,resp
O1,O2,O3 = the circumcenters of N1BC, N2CA, N3AB, N1N2N3, resp.

The triangles OaObOc, O1O2O3 are circumcyclologic.

[Peter Moses]:

Hi Antreas,

>Cyclologic centers? 

X(6584) & X(484).

>The triangles OaObOc, O1O2O3 are circumcyclologic. 

At X(110) and the slightly unwieldly:

a (a^18+a^17 b-6 a^16 b^2-6 a^15 b^3+14 a^14 b^4+14 a^13 b^5-14 a^12 b^6-14 a^11 b^7+14 a^8 b^10+14 a^7 b^11-14 a^6 b^12-14 a^5 b^13+6a^4 b^14+6 a^3 b^15-a^2 b^16-a b^17+a^17 c+5 a^16 b c-2 a^15 b^2 c-18 a^14 b^3 c-4 a^13 b^4 c+22 a^12 b^5 c+14 a^11 b^6 c-14 a^10 b^7 c-10 a^9 b^8 c+20 a^8 b^9 c-6 a^7 b^10 c-30 a^6 b^11 c+12 a^5 b^12 c+18 a^4 b^13 c-6 a^3 b^14 c-2 a^2 b^15 c+a b^16 c-b^17 c-6 a^16 c^2-2 a^15 b c^2+26 a^14 b^2 c^2+12 a^13 b^3 c^2-44 a^12 b^4 c^2-30 a^11 b^5 c^2+38 a^10 b^6 c^2+42 a^9 b^7 c^2-22 a^8 b^8 c^2-38 a^7 b^9 c^2+14 a^6 b^10 c^2+24 a^5 b^11 c^2-8 a^4 b^12 c^2-10 a^3 b^13 c^2+2 a^2 b^14 c^2+2 a b^15 c^2-6 a^15 c^3-18 a^14 b c^3+12 a^13 b^2 c^3+50 a^12 b^3 c^3+2 a^11 b^4 c^3-44 a^10 b^5 c^3-24 a^9 b^6 c^3-8 a^8 b^7 c^3+32 a^7 b^8 c^3+50 a^6 b^9 c^3-26 a^5 b^10 c^3-34 a^4 b^11 c^3+12 a^3 b^12 c^3-4 a^2 b^13 c^3-2 a b^14 c^3+8 b^15 c^3+14 a^14 c^4-4 a^13 b c^4-44 a^12 b^2 c^4+2 a^11 b^3 c^4+53 a^10 b^4 c^4+9 a^9 b^5 c^4-28 a^8 b^6 c^4-8 a^7 b^7 c^4+3 a^6 b^8 c^4+3 a^5 b^9 c^4+2 a^4 b^10 c^4-4 a^3 b^11 c^4+2 a b^13 c^4+14 a^13 c^5+22 a^12 b c^5-30 a^11 b^2 c^5-44 a^10 b^3 c^5+9 a^9 b^4 c^5+45 a^8 b^5 c^5+2 a^7 b^6 c^5-28 a^6 b^7 c^5+9 a^5 b^8 c^5+9 a^4 b^9 c^5-2 a^3 b^10 c^5+24 a^2 b^11 c^5-2 a b^12 c^5-28 b^13 c^5-14 a^12 c^6+14 a^11 b c^6+38 a^10 b^2 c^6-24 a^9 b^3 c^6-28 a^8 b^4 c^6+2 a^7 b^5 c^6+10 a^6 b^6 c^6-8 a^5 b^7 c^6+18 a^3 b^9 c^6-2 a^2 b^10 c^6-6 a b^11 c^6-14 a^11 c^7-14 a^10 b c^7+42 a^9 b^2 c^7-8 a^8 b^3 c^7-8 a^7 b^4 c^7-28 a^6 b^5 c^7-8 a^5 b^6 c^7+14 a^4 b^7 c^7-14 a^3 b^8 c^7-18 a^2 b^9 c^7+6 a b^10 c^7+56 b^11 c^7-10 a^9 b c^8-22 a^8 b^2 c^8+32 a^7 b^3 c^8+3 a^6 b^4 c^8+9 a^5 b^5 c^8-14 a^3 b^7 c^8+2 a^2 b^8 c^8+20 a^8 b c^9-38 a^7 b^2 c^9+50 a^6 b^3 c^9+3 a^5 b^4 c^9+9 a^4 b^5 c^9+18 a^3 b^6 c^9-18 a^2 b^7 c^9-70 b^9 c^9+14 a^8 c^10-6 a^7 b c^10+14 a^6 b^2 c^10-26 a^5 b^3 c^10+2 a^4 b^4 c^10-2 a^3 b^5 c^10-2 a^2 b^6 c^10+6 a b^7 c^10+14 a^7 c^11-30 a^6 b c^11+24 a^5 b^2 c^11-34 a^4 b^3 c^11-4 a^3 b^4 c^11+24 a^2 b^5 c^11-6 a b^6 c^11+56 b^7 c^11-14 a^6 c^12+12 a^5 b c^12-8 a^4 b^2 c^12+12 a^3 b^3 c^12-2 a b^5 c^12-14 a^5 c^13+18 a^4 b c^13-10 a^3 b^2 c^13-4 a^2 b^3 c^13+2 a b^4 c^13-28 b^5 c^13+6 a^4 c^14-6 a^3 b c^14+2 a^2 b^2 c^14-2 a b^3 c^14+6 a^3 c^15-2 a^2 b c^15+2 a b^2 c^15+8 b^3 c^15-a^2 c^16+a b c^16-a c^17-b c^17):: 

on lines {{1,2687},{1325,2771}}.

Best regards,
Peter Moses.

HYACINTHOS 27760

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

Na,Nb,Nc = the NPC centers of NBC, NCA, NAB, resp.

A'B'C' = the pedal triangle of N wrt triangle NaNbNc

 

The reflections of NA', NB', NC' in BC,CA, AB, resp. bound triangle A*B*C*

ABC, A*B*C* are parallelogic.

Parallelogic centers?

[Peter Moses]:

Hi Antreas,

(ABC, A*B*C*):

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

5 X[1656] - 4 X[10615], 3 X[381] - 2 X[16337].

= lies on the cubics K025, K465, the sextic Q114 and the lines: {2,6150},{3,3432},{4,93},{5,252},{30,930},{265,14859},{381,16337},{571,2963},{1487,3850},{1510,11583},{1656,10615},{3447,5899},{3627,14141},....

= anticomplement X(6150).

= midpoint of X(3627) and X(14141).

= reflection of X(i) in X(j) for these {i,j}: {{3, 16336}, {1157, 5}, {14980, 18403}}.

= {X(4),X(3519)}-harmonic conjugate of X(18370).

= polar circle inverse of X(6152).

= (anticomplement of circumcirlce) inverse of X(12325).

= antigonal image of X(2070).

= barycentric product X(2070)X(11140).

= barycentric quotient X(i)/X(j) for these {i,j}: {{93, 9381}, {2070, 1994}, {9380, 49}}.

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

(A*B*C*, ABC):

a^16-5 a^14 b^2+10 a^12 b^4-10 a^10 b^6+5 a^8 b^8-a^6 b^10-5 a^14 c^2+14 a^12 b^2 c^2-10 a^10 b^4 c^2-5 a^8 b^6 c^2+11 a^6 b^8 c^2-8 a^4 b^10 c^2+4 a^2 b^12 c^2-b^14 c^2+10 a^12 c^4-10 a^10 b^2 c^4-3 a^8 b^4 c^4-a^6 b^6 c^4+10 a^4 b^8 c^4-12 a^2 b^10 c^4+6 b^12 c^4-10 a^10 c^6-5 a^8 b^2 c^6-a^6 b^4 c^6-4 a^4 b^6 c^6+8 a^2 b^8 c^6-15 b^10 c^6+5 a^8 c^8+11 a^6 b^2 c^8+10 a^4 b^4 c^8+8 a^2 b^6 c^8+20 b^8 c^8-a^6 c^10-8 a^4 b^2 c^10-12 a^2 b^4 c^10-15 b^6 c^10+4 a^2 b^2 c^12+6 b^4 c^12-b^2 c^14::

= lies on the lines: {5,195},{30,11671},{252,15345},{382,15619},{523,2070},{539,16337},{930,6150},{1141,1154},....

= reflection of X(i) in X(j) for these {i,j}: {{930,6150},{14140,12026}}.

Best regards,
Peter Moses.

 

HYACINTHOS 27747

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and A'B'C', A"B"C" the cevian,
circumcevian triangles of P, resp.

Denote:

(Oab), (Oac) = the circumcircles of AB'B", AC'C", resp.

(Obc), (Oba) = the circumcircles of BC'C", BA'A", resp.

(Oca), (Ocb) = the circumcircles of CA'A", CB'B", resp.

Ra = the radical axis of (Oab), (Oac)
Rb = the radical axis of (Obc), (Oba)
Rc = the radical axis of (Oca), (Ocb)

Sa = the radical axis of (Obc), (Ocb)
Sb = the radical axis of (Oca), (Oac)
Sc = the radical axis of (Oab), (Oba)

1. The locus of P such that Ra, Rb, Rc are concurrent is Stammler Quartic (*)

(*) See Angel Montesdeoca HG050618
http://amontes.webs.ull.es/otrashtm/HGT2018.htm#HG050618

2. Which is the locus of P such that Sa, Sb, Sc are concurrent?

Note: For which P's the six circumcenters are concyclic or in general conconic?

 

[César Lozada]:

 

>>2. Which is the locus of P such that Sa, Sb, Sc are concurrent?

 

Locus = {circumcircle} ∪ {circum-quartic q4 through ETC’s 2, 4, 20, 23 and vertices of anticomplementary triangle}

q4 =  ∑ [a^2*y*z*((b^2-c^2)*x^2-2*(b^2-c^2)*y*z-b^2*y^2+c^2*z^2)]=0

 

ETC pairs (P, S*(P)=point of concurrence) = (2, 2), (4,54), (20,3)

 

S*(X(23)) = X(23)X(895) ∩ X(111)X(468)

= (SB+SC)*(3*SB-SW)*(3*SC-SW)*((9*R^2*(2*R^2-SW)+SW*(SB+SC))*S^2+(R^2*(SW+3*SA)-SA^2+SB*SC)*SA*SW) : : (barys)

= on lines: {3, 15899}, {23, 895}, {111, 468}, {5486, 13574}, {7493, 10416}

= [ 1.2080038377115620, 4.1242319473470940, 0..2278867471079780 ]

 

>>For which P's the six circumcenters are concyclic or in general conconic?

Locus = {circumcircle} ∪ {q10=degree-10-circumcurve through ETC’s 2, 523 and vertices of anticomplementary triangle}

q10 = ∑ [y*z*((-((5*b^2-c^2)*a^4-(12*b^4+9*b^2*c^2-c^4)*a^2+(7*b^4+10*b^2*c^2+7*c^4)*b^2)*c^2*y^2+(b^2-c^2)*(a^6-4*(b^2+c^2)*a^4+(5*b^4+12*b^2*c^2+5*c^4)*a^2-2*(b^2+c^2)^3)*z*y-((b^2-5*c^2)*a^4-(b^4-9*b^2*c^2-12*c^4)*a^2-(7*b^4+10*b^2*c^2+7*c^4)*c^2)*b^2*z^2)*x^6+(-(5*a^6-2*(7*b^2+8*c^2)*a^4+(7*b^4+3*b^2*c^2+9*c^4)*a^2+(2*b^2+c^2)*(b^4+6*b^2*c^2-c^4))*c^2*y^3+(a^8-2*(3*b^2+2*c^2)*a^6+2*(5*b^4+6*b^2*c^2+3*c^4)*a^4-(6*b^6+10*b^2*c^4+3*c^6)*a^2+(b^2+c^2)*(b^4-5*b^2*c^2-2*c^4)*b^2)*z*y^2+(-a^8+2*(2*b^2+3*c^2)*a^6-2*(3*b^4+6*b^2*c^2+5*c^4)*a^4+(3*b^6+10*b^4*c^2+6*c^6)*a^2+(b^2+c^2)*(2*b^4+5*b^2*c^2-c^4)*c^2)*z^2*y+(5*a^6-2*(8*b^2+7*c^2)*a^4+(9*b^4+3*b^2*c^2+7*c^4)*a^2-(b^2+2*c^2)*(b^4-6*b^2*c^2-c^4))*b^2*z^3)*x^5+(b^2-c^2)*(3*a^6-5*(b^2+c^2)*a^4+(3*b^4+5*b^2*c^2+3*c^4)*a^2-(b^2+c^2)*(b^4-4*b^2*c^2+c^4))*z^2*y^2*x^4+(b^2-c^2)*(a^6+4*(b^2+c^2)*a^4+(3*b^4+4*b^2*c^2+3*c^4)*a^2-2*(b^4-c^4)*(b^2-c^2))*z^3*y^3*x^2-2*(a^4-(b^2-2*c^2)*a^2-3*(b^2-c^2)*c^2)*a^4*z^3*y^5+6*(b^4-c^4)*a^4*y^4*z^4+2*(a^4+(2*b^2-c^2)*a^2+3*(b^2-c^2)*b^2)*a^4*z^5*y^3)] = 0

 

For P=X(2), the conic q(X(2)) through the circumcenters have center X(140). No ETC centers lie on q(X(2)) and its perspector is not interesting.

 

César Lozada

HYACINTHOS 27743

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

N1, N2, N3 = the NPC centers of PNbNc, PNcNa, PNaNb, resp.

For P = N:

NaNbNc, N1N2N3 are perspective at N.

1. The reflections of NaNN1, NbNN2, NcNN3 in the sidelines of NaNbNc : NbNc, NcNa, NaNb, resp. are concurrent.

Point of concurrence?

 

2.  Let A'B'C' be the pedal triangle of N.

Denote:

La, Lb, Lc = the reflections of NNa, NNb, NNc in NbNc, NcNa, NaNb, resp.

The parallels to La, Lb, Lc through A', B', C', resp. are concurrent.

Point of concurrence?


[Peter Moses]:


Hi Antreas,

1). X(14051).

2). (a^2-b^2-b c-c^2) (a^2-b^2+b c-c^2) (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^4-a^2 b^2+b^4-2 a^2 c^2-2 b^2 c^2+c^4) (a^4-2 a^2 b^2+b^4-a^2 c^2-2 b^2 c^2+c^4) (3 a^12-12 a^10 b^2+19 a^8 b^4-16 a^6 b^6+9 a^4 b^8-4 a^2 b^10+b^12-12 a^10 c^2+20 a^8 b^2 c^2-4 a^6 b^4 c^2-10 a^4 b^6 c^2+12 a^2 b^8 c^2-6 b^10 c^2+19 a^8 c^4-4 a^6 b^2 c^4+5 a^4 b^4 c^4-8 a^2 b^6 c^4+15 b^8 c^4-16 a^6 c^6-10 a^4 b^2 c^6-8 a^2 b^4 c^6-20 b^6 c^6+9 a^4 c^8+12 a^2 b^2 c^8+15 b^4 c^8-4 a^2 c^10-6 b^2 c^10+c^12)::
on lines {{5,930},{128,1154},{10615,14071}}.

Best regards,
Peter Moses.

 

HYACINTHOS 27722

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

The parallels through Na, Nb, Nc to BC, CA, AB, resp. bound a triangle
A*B*C* homothetic to ABC.

Which is the locus of P such that the homothetic center of ABC, A*B*C*
lies on the OP line?

 

[Angel Montesdeoca]:

*** The locus of P such that the homothetic center of ABC, A*B*C* lies
on the OP line is the McCay cubic.

*** If P=u:v:w (barycentric), then the homothetic center of ABC, A*B*C*
is
Q = v w (-(b^2-c^2)^2 u^2+a^2 u (c^2 (u+v-w)+b^2 (u-v+w))+a^4 (2 v
w+u (v+w))) : ... : ...

*** If Q is the homothetic center of ABC, A*B*C*, pairs {P=X(i), Q=X(j)}, for pairs {i,j }: {1,65}, {3,4}, {4,5}, {6,5640},
{13,5472}, {14,5471}, {54,10095}, {56,14923}, {64,12111}, {84,14872},
{511,13137}, {512,12833}, {523,7471}, {947,10110}.

If P=X(2), then Q = X(5)X(32) /\ X(6)X(30)

Q = 4 a^4+a^2 (b^2+c^2)-(b^2-c^2)^2 : ... : ....

Q is the midpoint of X(i) and X(j), for these {i, j}: {6,7737},
{1992,11159}.

Q is the reflection of X(i) in X(j), for these {i, j}: {5,10796},
{141,7804}, {7761,3589}, {14929,141}, {15048,6}.

Q lies on lines X(i)X(j) for these {i, j}: {2,1285}, {3,7736},
{4,3172}, {5,32}, {6,30}, {11,609}, {12,7031}, {20,9605}, {39,550},
{51,15510}, {53,14581}, {69,11286}, {83,7750}, {98,14485}, {99,12156},
{112,251}, {115,3845}, {140,2548}, {141,754}, {172,496}, {183,3793},
{187,549}, {193,14033}, {235,10312}, {315,7819}, {316,7792}, {325,3972},
{376,5024}, {381,7735}, {382,5286}, {384,3933}, {385,8370}, {393,18494},
{428,5359}, {468,9745}, {495,1914}, {524,3734}, {538,3629}, {543,8584},
{546,3767}, {548,5013}, {574,8703}, {597,2030}, {598,3363}, {632,1506},
{952,1572}, {966,11354}, {1003,6390}, {1007,11288}, {1316,6792},
{1353,2782}, {1383,7426}, {1500,10386}, {1503,5039}, {1513,10788},
{1555,8779}, {1595,1968}, {1596,10311}, {1597,3087}, {1609,7514},
{1611,10128}, {1625,3051}, {1657,7738}, {1901,5037}, {1992,11159},
{1995,16317}, {2207,6756}, {2386,9969}, {2393,16983}, {2420,11007},
{2794,5480}, {2896,16988}, {3054,7603}, {3055,11539}, {3058,16785},
{3199,7715}, {3314,6661}, {3329,8356}, {3524,15655}, {3530,5023},
{3541,8778}, {3552,7921}, {3575,8743}, {3589,7761}, {3618,11287},
{3627,5007}, {3820,4386}, {3850,13881}, {3853,5319}, {3858,7755},
{3934,15598}, {5041,7756}, {5103,16385}, {5206,15712}, {5210,12100},
{5276,11113}, {5277,17527}, {5280,6284}, {5299,7354}, {5309,15687},
{5434,16784}, {5471,14137}, {5472,14136}, {5585,14891}, {6423,7584},
{6424,7583}, {6656,7787}, {6658,7839}, {6680,7843}, {6772,9112},
{6775,9113}, {6823,10316}, {7575,9699}, {7576,8744}, {7754,14035},
{7759,7789}, {7766,11361}, {7767,7770}, {7772,15704}, {7773,8361},
{7776,14001}, {7778,8368}, {7784,8364}, {7785,7807}, {7802,7878},
{7803,8357}, {7816,7838}, {7820,7845}, {7829,7842}, {7841,16989},
{7846,7860}, {7873,7889}, {7879,16898}, {7881,14037}, {7885,8363},
{7892,7900}, {7897,14036}, {7929,16895}, {8359,11174}, {8367,15271},
{8573,9818}, {8588,17504}, {8981,12963}, {9465,10301}, {9575,18481},
{9599,15325}, {9698,15513}, {9939,16986}, {10109,18584}, {10317,15760},
{10547,11380}, {11001,14482}, {11163,12040}, {11185,14614},
{11297,11488}, {11298,11489}, {11646,12212}, {11648,14075},
{11842,15980}, {12006,15575}, {12968,13966}, {13357,14881},
{13785,18539}, {15480,17131}, {15603,15700}, {15809,17409},
{16306,18572}.

(6 - 9 - 13) - search numbers of Q: (0.310524995543071,
0.808779387977605, 2.93742106151847).

Angel Montesdeoca

HYACINTHOS 27718

[Antreas P.  Hatzipolakis]:

 

Let ABC be a triangle, A'B'C' the pedal triangle of O and and A1B1C1 the pedal triangle of N.

Denote:

A2B2C2 = the pedal triangle of N wrt triangle A1B1C1

The circumcircles of NA'A2, NB'B2, NC'C2 are coaxial,

Second (other than N) intersection?

[Peter Moses]:

Hi Antreas,

a^2 (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^10 b^2-4 a^8 b^4+6 a^6 b^6-4 a^4 b^8+a^2 b^10+a^10 c^2-4 a^8 b^2 c^2+2 a^6 b^4 c^2+9 a^4 b^6 c^2-13 a^2 b^8 c^2+5 b^10 c^2-4 a^8 c^4+2 a^6 b^2 c^4+2 a^4 b^4 c^4+3 a^2 b^6 c^4-9 b^8 c^4+6 a^6 c^6+9 a^4 b^2 c^6+3 a^2 b^4 c^6+8 b^6 c^6-4 a^4 c^8-13 a^2 b^2 c^8-9 b^4 c^8+a^2 c^10+5 b^2 c^10:: 

on line: {5,51}.
midpoint of X(11583) and X(16336).

Best regards,
Peter Moses.

HYACINTHOS 27715

[Antreas P.  Hatzipolakis]:

 

Let ABC be a triangle and P a point.

The perpendicular bisectors of BC, CA, AB intersect the circumcircle at (A1, A2), (B1, B2), (C1, C2) ,resp.

 

The circumcircles of PA1A2, PB1B2, PC1C2 are coaxial.

 

Now, let (A'1, A'2), (B'1, B'2), (C'1, C'2) be the reflections of  (A1, A2), (B1, B2), (C1, C2) in BC, CA, AB, resp.

Which is the locus of P such that the circumcircles of  PA'1A'2, PB'1B'2, PC'1C'2 are coaxial ?
N lies on the locus.

 

[César Lozada]:
 

> Now, let (A'1, A'2), (B'1, B'2), (C'1, C'2) be the reflections of  (A1, A2), (B1, B2), (C1, C2) in BC, CA, AB, resp.

> Which is the locus of P such that the circumcircles of  PA'1A'2, PB'1B'2, PC'1C'2 are coaxial ?
> N lies on the locus.

The locus is {Linf} ∪ {Gibert’s K800, through ETC’s 1, 3, 5, 30, 191, 399, 5127, 5535, 6326, 7701, 13513 and vertices of triangles ABC, EXCENTRAL, FUHRMANN, FUHRMANN2}

 

Let P ∈ K800 and P’ be the other point of intersection.

ETC pairs (P,P’): (1, 6326), (5, 399), (191, 5535), (399, 5), (5535, 191), (6326, 1), (7701, 13513), (13513, 7701)

 

Missing point in the above list:

P’(X(5127)) = X(4)X(5535) ∩ X(5)X(580)

= a^10-(b+c)*a^9-2*(b^2+b*c+c^2)*a^8+(b+c)*(3*b^2-2*b*c+3*c^2)*a^7+(b^4+c^4+(4*b^2+7*b*c+4*c^2)*b*c)*a^6-3*(b+c)*(b^4+c^4-(b^2-b*c+c^2)*b*c)*a^5-(b^3+c^3)*(b+c)*(b^2+3*b*c+c^2)*a^4+(b+c)*(b^6+c^6-2*(b^4+b^2*c^2+c^4)*b*c)*a^3+2*(b^2-c^2)^2*(b+c)*(b^3+c^3)*a^2+(b^4-c^4)*(b^2-c^2)*(b+c)*b*c*a-(b^4-c^4)*(b^2-c^2)^3 : : (barys)

= 2*X(502)-3*X(5587)

= on the Fuhrmann circle, the cubic K800 and these lines: {4, 5535}, {5, 580}, {355, 13514}, {502, 5587}

= antipode of X(13514) in the Fuhrmann circle

= X(6798) of excentral triangle

= [ 4.8220713398357920, 1.0085543351807640, 0.7168631699350233 ]

 

P->P’ seems to map K800 -> K800, but I have not proved it.

 

César Lozada