Πέμπτη, 20 Οκτωβρίου 2016

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

(Antreas 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(10282) = X(3)X(64)∩X(51)X(54)

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

Let ABC be a triangle.

Denote:

Oa, Ob, Oc = the circumcenters of OBC, OCA, OAB, resp.

N1, N2, N3 = the NPC centers of OObOc, OOcOa, OOaOb, resp.

ABC, N1N2N3 are orthologic. The orthologic center (ABC, N1N2N3) is X74

The point is the orthologic center (N1N2N3, ABC)

(Antreas Hatzipolakis and Angel Montesdeoca, Oct. 20, 2016. See: Hyacinthos #24665)

The point lies on these lines: {2, 9833}, {3, 64}, {4, 1495}, {5, 5944}, {6, 3517}, {22, 1092}, {24, 184}, {25, 578}, {26, 206}, {30, 5448}, {39, 1971}, {49, 52}, {51, 54}, {110, 5562}, {125, 10018}, {140, 1503}, {143, 5097}, {156, 1658}, {159, 182}, {161, 569}, {185, 186}, {216, 3463}, {376, 5878}, {394, 9715}, {436, 8884}, {468, 6146}, {549, 6247}, {550, 1511}, {567, 9920}, {568, 9704}, {575, 2393}, {1181, 3515}, {1216, 7502}, {1660, 6644}, {1853, 3526}, {1899, 3147}, {1970, 3199}, {1994, 9706}, {2781, 7555}, {3060, 9545}, {3270, 9638}, {3292, 7556}, {3522, 5656}, {3528, 6225}, {3530, 6696}, {3534, 5895}, {3574, 7576}, {3917, 7512}, {5010, 6285}, {5050, 9924}, {5447, 7525}, {5449, 10020}, {5480, 7715}, {5651, 7509}, {5889, 9544}, {5894, 8703}, {6001, 7508}, {6102, 7575}, {6243, 9703}, {7280, 7355}, {8681, 9937}, {8718, 9934}.

H003 = HATZIPOLAKIS - MOSES

X(10283) = REFLECTION OF X(5) IN X(5886)

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

Let ABC be a triangle.

Denote:

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

Oib, Oib, Oic = the circumcenters of INbNc, INcNa, INaNb, resp.

The point is the centroid of OiaOibOic lying on the IN line.

(Antreas Hatzipolakis and Peter Moses, Oct. 20, 2016. See: Hyacinthos #24667)

The point lies on these lines: {1,5},{2,5844},{3,3622},{8, 3628},{30,5603},{140,1482},{ 145,1656},{165,3653},...

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,5,1483),(1,5901,5),(1,7951, 1317),(1,9624,355),(355,5886, 7988),(1482,3616,140),(7988, 9624,5886).

= Reflection of X(i) and X(j) for these {i,j}: {{5,5886},{5657,140},{5790, 547},{5886,5901},{8703,3576}}.

= Midpoint of X(i) and X(j) for these {i,j}: {{1,5886},{2,10247},{381,7967} ,{1482,5657},{1699,3655},{ 3241,5790},{3576,3656},{5603, 10246}}.

= 2 X[1] + X[5], 5 X[5] - 2 X[355], 5 X[1] + X[355], 2 X[140] + X[1482], 4 X[1] - X[1483], 2 X[5] + X[1483], 4 X[355] + 5 X[1483], 4 X[1387] - X[1484], X[145] + 5 X[1656], 2 X[140] - 5 X[3616], X[1482] + 5 X[3616], X[3] - 7 X[3622], X[8] - 4 X[3628], X[165] - 3 X[3653], 3 X[355] - 5 X[5587], 3 X[5] - 2 X[5587], 3 X[1] + X[5587], 3 X[1483] + 4 X[5587], 5 X[3616] - X[5657], 3 X[5603] + X[5731], 11 X[355] - 5 X[5881], 11 X[5587] - 3 X[5881], 11 X[5] - 2 X[5881], 11 X[1] + X[5881], 11 X[1483] + 4 X[5881], X[5881] - 11 X[5886], X[355] - 5 X[5886], X[5587] - 3 X[5886], X[1483] + 4 X[5886], X[355] - 10 X[5901], X[5587] - 6 X[5901], X[5] - 4 X[5901], X[1] + 2 X[5901], X[1483] + 8 X[5901], 5 X[5587] - 9 X[7988], 5 X[5] - 6 X[7988], X[355] - 3 X[7988], 5 X[5886] - 3 X[7988], 10 X[5901] - 3 X[7988], 5 X[1] + 3 X[7988], 5 X[1483] + 12 X[7988], 17 X[5] - 14 X[7989], 17 X[5886] - 7 X[7989], 17 X[1] + 7 X[7989], 7 X[5587] - 15 X[8227], 7 X[5] - 10 X[8227], 7 X[5886] - 5 X[8227], 14 X[5901] - 5 X[8227], 7 X[1] + 5 X[8227], 5 X[7989] - 17 X[9624], 5 X[5] - 14 X[9624], X[355] - 7 X[9624], 5 X[5886] - 7 X[9624], 10 X[5901] - 7 X[9624], 3 X[7988] - 7 X[9624], 5 X[1] + 7 X[9624], 5 X[5603] - X[9812], 5 X[5731] + 3 X[9812], X[5731] - 3 X[10246], X[9812] + 5 X[10246],...

H004 = 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},....

H005 = HUNG - MOSES - EULER 1

X(10285) = EULER-LINE INTERCEPT OF X(54)X(1263)

Barycentrics 2 a^16-9 a^14 b^2+15 a^12 b^4-9 a^10 b^6-5 a^8 b^8+13 a^6 b^10-11 a^4 b^12+5 a^2 b^14-b^16-9 a^14 c^2+22 a^12 b^2 c^2-13 a^10 b^4 c^2-15 a^6 b^8 c^2+34 a^4 b^10 c^2-27 a^2 b^12 c^2+8 b^14 c^2+15 a^12 c^4-13 a^10 b^2 c^4+4 a^8 b^4 c^4-7 a^6 b^6 c^4-22 a^4 b^8 c^4+51 a^2 b^10 c^4-28 b^12 c^4-9 a^10 c^6-7 a^6 b^4 c^6-2 a^4 b^6 c^6-29 a^2 b^8 c^6+56 b^10 c^6-5 a^8 c^8-15 a^6 b^2 c^8-22 a^4 b^4 c^8-29 a^2 b^6 c^8-70 b^8 c^8+13 a^6 c^10+34 a^4 b^2 c^10+51 a^2 b^4 c^10+56 b^6 c^10-11 a^4 c^12-27 a^2 b^2 c^12-28 b^4 c^12+5 a^2 c^14+8 b^2 c^14-c^16::

Let ABC be a triangle.

Denote: Ha, Hb, Hc = the orthocenters of NBC, NCA, NAB, resp.

Oa, Ob, Oc = the circumcenters of NBC, NCA, NAB, resp.

Nha,Nhb,Nhc = the NPC centers of NHbHc,NHcHa,NHaHb, resp.

The point is the NPC center of NhaNhbNhc lying on the Euler line of ABC.

(Tran Quang Hung and Peter Moses, Oct. 20, 2016. See: Hyacinthos #24664)

The point lies on these lines:{2,3},{54,1263}.

= Anticomplement X[10126].

= Reflection of X(i) in X(j) for these {i,j}: {{5, 5501}, {10205, 140}}.

X(54) {Nha,Nhb,Nhc} = N ABC.

X(1141) {Nha,Nhb,Nhc} = H ABC.

X(1157) {Nha,Nhb,Nhc} = X(1263) ABC.

X(8254) {Nha,Nhb,Nhc} = X(5501) ABC.

H006 = HUNG - MOSES - EULER 2

X(10286) = MIDPOINT OF X(5) AND X(5500)

Barycentrics 2 a^22-15 a^20 b^2+50 a^18 b^4-93 a^16 b^6+92 a^14 b^8-14 a^12 b^10-84 a^10 b^12+110 a^8 b^14-62 a^6 b^16+13 a^4 b^18+2 a^2 b^20-b^22-15 a^20 c^2+82 a^18 b^2 c^2-172 a^16 b^4 c^2+139 a^14 b^6 c^2+41 a^12 b^8 c^2-125 a^10 b^10 c^2+3 a^8 b^12 c^2+97 a^6 b^14 c^2-54 a^4 b^16 c^2-a^2 b^18 c^2+5 b^20 c^2+50 a^18 c^4-172 a^16 b^2 c^4+160 a^14 b^4 c^4+52 a^12 b^6 c^4-94 a^10 b^8 c^4-65 a^8 b^10 c^4+58 a^6 b^12 c^4+32 a^4 b^14 c^4-14 a^2 b^16 c^4-7 b^18 c^4-93 a^16 c^6+139 a^14 b^2 c^6+52 a^12 b^4 c^6-72 a^10 b^6 c^6-39 a^8 b^8 c^6-84 a^6 b^10 c^6+98 a^4 b^12 c^6+4 a^2 b^14 c^6-5 b^16 c^6+92 a^14 c^8+41 a^12 b^2 c^8-94 a^10 b^4 c^8-39 a^8 b^6 c^8-18 a^6 b^8 c^8-89 a^4 b^10 c^8+76 a^2 b^12 c^8+22 b^14 c^8-14 a^12 c^10-125 a^10 b^2 c^10-65 a^8 b^4 c^10-84 a^6 b^6 c^10-89 a^4 b^8 c^10-134 a^2 b^10 c^10-14 b^12 c^10-84 a^10 c^12+3 a^8 b^2 c^12+58 a^6 b^4 c^12+98 a^4 b^6 c^12+76 a^2 b^8 c^12-14 b^10 c^12+110 a^8 c^14+97 a^6 b^2 c^14+32 a^4 b^4 c^14+4 a^2 b^6 c^14+22 b^8 c^14-62 a^6 c^16-54 a^4 b^2 c^16-14 a^2 b^4 c^16-5 b^6 c^16+13 a^4 c^18-a^2 b^2 c^18-7 b^4 c^18+2 a^2 c^20+5 b^2 c^20-c^22::

Let ABC be a triangle.

Denote:

Oa, Ob, Oc = the circumcenters of NBC, NCA, NAB, resp.

Noa,Nob,Noc = the NPC centers of NObOc,NOcOa,NOaOb, resp.

The point is the NPC center of NoaNobNoc lying on the Euler line of ABC.

(Tran Quang Hung and Peter Moses, Oct. 20, 2016. See: Hyacinthos #24664)

The point lies on these lines:{2,3} = Midpoint of X[5] and X[5500].

H007 = KIRIKAMI - MONTESDEOCA - EULER 1

X(10287) = X(3)X(2575)∩X(5)X(523)

Barycentrics (b^2+c^2-a^2) (R F1 + a^2 G1 |OH|)::

where F1 = a^30 (b^2+c^2)-6 a^28 (2 b^4+b^2 c^2+2 c^4)+9 a^26 (6 b^6+5 b^4 c^2+5 b^2 c^4+6 c^6)-a^24 (111 b^8+245 b^6 c^2+42 b^4 c^4+245 b^2 c^6+111 c^8)+3 a^22 (21 b^10+242 b^8 c^2+41 b^6 c^4+41 b^4 c^6+242 b^2 c^8+21 c^10)+a^20 (174 b^12-1145 b^10 c^2-856 b^8 c^4+750 b^6 c^6-856 b^4 c^8-1145 b^2 c^10+174 c^12)-8 a^18 (47 b^14-81 b^12 c^2-306 b^10 c^4+150 b^8 c^6+150 b^6 c^8-306 b^4 c^10-81 b^2 c^12+47 c^14)+a^16 (207 b^16+982 b^14 c^2-4004 b^12 c^4+410 b^10 c^6+2794 b^8 c^8+410 b^6 c^10-4004 b^4 c^12+982 b^2 c^14+207 c^16)+a^14 (207 b^18-2387 b^16 c^2+4142 b^14 c^4+498 b^12 c^6-2076 b^10 c^8-2076 b^8 c^10+498 b^6 c^12+4142 b^4 c^14-2387 b^2 c^16+207 c^18)-4 a^12 (94 b^20-498 b^18 c^2+441 b^16 c^4+582 b^14 c^6-615 b^12 c^8+24 b^10 c^10-615 b^8 c^12+582 b^6 c^14+441 b^4 c^16-498 b^2 c^18+94 c^20)+a^10 (b^2-c^2)^4 (174 b^14+493 b^12 c^2-1435 b^10 c^4-1232 b^8 c^6-1232 b^6 c^8-1435 b^4 c^10+493 b^2 c^12+174 c^14)+a^8 (b^2-c^2)^4 (63 b^16-845 b^14 c^2+1176 b^12 c^4-123 b^10 c^6+322 b^8 c^8-123 b^6 c^10+1176 b^4 c^12-845 b^2 c^14+63 c^16)-a^6 (b^2-c^2)^6 (111 b^14-402 b^12 c^2-36 b^10 c^4+7 b^8 c^6+7 b^6 c^8-36 b^4 c^10-402 b^2 c^12+111 c^14)+a^4 (b^2-c^2)^8 (b^2+c^2)^2 (54 b^8-149 b^6 c^2+124 b^4 c^4-149 b^2 c^6+54 c^8)-6 a^2 (b^2-c^2)^10 (b^2+c^2)^3 (2 b^4-3 b^2 c^2+2 c^4)+(b^2-c^2)^12 (b^2+c^2)^4,

and G1 = a^26 (b^4+c^4)-7 a^24 (b^6+b^4 c^2+b^2 c^4+c^6)+a^22 (18 b^8+59 b^6 c^2+16 b^4 c^4+59 b^2 c^6+18 c^8)-a^20 (14 b^10+205 b^8 c^2+73 b^6 c^4+73 b^4 c^6+205 b^2 c^8+14 c^10)+a^18 (-25 b^12+349 b^10 c^2+363 b^8 c^4-134 b^6 c^6+363 b^4 c^8+349 b^2 c^10-25 c^12)+a^16 (63 b^14-194 b^12 c^2-992 b^10 c^4+291 b^8 c^6+291 b^6 c^8-992 b^4 c^10-194 b^2 c^12+63 c^14)-2 a^14 (18 b^16+169 b^14 c^2-808 b^12 c^4+31 b^10 c^6+492 b^8 c^8+31 b^6 c^10-808 b^4 c^12+169 b^2 c^14+18 c^16)-2 a^12 (18 b^18-371 b^16 c^2+765 b^14 c^4+191 b^12 c^6-443 b^10 c^8-443 b^8 c^10+191 b^6 c^12+765 b^4 c^14-371 b^2 c^16+18 c^18)+a^10 (63 b^20-518 b^18 c^2+319 b^16 c^4+1488 b^14 c^6-1406 b^12 c^8+236 b^10 c^10-1406 b^8 c^12+1488 b^6 c^14+319 b^4 c^16-518 b^2 c^18+63 c^20)-a^8 (b^2-c^2)^2 (25 b^18+109 b^16 c^2-1104 b^14 c^4+1160 b^12 c^6-126 b^10 c^8-126 b^8 c^10+1160 b^6 c^12-1104 b^4 c^14+109 b^2 c^16+25 c^18)-a^6 (b^2-c^2)^4 (14 b^16-303 b^14 c^2+608 b^12 c^4-57 b^10 c^6+148 b^8 c^8-57 b^6 c^10+608 b^4 c^12-303 b^2 c^14+14 c^16)+a^4 (b^2-c^2)^6 (18 b^14-157 b^12 c^2+39 b^10 c^4+52 b^8 c^6+52 b^6 c^8+39 b^4 c^10-157 b^2 c^12+18 c^14)-a^2 (b^2-c^2)^8 (b^2+c^2)^2 (7 b^8-47 b^6 c^2+38 b^4 c^4-47 b^2 c^6+7 c^8)+(b^2-c^2)^10 (b^2+c^2)^3 (b^4-5 b^2 c^2+c^4)

The point is the point of concurrence of the Euler lines of AHX(1113), BHX(1113), CHX(1113)

(Seiichi Kirikami and Angel Montesdeoca, Oct 6, 2016. See: Hyacinthos #24541) and #24545)

= X(3)X(2575) /\ X(5)X(523)

H008 = KIRIKAMI - MONTESDEOCA - EULER 2

X(10288) = X(3)X(2574)∩X(5)X(523)

Barycentrics (b^2+c^2-a^2) (R F1 - a^2 G1 |OH|)::

where F1 = a^30 (b^2+c^2)-6 a^28 (2 b^4+b^2 c^2+2 c^4)+9 a^26 (6 b^6+5 b^4 c^2+5 b^2 c^4+6 c^6)-a^24 (111 b^8+245 b^6 c^2+42 b^4 c^4+245 b^2 c^6+111 c^8)+3 a^22 (21 b^10+242 b^8 c^2+41 b^6 c^4+41 b^4 c^6+242 b^2 c^8+21 c^10)+a^20 (174 b^12-1145 b^10 c^2-856 b^8 c^4+750 b^6 c^6-856 b^4 c^8-1145 b^2 c^10+174 c^12)-8 a^18 (47 b^14-81 b^12 c^2-306 b^10 c^4+150 b^8 c^6+150 b^6 c^8-306 b^4 c^10-81 b^2 c^12+47 c^14)+a^16 (207 b^16+982 b^14 c^2-4004 b^12 c^4+410 b^10 c^6+2794 b^8 c^8+410 b^6 c^10-4004 b^4 c^12+982 b^2 c^14+207 c^16)+a^14 (207 b^18-2387 b^16 c^2+4142 b^14 c^4+498 b^12 c^6-2076 b^10 c^8-2076 b^8 c^10+498 b^6 c^12+4142 b^4 c^14-2387 b^2 c^16+207 c^18)-4 a^12 (94 b^20-498 b^18 c^2+441 b^16 c^4+582 b^14 c^6-615 b^12 c^8+24 b^10 c^10-615 b^8 c^12+582 b^6 c^14+441 b^4 c^16-498 b^2 c^18+94 c^20)+a^10 (b^2-c^2)^4 (174 b^14+493 b^12 c^2-1435 b^10 c^4-1232 b^8 c^6-1232 b^6 c^8-1435 b^4 c^10+493 b^2 c^12+174 c^14)+a^8 (b^2-c^2)^4 (63 b^16-845 b^14 c^2+1176 b^12 c^4-123 b^10 c^6+322 b^8 c^8-123 b^6 c^10+1176 b^4 c^12-845 b^2 c^14+63 c^16)-a^6 (b^2-c^2)^6 (111 b^14-402 b^12 c^2-36 b^10 c^4+7 b^8 c^6+7 b^6 c^8-36 b^4 c^10-402 b^2 c^12+111 c^14)+a^4 (b^2-c^2)^8 (b^2+c^2)^2 (54 b^8-149 b^6 c^2+124 b^4 c^4-149 b^2 c^6+54 c^8)-6 a^2 (b^2-c^2)^10 (b^2+c^2)^3 (2 b^4-3 b^2 c^2+2 c^4)+(b^2-c^2)^12 (b^2+c^2)^4,

and G1 = a^26 (b^4+c^4)-7 a^24 (b^6+b^4 c^2+b^2 c^4+c^6)+a^22 (18 b^8+59 b^6 c^2+16 b^4 c^4+59 b^2 c^6+18 c^8)-a^20 (14 b^10+205 b^8 c^2+73 b^6 c^4+73 b^4 c^6+205 b^2 c^8+14 c^10)+a^18 (-25 b^12+349 b^10 c^2+363 b^8 c^4-134 b^6 c^6+363 b^4 c^8+349 b^2 c^10-25 c^12)+a^16 (63 b^14-194 b^12 c^2-992 b^10 c^4+291 b^8 c^6+291 b^6 c^8-992 b^4 c^10-194 b^2 c^12+63 c^14)-2 a^14 (18 b^16+169 b^14 c^2-808 b^12 c^4+31 b^10 c^6+492 b^8 c^8+31 b^6 c^10-808 b^4 c^12+169 b^2 c^14+18 c^16)-2 a^12 (18 b^18-371 b^16 c^2+765 b^14 c^4+191 b^12 c^6-443 b^10 c^8-443 b^8 c^10+191 b^6 c^12+765 b^4 c^14-371 b^2 c^16+18 c^18)+a^10 (63 b^20-518 b^18 c^2+319 b^16 c^4+1488 b^14 c^6-1406 b^12 c^8+236 b^10 c^10-1406 b^8 c^12+1488 b^6 c^14+319 b^4 c^16-518 b^2 c^18+63 c^20)-a^8 (b^2-c^2)^2 (25 b^18+109 b^16 c^2-1104 b^14 c^4+1160 b^12 c^6-126 b^10 c^8-126 b^8 c^10+1160 b^6 c^12-1104 b^4 c^14+109 b^2 c^16+25 c^18)-a^6 (b^2-c^2)^4 (14 b^16-303 b^14 c^2+608 b^12 c^4-57 b^10 c^6+148 b^8 c^8-57 b^6 c^10+608 b^4 c^12-303 b^2 c^14+14 c^16)+a^4 (b^2-c^2)^6 (18 b^14-157 b^12 c^2+39 b^10 c^4+52 b^8 c^6+52 b^6 c^8+39 b^4 c^10-157 b^2 c^12+18 c^14)-a^2 (b^2-c^2)^8 (b^2+c^2)^2 (7 b^8-47 b^6 c^2+38 b^4 c^4-47 b^2 c^6+7 c^8)+(b^2-c^2)^10 (b^2+c^2)^3 (b^4-5 b^2 c^2+c^4)

The point is the point of concurrence of the Euler lines of AHX(1114), BHX(1114), CHX(1114)

(Seiichi Kirikami and Angel Montesdeoca, Oct 6, 2016. See: Hyacinthos #24541) and #24545)

= X(3)X(2574) /\ X(5)X(523)

H009 = HATZIPOLAKIS - MOSES

X(10289) = 6th HATZIPOLAKIS-MOSES-EULER POINT

Barycentrics 2 a^16-13 a^14 b^2+43 a^12 b^4-89 a^10 b^6+115 a^8 b^8-87 a^6 b^10+33 a^4 b^12-3 a^2 b^14-b^16-13 a^14 c^2+62 a^12 b^2 c^2-113 a^10 b^4 c^2+64 a^8 b^6 c^2+61 a^6 b^8 c^2-94 a^4 b^10 c^2+33 a^2 b^12 c^2+43 a^12 c^4-113 a^10 b^2 c^4+68 a^8 b^4 c^4+17 a^6 b^6 c^4+46 a^4 b^8 c^4-81 a^2 b^10 c^4+20 b^12 c^4-89 a^10 c^6+64 a^8 b^2 c^6+17 a^6 b^4 c^6+30 a^4 b^6 c^6+51 a^2 b^8 c^6-64 b^10 c^6+115 a^8 c^8+61 a^6 b^2 c^8+46 a^4 b^4 c^8+51 a^2 b^6 c^8+90 b^8 c^8-87 a^6 c^10-94 a^4 b^2 c^10-81 a^2 b^4 c^10-64 b^6 c^10+33 a^4 c^12+33 a^2 b^2 c^12+20 b^4 c^12-3 a^2 c^14-c^16::

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

Denote:

Oa, Ob, Oc = the circumcenters of NB'C', NC'A', NA'B', resp.

Ooa, Oob, Ooc = the circumcenters of NObOc,NOcOa,NOaOb, resp.

The point is the NPC center of OoaOobOoc lying on the Euler line of ABC.

(Antreas Hatzipolakis and Peter Moses, Oct 21, 2016. See: Hyacinthos #24670)

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


Κυριακή, 16 Οκτωβρίου 2016

TRIANGLE CENTERS FROM HYACINTHOS H011 - H020

H011 = HATZIPOLAKIS - MOSES

Barycentrics 2 a^9-2 a^8 b-3 a^7 b^2+a^6 b^3+a^5 b^4+5 a^4 b^5-a^3 b^6-5 a^2 b^7+a b^8+b^9-2 a^8 c+8 a^7 b c-a^6 b^2 c-2 a^5 b^3 c-3 a^4 b^4 c-12 a^3 b^5 c+9 a^2 b^6 c+6 a b^7 c-3 b^8 c-3 a^7 c^2-a^6 b c^2+2 a^5 b^2 c^2-2 a^4 b^3 c^2+a^3 b^4 c^2+19 a^2 b^5 c^2-16 a b^6 c^2+a^6 c^3-2 a^5 b c^3-2 a^4 b^2 c^3+24 a^3 b^3 c^3-23 a^2 b^4 c^3-6 a b^5 c^3+8 b^6 c^3+a^5 c^4-3 a^4 b c^4+a^3 b^2 c^4-23 a^2 b^3 c^4+30 a b^4 c^4-6 b^5 c^4+5 a^4 c^5-12 a^3 b c^5+19 a^2 b^2 c^5-6 a b^3 c^5-6 b^4 c^5-a^3 c^6+9 a^2 b c^6-16 a b^2 c^6+8 b^3 c^6-5 a^2 c^7+6 a b c^7+a c^8-3 b c^8+c^9::

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

Denote:

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

M1, M2, M3 = the midpoints of IA', IB', IC', resp.

MaMbMc, M1M2M3 are cyclologic. The point is the cyclologic center (MaMbMc, M1M2M3). The other cyclologic center (M1M2M3, MaMbMc) is the point X(1387)

(Antreas Hatzipolakis and Peter Moses, Sept. 20, 2016. See: Hyacinthos #24436)

The point lies on these lines: {1,1537},{55,108},{123,3816}, ...

H012 = HATZIPOLAKIS - MOSES

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

3 X[3] + X[146], 3 X[5] - X[265], 3 X[110] + X[265], 3 X[2] + X[399], X[74] - 3 X[549], 3 X[113] - X[1539], 3 X[1511] + X[1539], 5 X[1656] - X[3448], 2 X[3628] + X[5609], X[1511] - 3 X[5642], X[113] + 3 X[5642], X[1539] + 9 X[5642], X[2931] + 3 X[5654], X[74] + 3 X[5655], X[2948] + 3 X[5886], 3 X[5972] + X[6053], 3 X[140] + 2 X[6053], 3 X[140] - 2 X[6699], 3 X[5972] - X[6699], 3 X[5066] - 2 X[7687], 3 X[5055] + X[9143], 3 X[597] - X[9976].

Let ABC be a triangle, NaNbNc the pedal triangle of N and OaObOc the pedal triangle of O.

Denote:

N1, N2, N3 = the reflections of N in BC, CA, AB, resp.

O1,O2, O3 = the reflections of O in BC, CA, AB, resp.

The point is the intersection of the parallels to O1N1, O2N2, O3N3 through Na,Nb,Nc, resp. The parallels to O1N1, O2N2, O3N3 through Oa,Ob,Oc, resp. concur at X(1511). The lines O1N1, O2N2, O3N3 concur at X(110)

(Antreas Hatzipolakis and Peter Moses, Sept. 21, 2016. See: Hyacinthos #24449)

The point lies on these lines:{2,399},{3,146},{4,7666},{5,4 9},{30,113},{69,10201},{74,549 },{125,3628},{140,5663},{403,3 043},{468,1986},{495,10091},{4 96,10088},{542,547},{546,9820} ,{548,2777},{550,7728},{597, 9976},{1125,2771},{1154,10096} ,{1656,3448},{2931,5654},{ 2948,5886},{3564,6593},{3582, 6126},{3584,7343},{3850,10113} ,{5055,9143},{5066,7687},{ 5432,7727},{5876,10125},{5898, 7693},{6140,6592},{6153,10095} ,{6677,9826},{7525,10117},{754 2,7723},{7722,10018} = Complement of the complement of X(399).

= Midpoint of X(i) and X(j) for these {i,j}: {5,110},{113,1511},{125,5609} ,{549,5655},{550,7728},{6053,6 699}}. Reflection of X(i) in X(j) for these {i,j}: {{125,3628},{140,5972},{10113, 3850}.

= Crossdifference of every pair of points on line {2081,2433}.

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (113,5642,1511), (5972,6053,6699).

H013 = HATZIPOLAKIS - MOSES

Barycentrics a (3 a^5 b-3 a^4 b^2-6 a^3 b^3+6 a^2 b^4+3 a b^5-3 b^6+3 a^5 c-6 a^4 b c+8 a^3 b^2 c-11 a b^4 c+6 b^5 c-3 a^4 c^2+8 a^3 b c^2-16 a^2 b^2 c^2+8 a b^3 c^2+3 b^4 c^2-6 a^3 c^3+8 a b^2 c^3-12 b^3 c^3+6 a^2 c^4-11 a b c^4+3 b^2 c^4+3 a c^5+6 b c^5-3 c^6)::

(3 r + 2 R) X[1] - (3 r - R) X[3]. 2 X[942]-X[10247],X[3057]-4 X[5885],X[3576]+X[5903],3 X[10202]-2 X[10246],2 X[5]-5 X[4004],X[355]-4 X[10107],X[3817]-3 X[3919],5 X[3698]-2 X[5694],4 X[3754]-X[5887],2 X[3754]-X[10175],X[5887]-2 X[10175],7 X[3922]-4 X[9956],X[4018]+2 X[5690].

Let ABC be a triangle.

Denote:

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

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

N1, N2, N3 = the reflections of Na, Nb, Nc in B'C', C'A', A'B', resp.

The point is the centroid of N1N2N3 lying on the OI line of ABC.

(Antreas Hatzipolakis and Peter Moses, Oct. 13, 2016. See: Hyacinthos #24611)

The point lies on these lines: {1,3},{5,4004},{355,10107},{1 864,6797},{2800,3817},{3698,56 94},{3754,5887},{3922,9956},{ 4018,5690},{4323,6961},{4848,6 842},{5927,9952}

= Midpoint of X(3576) and X(5903).

= Reflection of X(i) in X(j) for these {i,j}: {{5887, 10175}, {10175, 3754}, {10247, 942}}.

H014 = HATZIPOLAKIS - LOZADA - MOSES

Barycentrics a^4 (a^12-4 a^10 b^2+5 a^8 b^4-5 a^4 b^8+4 a^2 b^10-b^12-4 a^10 c^2+9 a^8 b^2 c^2-5 a^6 b^4 c^2+a^4 b^6 c^2-3 a^2 b^8 c^2+2 b^10 c^2+5 a^8 c^4-5 a^6 b^2 c^4+2 a^4 b^4 c^4-a^2 b^6 c^4-b^8 c^4+a^4 b^2 c^6-a^2 b^4 c^6-5 a^4 c^8-3 a^2 b^2 c^8-b^4 c^8+4 a^2 c^10+2 b^2 c^10-c^12)::

Trilinears cos(2*A)*cos(3*A)-cos(4*A)* cos(B-C) ::

R^2*X(4)+(7*R^2-2*SW)*X(54)

2 X[54] + X[6759], 3 X[154] - X[9920].

X[4] + (J^2 - 2) X[54], (J = OH/R).

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. Denote: O', Oa, Ob, Oc = the circumcenters of A'B'C', PBC, PCA, PAB, resp. The reflections of O'Oa, O'Ob, O'Oc in BC, CA, AB, resp. are concurrent for P = O. The point is the point of concurrence for P = O

(Antreas P. Hatzipolakis, Peter Moses, Cesar Lozada, Oct. 8, 2016. See: Hyacinthos #24566 and #24574)

The point lies on these lines:{3,8157},{4,54},{49,52},{110, 2888},{154,9704},{156,9927},{ 182,6689},{206,576},{539, 10201},{569,6145},{1092,7691}, {1147,1154},{1209,6639},{1971, 9697},{2904,9707},{3518,7730}, {6288,10254},{9813,9827},{ 10182,10203}

= Midpoint of X(195) and X(2917)

= X(324)-Ceva conjugate of X(571).

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (54,3574,578).

H015 = HATZIPOLAKIS - MOSES

Barycentrics 14 a^16-103 a^14 b^2+335 a^12 b^4-633 a^10 b^6+765 a^8 b^8-609 a^6 b^10+313 a^4 b^12-95 a^2 b^14+13 b^16-103 a^14 c^2+454 a^12 b^2 c^2-729 a^10 b^4 c^2+342 a^8 b^6 c^2+459 a^6 b^8 c^2-786 a^4 b^10 c^2+469 a^2 b^12 c^2-106 b^14 c^2+335 a^12 c^4-729 a^10 b^2 c^4+360 a^8 b^4 c^4+15 a^6 b^6 c^4+480 a^4 b^8 c^4-837 a^2 b^10 c^4+376 b^12 c^4-633 a^10 c^6+342 a^8 b^2 c^6+15 a^6 b^4 c^6-14 a^4 b^6 c^6+463 a^2 b^8 c^6-758 b^10 c^6+765 a^8 c^8+459 a^6 b^2 c^8+480 a^4 b^4 c^8+463 a^2 b^6 c^8+950 b^8 c^8-609 a^6 c^10-786 a^4 b^2 c^10-837 a^2 b^4 c^10-758 b^6 c^10+313 a^4 c^12+469 a^2 b^2 c^12+376 b^4 c^12-95 a^2 c^14-106 b^2 c^14+13 c^16::

Let A be a triangle.

Denote:

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

Denote:

Nab, Nac = the orthogonal projections of Na on BNb, CNc, resp.

Nbc, Nba = the orthogonal projections of Nb on CNc, ANa, resp.

Nca, Ncb = the orthogonal projections of Nc on ANa, BNb, resp.

Let Oa, Ob, Oc be the circumcenters of NaNabNac, NbNbcNba, NcNcaNc, resp.

ABC, OaObOc are orthologic.

ABC, OaObOc are orthologic at X(1263)

(Antreas Hatzipolakis and Peter Moses, Sept. 30, 2016. See: Hyacinthos #24515)

The point is the orthologic center (OaObOc,ABC)

The point lies on these lines: {140, 930}.

= Midpoint of X(140), X(1487).

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},....

Τετάρτη, 22 Ιουνίου 2016

CONCENTRIC CIRCLES

Let ABC be a triangle.

Denote: A1, B1, C1 = The NPC centers of NBC, NCA, NAB, resp.

A2, B2 ,C2 = The NPC centers of A1BC, B1CA, C1AB, resp.

A3, B3, C3 = The NPC centers of A2BC, B2CA, C2AB

An, Bn, Cn = The NPC centers of A_n-1BC, B_n-1CA, C_n-1AB.

On = the circumcenter of the triangle AnBnCn.

The points O1, O2, O3, .... On,.... lie on the Euler line (see HERE)

Denote:

O11 = the reflection of O1 in BC

O111 = the reflection of O11 in AN

O112 = the reflection of O111 in BC

Similarly:

O12 = the reflection of O1 in CA

O121 = the reflection of O12 in BN

O122 = the reflection of O121 in CA

and

O13 = the reflection of O1 in AB

O131 = the reflection of O13 in CN

O132 = the reflection of O131 in AB

The circumcenter of O112O122O132 coincides with the cirumcenter of ABC

The same if we take O2:

O21 = the reflection of O2 in BC

O211 = the reflection of O21 in AN

O212 = the reflection of O211 in BC

Similarly:

O22 = the reflection of O2 in CA

O221 = the reflection of O22 in BN

O222 = the reflection of O221 in CA

and

O23 = the reflection of O2 in AB

O231 = the reflection of O23 in CN

O232 = the reflection of O231 in AB

The circumcenter of O212O222O232 coincides with the cirumcenter of ABC

and so on......

The circumcenter of On12On22On32 coincides with the cirumcenter of ABC

So we have a sequence of concentric circles.

Antreas P. Hatzipolakis, 22 June 2016

Παρασκευή, 3 Ιουνίου 2016

MAPPINGS OF THE EULER LINE ONTO ITSELF

NPC centers

1. Let ABC be a triangle and N1, N2, N3 = the NPC centers of OBC, OCA, OAB, resp.

The NPC center of N1N2N3 lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and M1, M2, M3 the midpoints of PN1, PN2, PN3, resp.

The NPC center of M1M2M3 lies on the Euler line of ABC. (ie it is the intersection of the Euer lines of ABC and M1M2M3).

2. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

The circumcenter of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc.)

3. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

The circumcenter of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Which other (than N,I) points have that property?

4. Let ABC be a triangle and Na, Nb, Nc = the NPC centers of GBC, GCA, GAB, resp.

The centroid of NaNbNc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The G (centroid) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Circumcenters

1. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of IBC, ICA, IAB, resp.

The circumcenter of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The O (circumcenter) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

2. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of NBC, NCA, NAB, resp.

The NPC center of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The NPC center of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

3. Let ABC be a triangle and Oa, Ob, Oc = the circumcenters of GBC, GCA, GAB, resp.

The centroid of OaObOc lies on the Euler line of ABC

Let P be a point on the Euler line of ABC and Ma, Mb, Mc the midpoints of PNa, PNb, PNc, resp.

The G (centroid) of MaMbMc lies on the Euler line of ABC. (ie it is the intersection of the Euler lines of ABC and MaMbMc).

Antreas P. Hatzipolakis, 3 June 2016

Σάββατο, 27 Δεκεμβρίου 2014

ANOPOLIS PRIMES

Pn = n + Qk, n is a natural number (including 0) ie n belongs to No = {0, 1, 2, 3, ....} with n > 0

where Qk is the smallest number in the set No - {Q1, Q2,.... Qk-1} such that Pn is prime

n = 1

Q1 = the smallest number in the set No - {Q0} = No - {0} such that 1 + Q1 is prime = 1

P1 = 1 + 1 = 2

n = 2

Q2 = the smallest number in the set No - {Q0, Q1} = No - {0, 1}such that 2 + Q2 is prime = 3

P2 = 2 + 3 = 5

n = 3

Q3 = the smallest number in the set No - {Q0, Q1, Q2} = No - {0, 1, 3} such that 3 + Q3 is prime = 2

P3 = 3 + 2 = 5

n = 4

Q4 = the smallest number in the set No - {Q0, Q1, Q2, Q3} = No - {0, 1, 3, 2} such that 4 + Q4 is prime = 7

P4 = 4 + 7 = 11

and so on.

Pn : 2, 5, 5, 11, 11, 11, 11, 17, 17, 23, 23, 23, 23, 29, 29, 37, 37, 37, 37, 37, 37, 47, 47, 47, 47, 53, 53, 59, 59, 59, 59, 59, 67, 67, 67, 67, 73, 73, 79, 79, 83, 83, 83, 83, 89, 89, 97, 97, 97, 97, 97, 97,.....

Anopolis Prime Numbers Sequence:

An = the missing primes from Pn: 3, 7, 13, 19, 31, 41, 43, 61, 71,.............................

Antreas P Hatzipolakis, 27 Dec. 2014

Παρασκευή, 3 Οκτωβρίου 2014

I - CONCURRENT EULER LINES

Let ABC be a triangle and A'B'C' the cevian triangle of P = I (incenter).

Denote:

Oab, Oac = the circumcenters of ABA', ACA', resp.

Oa = the circumcenter of AOabOac. Similarly Ob, Oc

1. The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent.

2. The Orthocenter of OaObOc is lying on the Euler Line of ABC.

Note:

For A'B'C' = cevian triangle of a point P, the circumcircles of AOabOac, BObcOba, COcaOcb are concurrent at O of ABC.

By Miquel theorem in the quadrilateral (AB,BC,CA,AA'), O is lying on the circumcircle of AOabOac. Similarly O is lying on the circumcircles of BObcOba, COcaOcb

Antreas P. Hatzipolakis, 3 October 2014

*** The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent in X(186) = Inverse in circumcircle of orthocenter.

*** The Orthocenter of OaObOc is

X=(a^2 (a^2 - b^2 - b c - c^2) (a^5 (b + c) - 2 a^3 (b^3 + c^3) - a^2 b c (b^2 + c^2) + a (b^5 - b^4 c - b c^4 + c^5)+ b c (b^2 - c^2)^2) : ... :...)

lying on the Euler Line of ABC, and with (6-9-13)-search number 1.269044824987441168348286504.

X = (r^2 + 2 r R - R^2 + s^2)X(3) + R^2 X(4)

Angel Montesdeoca. Hyacinthos #22604

1) X(186)

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

on lines {{2,3},{35,500},{55,5453},{511,5495},{3724,5492}},

Search = 3.7222371386671506172

Agree with barys, however I think the search number is suspect. Reckon it should be 3.7222371386671506172.

Peter Moses


 

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