## Τετάρτη, 21 Σεπτεμβρίου 2016

### TRIANGLE CENTERS FROM HYACINTHOS H001 - H010

Triangle centers from Hyacinthos

H001 = HATZIPOLAKIS - MONTESDEOCA

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

Let ABC be a triangle.

Denote:

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

A1, B1, C1 = the orthogonal projections of Na, Nb, Nc on IA, IB, IC, resp.

A2, B2, C2 = the reflections of Na, Nb, Nc, in IA, IB, IC, resp.

The Euler line of A2B2C2 is the OI line of ABC.

The point is the O of ABC wrt the triangle A2B2C2

The orthocenter of A1B1C1 is the point X942

(Atreas Hatzipolakis and Angel Montesdeoca, Sept. 13, 2016. See: Hyacinthos #24380)

The point lies on these lines: {1, 3}, {4, 1392}, {5, 519}, {8, 3090}, {10, 3628}, {20, 3655}, {30, 4301}, {72, 1173}, {140, 551}, {145, 355}, {381, 5881}, {392, 5047}, {515, 1483}, {518, 576}, {546, 946}, {547, 4669}, {548, 5493}, {573, 3723}, {575, 1386}, {631, 3654}, {632, 1125}, {944, 3146}, {956, 3951}, {962, 3529}, {1000, 5703}, {1056, 4323}, {1058, 4345}, {1210, 1387}, {1320, 1389}, {1339, 6048}, {1457, 5399}, {1656, 3679}, {1657, 9589}, {1837, 7743}, {1870, 1872}, {2771, 7984}, {2800, 3881}, {3058, 7491}, {3419, 6984}, {3485, 6982}, {3488, 5812}, {3523, 3653}, {3525, 3616}, {3555, 5887}, {3584, 5559}, {3585, 7972}, {3621, 5818}, {3622, 5657}, {3632, 5079}, {3633, 5072}, {3636, 6684}, {3680, 6918}, {3872, 3984}, {3892, 5884}, {3913, 6911}, {3915, 5398}, {3940, 4853}, {3962, 5288}, {3991, 4919}, {4004, 5253}, {4511, 6946}, {4677, 5055}, {4870, 6980}, {4902, 5059}, {4930, 6913}, {5044, 5289}, {5054, 9588}, {5076, 5691}, {5258, 7489}, {5722, 5761}, {5727, 9669}, {6419, 7969}, {6420, 7968}, {6447, 9583}, {6519, 9616}, {6863, 10056}, {6914, 8666}, {6924, 8715}, {6958, 10072}, {6988, 7320}

= Midpoint of X(i) and X(j) for these {i,j}: {1, 1482}, {3, 7982}, {40, 8148}, {145, 355}, {381, 2487}, {946, 3244}, {1320, 6265}, {1657, 9589}, {3241, 3656}, {3555, 5887}, {4301, 5882}.

= Reflection of X(i) in X(j) for these {i,j}: {8, 9956}, {10, 5901}, {65, 6583}, {355, 9955}, {1385, 1}, {1483, 3635}, {3579, 1385}, {4669, 547}, {5493, 548}, {5690, 1125}, {6684, 3636}.

H002 = HATZIPOLAKIS - MONTESDEOCA

X(10223) = 37th HATZIPOLAKIS-MONTESDEOCA POINT

Barycentrics (2 a^14 (b^2+c^2)-3 a^12 (3 b^4+2 b^2 c^2+3 c^4)+5 a^10 (3 b^6+b^4 c^2+b^2 c^4+3 c^6)-a^8 (10 b^8+3 b^6 c^2-2 b^4 c^4+3 b^2 c^6+10 c^8)+2 a^6 (5 b^8 c^2-4 b^6 c^4-4 b^4 c^6+5 b^2 c^8)+a^4 (b^2-c^2)^2 (3 b^8-8 b^6 c^2-8 b^2 c^6+3 c^8)-a^2 (b^2-c^2)^4 (b^6-3 b^4 c^2-3 b^2 c^4+c^6)-b^2 c^2 (b^2-c^2)^6+::

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

Denote:

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

Na, Nb, Nc = the NPC centers of A"B'C', B"C'A', C"A'B', resp.

The locus of P such that Na, Nb, Nc are collinear is the cubic K187 and a circum-quintic through the points X(74), X(1304)

The point is the intersection of the Euler line and the line NaNbNc for P = H (For P = O is X(140))

(Antreas Hatzipolakis and Angel Montesdeoca, Sept. 13, 2016. See: Hyacinthos #24377)

The point lies on these lines: {2,3},{143,523}

H003 = HATZIPOLAKIS - LOZADA

X(10222) = CENTER OF HATZIPOLAKIS-LOZADA CIRCLE

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

= 3*X(1)-X(3)

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

Denote:

N1, N2, N3 = the NPC centers of IBC, ICA, IAB, resp.

Na, Nb, Nc = the reflections of N1, N2, N3 in IA, IB, IC, resp.

N'a, N'b, N'c = the reflections of N1, N2, N3 in IA', IB', IC', resp.

Na, N'a, Nb, N'b, Nc, N'c are concyclic. The point is the center of the circle.

(Antreas Hatzipolakis and Cesar Lozada, Sept. 7, 2016. See: Hyacinthos #24303)

The point lies on these lines:{1,3}, {4,1392}, {5,519}, {8,3090}, {10,3628}, {20,3655}, {30,4301}, {72,1173}, {140,551}, {145,355}, {381,5881}, {392,5047}, {515,1483}, {518,576}, {546,946}, {547,4669}, {548,5493}, {573,3723}, {575,1386}, {631,3654}, (632,1125), (944,3146), (956,3951), (962,3529), (1000,5703), (1056,4323), (1058,4345), (1210,1387), (1320,1389), (1457,5399), {1656,3679}, {1657,9589}, {1837,7743}, {1870,1872}, {2771,7984}, {2800,3881}, {3058,7491}, {3419,6984}, {3485,6982}, {3488,5812}, {3523,3653}, {3525,3616}, {3555,5887}, {3584,5559}, {3585,7972}, {3621,5818}, {3622,5657}, {3625,10175}, {3632,5079}, {3633,5072}, {3636,6684}, {3680,6918}, {3892,5884}, {3913,6911}, {3915,5398}, {3940,4853}, {3962,5288}, {3991,4919}, {4004,5253}, {4511,6946}, {4677,5055}, {4691,10172}, {4701,10171}, {4870,6980}, {4930,6913}, {5044,5289}, {5054,9588}, {5076,5691}, {5258,7489}, {5722,5761}, {5727,9669}, {6419,7969}, {6420,7968}, {6447,9583}, {6519,9616}, {6863,10056}, {6914,8666}, {6924,8715}, {6958,10072}, {6988,7320}

= Midpoint of X(i),X(j) for these {i,j}: {1,1482}, {3,7982}, {40,8148}, {145,355}, {946,3244}, {1657,9589}, {3241,3656}, {3555,5887}, {4301,5882}

Reflection of X(i) in X(j) for these {i,j}: {8,9956}, {10,5901}, {65,6583}, {355,9955}, {1385,1}, {1483,3635}, {3579,1385}, {4669,547}, {5493,548}, {5690,1125}, {6684,3636}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): {1,46,1388}, {1,2098,9957}, {1,2099,942}, {1,3340,999}, {1,5697,2646}, {1,7962,3295}, {1,7982,3}, {3,1482,7982}, {4,5734,3656}, {8,5886,9956}, {46,1388,5126}, {145,5603,355}, {355,5603,9955}, {1466,2099,3340}, {3241,5734,4}, {3632,8227,5790}, {3679,9624,1656}

H004 = HATZIPOLAKIS - LOZADA

X(10224) = COMPLEMENT OF X(1658)

Trilinears (cos(2*A)+1/2)*cos(B-C)-cos(A)*cos(2*(B-C)) ::

= (9*R^2-2*SW)*X(3)+(7*R^2-2*SW) *X(4)

= [E-8*F, -3*E-8*F]

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

Denote:

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

N1, N2, N3 = the reflections of Na, Nb, Nc in NA', NB', NC', resp.

The locus of P such that the circumcenter of N1N2N3 lies on the Euler line is the excentral circum-quintic with barycentrics equation:

a^2*y*z*(((a^8-4*a^6*b^2+(6*b^ 4-2*b^2*c^2-c^4)*a^4-(b^2-c^2) *a^2*((2*b^2+c^2)^2-4*b^2*c^2) +(b^2+c^2)*(b^2-c^2)^3)*b^2*y- (a^8-4*a^6*c^2+(6*c^4-b^4-2*b^ 2*c^2)*a^4+(b^2-c^2)*a^2*((b^ 2+2*c^2)^2-4*b^2*c^2)-(b^2+c^ 2)*(b^2-c^2)^3)*c^2*z)*a^2*y* z+3*(b^2-c^2)*b^4*c^4*(a^2-b^ 2-c^2)*x^3+(b^2-c^2)*(a^8+b^8+ 2*b^6*c^2+2*b^2*c^6+c^8-4*(b^ 2+c^2)*a^6+2*(3*b^4+5*b^2*c^2+ 3*c^4)*a^4-4*(-b^2*c^2+(b^2+c^ 2)^2)*(b^2+c^2)*a^2)*a^2*x*y* z)+… = 0

The point is for P = O.

(Antreas Hatzipolakis and Cesar Lozada, Sept. 10, 2016. See: Hyacinthos #24351)

The point lies on these lines: {2,3}, {125,6102}, {569,8254}, {1154,5449}, {1568,5876}, {3574,5946}, {5448,5663}, {7741,8144}

= Anticomplement of X(10125)

= Complement of X(1658)

= Reflection of X(i) in X(j) for these {i,j}: {3,5498}, {1658,10125}, {10020,3628}

H005 = HATZIPOLAKIS - LOZADA

X(10225) = MIDPOINT OF X(3) AND X(484)

Trilinears (2*sin(A/2)+sin(3*A/2))*cos(( B-C)/2)+(cos(A)-1)*cos(B-C)- cos(A)+cos(2*A)-1/2 ::

Let ABC be a triangle and P a point.

Denote:

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

Nab, Nac = the reflections of Na in AC, AB, resp.

Nbc, Nba = the reflections of Nb in BA, BC, resp.

Nca, Ncb = the reflections of Nc in CB, CA, resp.

S1, S2, S3 = the perpendicular bisectors of NbaNca, NcbNab, NacNbc, resp. The locus of P such that S1, S2, S3 are concurrent is {sidelines} \/ {circum-quartic q21, equation y*z*(-a^4*y*z+2*(-SA^2+S^2)*x^ 2)+…=0, through H} \/ {excentral circum-septic q22 through I, O, H}

The point is for P = I

(Antreas Hatzipolakis ad Cesar Lozada, Sept. 8, 2016. See: Hyacinthos #24332)

The point lies on these lines: {1,3}, {631,5180}, {2475,9956}, {3814,4640}, {3916,5176}, {4973,5844}, {5057,6853}, {5080,6951}, {5499,6684}, {5886,9352}, {6952,9955}, {6972,7704}

= Midpoint of X(i),X(j) for these {i,j}: {3,484}

H006 = HATZIPOLAKIS - LOZADA

X(10216) = HATZIPOLAKIS-LOZADA-X(5) POINT

Trilinears (-1+2*cos(2*A))*cos(B-C)^3 ::

Let ABC be a triangle and P a point.

Denote:

Ba, Ca = the orthogonal projections of B, C on AP, resp.

R1 = the radical axis of the NPCs of ABaC, ACaB

Similarly R2, R3

The locus of P such that R1, R2, R3 are concurrent is Q038

The point is the point of concurrence of R1, R2, R3 for P = N

(Antreas Hatzipolakis and Cesar Lozada, Sept. 22, 2016. See: Hyacinthos #24453)

The point lies on these lines: {4,250}, {137,143},…

H007 = HATZIPOLAKIS - MOSES

X(10226) = 6th HATZIPOLAKIS-MOSES POINT

Barycentrics a^2 (2 a^8-4 a^6 b^2+4 a^2 b^6-2 b^8-4 a^6 c^2+10 a^4 b^2 c^2-5 a^2 b^4 c^2-b^6 c^2-5 a^2 b^2 c^4+6 b^4 c^4+4 a^2 c^6-b^2 c^6-2 c^8)::

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of OBC, OCA, OAB, resp.

A', B', C' = the reflections of Na, Nb, Nc in OA, OB, OC, resp.

The point is the circumcenter of A'B'C' lying on the Euler line of ABC.

(Antreas Hatzipolakis and Peter Moses, Sept. 13, 2016. See: Hyacinthos #24376)

The point lies on these lines: {2,3},{49,74},{156,3357},... .

H008 = HUNG - MONTESDEOCA

X(10227) = HUNG-MONTESDEOCA-EULER POINT

Barycentrics (2 a^28-19 a^26 (b^2+c^2)+a^24 (77 b^4+142 b^2 c^2+77 c^4)-2 a^22 (83 b^6+215 b^4 c^2+215 b^2 c^4+83 c^6)+4 a^20 (44 b^8+161 b^6 c^2+221 b^4 c^4+161 b^2 c^6+44 c^8)+a^18 (11 b^10-421 b^8 c^2-744 b^6 c^4-744 b^4 c^6-421 b^2 c^8+11 c^10) +a^16 (-297 b^12-22 b^10 c^2+91 b^8 c^4+144 b^6 c^6+91 b^4 c^8-22 b^2 c^10-297 c^12)+2 a^14 (198 b^14+54 b^12 c^2+95 b^10 c^4+75 b^8 c^6+75 b^6 c^8+95 b^4 c^10+54 b^2 c^12+198 c^14)-2 a^12 (99 b^16-20 b^14 c^2+41 b^12 c^4-6 b^10 c^6+3 b^8 c^8-6 b^6 c^10+41 b^4 c^12-20 b^2 c^14+99 c^16)-a^10 (77 b^18-131 b^16 c^2+82 b^14 c^4+42 b^12 c^6+29 b^10 c^8+29 b^8 c^10+42 b^6 c^12+82 b^4 c^14-131 b^2 c^16+77 c^18)+a^8 (b^2-c^2)^2 (187 b^16-120 b^14 c^2+82 b^12 c^4+56 b^10 c^6+87 b^8 c^8+56 b^6 c^10+82 b^4 c^12-120 b^2 c^14+187 c^16)-2 a^6 (b^2-c^2)^4 (67 b^14-5 b^12 c^2+26 b^10 c^4+22 b^8 c^6+22 b^6 c^8+26 b^4 c^10-5 b^2 c^12+67 c^14)+2 a^4 (b^2-c^2)^6 (26 b^12+6 b^10 c^2+5 b^8 c^4+5 b^4 c^8+6 b^2 c^10+26 c^12)-a^2 (b^2-c^2)^8 (11 b^10+3 b^8 c^2-6 b^6 c^4-6 b^4 c^6+3b^2 c^8+11 c^10)+(b^2-c^2)^12 (b^2+c^2)^2 ) )::

Let ABC be a triangle. A'B'C' is the circumcevian triangle of N. A''B''C'' is the pedal triangle of N wrt A'B'C'. The point is the point of concurrence of the Euler lines of the triangles A'B''C'',B'C''A'',C'A''B''.

(Tran Quang Hung and Angel Montesdeoca, Sept. 15, 2016. See: Hyacinthos #24387)

The point lies on these lines: pending

H009 = HUNG - MONTESDEOCA

X(10228) = HUNG-MONTESDEOCA CENTER OF SIMILITUDE

Barycentrics ( a^2 (a^26-8 a^24 (b^2+c^2)+28 a^22 (b^2+c^2)^2 -6 a^20 (9 b^6+28 b^4 c^2+28 b^2 c^4+9 c^6)+a^18 (53 b^8+277 b^6 c^2+406 b^4 c^4+277 b^2 c^6+53 c^8)+a^16 (6 b^10-273 b^8 c^2-499 b^6 c^4-499 b^4 c^6-273 b^2 c^8+6 c^10)+a^14 (-96 b^12+184 b^10c^2+307 b^8 c^4+386 b^6 c^6+307 b^4 c^8+184 b^2c^10-96 c^12)+2 a^12 (66 b^14-56 b^12 c^2-9 b^10 c^4-38 b^8 c^6-38 b^6 c^8-9b^4c^10-56 b^2 c^12+66 c^14)-a^10 (69 b^16+6 b^14 c^2+97 b^12 c^4+43 b^10 c^6+5 b^8 c^8+43 b^6c^10+97 b^4 c^12+6 b^2c^14+69 c^16)-a^8 (28 b^18-238 b^16 c^2+176 b^14 c^4-11 b^12 c^6+63 b^10 c^8+63 b^8c^10-11 b^6 c^12+176 b^4 c^14-238 b^2 c^16+28 c^18)+a^6 (b^2-c^2)^2 (68 b^16-256 b^14 c^2+89 b^12 c^4+65 b^10 c^6+97 b^8c^8+65 b^6 c^10+89 b^4 c^12-256 b^2 c^14+68 c^16)-a^4 (b^2-c^2)^4 (46 b^14-120 b^12 c^2+26 b^10 c^4+73 b^8 c^6+73 b^6c^8+26 b^4 c^10-120 b^2 c^12+46 c^14)+a^2 (b^2-c^2)^6 (15 b^12-29 b^10 c^2+16 b^8 c^4+32 b^6 c^6+16 b^4c^8-29 b^2 c^10+15 c^12)-(b^2-c^2)^8 (2 b^10-3 b^8 c^2+5 b^6 c^4+5 b^4 c^6-3 b^2 c^8+2 c^10) )::

Let ABC be a triangle. A'B'C' is the circumcevian triangle of N. A''B''C'' is pedal triangle of N wrt A'B'C'.

The point is the center of similitude of ABC and A''B''C''

(Tran Quang Hung and Angel Montesdeoca, Sept. 15, 2016. See: Hyacinthos #24390)

The point lies on these lines: pending

H010 = HUNG - MONTESDEOCA

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

= (2 r – 3 R) X[1] - (2 r - R) X[3] =

Let ABC be a triangle.

A1B1C1 is pedal triangle of incenter I.

A2,B2,C2 are reflections of A1,B1,C1 through I.

A3,B3,C3 are reflections of A,B,C through A2,B2,C2, reps.

The point is the NPC center of A3B3C3 lying on the OI line of ABC.

(Tran Quang Hung and Angel Montesdeoca, Sept. 20, 2016. See: Hyacinthos #24438)

The point lies on these lines: {1,3},{5,2802},{8,6965},....