Let ABC be a triangle.
Denote:
Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.
N1N2N3 = the antireflection triangle of NaNbNc
ABC, N1N2N3 are orthologic.
[César Lozada]:
Replacing I with P, we could ask for the locus of P such that ABC, N1N2N3 are orthologic and the locus is the entire plane.
For P=x:y:z (barys), the orthologic centers are:
Q’(P) = ABC à N1N2N3 = x*((9*x+10*y+9*z)*y*SB-(x+y)*z*SC-(y+z)*x*SA)*((9*x+9*y+10*z)*z*SC-(x+z)*y*SB-(y+z)*x*SA) : :
Q”(P) = N1N2N3 à ABC = 3*(S^2-3*SB*SC)*x+(y+z)*(S^2-9*SB*SC) : :
If P lies on the circumcircle or P=H, then Q’(P) = H.
If P lies in the infinity then Q’(P)=Q”(P)=P.
Other ETC pairs (P, Q’(P)): No-ETC pairs (P=X(n), Q’(P) ) were found for other n<=5000.
Other ETC pairs (P=X(n), Q”(P)) (n<=5000): (3,382), (4,3627), (52,13474), (265,13202), (381,3543), (382,4), (399,10733), (1482,5691), (1657,3146), (3146,5), (3534,15682), (3543,15687), (3627,3853), (3830,3830)
Surprisingly, Q”(P) = homothecy( X(3830), P, 3/2) (The homothecy of X(3830) with center P and factor 3/2)
This fact surprises me because X(3830) =X(3543)com(Euler triangle) is a center created from algebraic abstractions and not from an euclidean geometric construction, ie., using points, lines, angles, reflections, etc.
Some related centers:
Q’(X(1)) = X(1)X(3845) ∩ X(8)X(4532)
= (5*a^2-c*a-5*b^2+5*c^2)*(5*a^2-b*a+5*b^2-5*c^2) : : (barys)
= lies on the Feuerbach hyperbola and these lines: {1, 3845}, {8, 4532}, {3583, 13606}
= [ -28.5047784985605800, -33.4133504417321300, 39.9290356332115000 ]
Q”(X(1)) = X(1)X(3830) ∩ X(3)X(7989)
= 6*a^4-(b+c)*a^3-2*(b^2-c*b+c^2)*a^2+(b^2-c^2)*(b-c)*a-4*(b^2-c^2)^2 : : (barys)
= X(1)-3*X(3830), 5*X(3)-7*X(7989), 3*X(3)-5*X(18492), 9*X(4)-5*X(3616), 7*X(4)-3*X(5731), 5*X(4)-3*X(5886), 13*X(4)-9*X(9779), 3*X(4)-2*X(9955), 3*X(4)-X(18481), 9*X(1385)-10*X(3616), 7*X(1385)-6*X(5731), 5*X(1385)-6*X(5886), 13*X(1385)-18*X(9779), 3*X(1385)-4*X(9955), 3*X(1385)-2*X(18481), 5*X(3616)-6*X(9955), 5*X(3616)-3*X(18481), 5*X(5731)-7*X(5886), 9*X(5731)-14*X(9955), 9*X(5731)-7*X(18481), 13*X(5886)-15*X(9779), 9*X(5886)-10*X(9955)
= lies on these lines: {1, 3830}, {3, 7989}, {4, 1385}, {5, 17502}, {8, 15682}, {10, 30}, {20, 9956}, {40, 5073}, {65, 16118}, {145, 3543}, {165, 17800}, {226, 15174}, {355, 3146}, {376, 19877}, {381, 3624}, {382, 517}, {388, 31795}, {515, 1483}, {516, 4746}, {519, 28645}, {546, 4297}, {548, 10175}, {549, 31253}, {550, 11231}, {551, 12101}, {942, 12943}, {944, 17578}, {946, 3853}, {1125, 3845}, {1319, 18514}, {1539, 11699}, {1657, 5587}, {1698, 3534}, {1699, 5076}, {2646, 18513}, {2771, 4018}, {3149, 23961}, {3529, 26446}, {3576, 3843}, {3583, 24928}, {3585, 5441}, {3586, 5045}, {3617, 15640}, {3634, 8703}, {3636, 15687}, {3648, 5086}, {3649, 10572}, {3656, 20057}, {3817, 3861}, {3822, 16160}, {3828, 19710}, {3850, 10165}, {3851, 7987}, {3857, 10171}, {3860, 19883}, {3922, 13145}, {4301, 28224}, {4316, 5442}, {4663, 11645}, {4668, 12702}, {4701, 28194}, {4743, 29040}, {5049, 9657}, {5059, 5818}, {5066, 19862}, {5122, 10826}, {5126, 10896}, {5251, 16117}, {5493, 28182}, {5586, 18541}, {5690, 28150}, {5698, 31672}, {5722, 31776}, {5901, 12102}, {5919, 26088}, {6583, 12680}, {6684, 15704}, {6851, 18516}, {7756, 31430}, {7967, 10248}, {9579, 31794}, {9613, 9668}, {9626, 14130}, {9778, 11541}, {9780, 11001}, {9957, 12953}, {10164, 12103}, {10172, 15712}, {11112, 25011}, {11113, 24564}, {11362, 28178}, {12262, 18383}, {12512, 31447}, {13369, 16616}, {13491, 31760}, {15685, 19875}, {15693, 19872}, {15695, 19876}, {15696, 31423}, {17609, 26089}, {18482, 25557}, {18526, 31162}, {18527, 18990}, {19541, 32612}, {22799, 22935}
= midpoint of X(i) and X(j) for these {i,j}: {40, 5073}, {355, 3146}, {382, 5691}, {9589, 12645}
= reflection of X(i) in X(j) for these (i,j): (20, 9956), (550, 19925), (551, 12101), (946, 3853), (1385, 4), (1657, 31663), (3579, 18480), (4297, 546), (5901, 12102), (11278, 12699), (11699, 1539), (12262, 18383), (12680, 6583), (13369, 16616), (13491, 31760), (15704, 6684), (18480, 31673), (18481, 9955), (19710, 3828), (22793, 3627), (22935, 22799), (24680, 22793), (31730, 18357)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (4, 18481, 9955), (9955, 18481, 1385)
= [ -15.5895745236797700, -16.8549582550649700, 22.5046699771123200 ]
Q’(X(2)) = X(98)X(15687) ∩ X(262)X(14269)
= (8*b^2-7*c^2-7*a^2)*(8*c^2-7*a^2-7*b^2) : :
= lies on the Kiepert hyperbola and these lines: {98, 15687}, {262, 14269}, {381, 11669}, {382, 7607}, {546, 7608}, {550, 10185}, {671, 3629}, {2996, 23334}, {5466, 32478}, {8352, 10302}, {10159, 33229}, {11668, 15681}, {33604, 33625}, {33605, 33623}
= isotomic conjugate of the anticomplement of X(20583)
= [ 68.7578054207365300, 85.1536734647325600, -87.0462504186319600 ]
Q”( X(2) ) = COMPLEMENT OF X(15685)
= 16*a^4-5*(b^2+c^2)*a^2-11*(b^2-c^2)^2 : :
= 11*X(2)-9*X(3), 5*X(2)-9*X(4), 8*X(2)-9*X(5), 17*X(2)-9*X(20), 19*X(2)-18*X(140), 13*X(2)-9*X(376), 7*X(2)-9*X(381), X(2)+9*X(382), 13*X(2)-18*X(546), 17*X(2)-18*X(547), 25*X(2)-18*X(548), 10*X(2)-9*X(549), 14*X(2)-9*X(550), 23*X(2)-9*X(1657), 7*X(2)+9*X(3146), 29*X(2)-9*X(3529), 5*X(2)-3*X(3534), X(2)-9*X(3543), 2*X(2)-9*X(3627), X(2)-3*X(3830)
= As a point on the Euler line, this center has Shinagawa coefficients (-5, 27)
= lies on these lines: {2, 3}, {302, 33610}, {303, 33611}, {395, 12817}, {396, 12816}, {519, 28645}, {1483, 31162}, {3654, 28178}, {3656, 28186}, {4654, 15935}, {4669, 28198}, {4745, 18480}, {4995, 18513}, {5298, 18514}, {5349, 16268}, {5350, 16267}, {5663, 21969}, {5690, 28202}, {5874, 13691}, {5875, 13810}, {7776, 32896}, {8584, 11645}, {10283, 28190}, {10386, 11237}, {10723, 14692}, {11648, 14075}, {12279, 16881}, {12290, 14449}, {12953, 15170}, {13391, 32062}, {13451, 15072}, {13607, 22793}, {13623, 14487}, {14537, 15048}, {14848, 14927}, {14915, 21849}, {15534, 31670}, {19116, 22615}, {19117, 22644}, {19924, 22165}, {22791, 28208}
= midpoint of X(i) and X(j) for these {i,j}: {4, 15684}, {376, 5073}, {381, 3146}, {382, 3543}, {3534, 15640}, {3830, 15682}
= reflection of X(i) in X(j) for these (i,j): (2, 12101), (3, 14893), (5, 15687), (20, 547), (376, 546), (381, 3853), (547, 12102), (549, 4), (550, 381), (1483, 31162), (3529, 15691), (3534, 5066), (3627, 3543), (3845, 3830), (8703, 3845), (11001, 12100), (12103, 11737), (15072, 13451), (15681, 140), (15683, 548), (15685, 15690), (15686, 5), (15687, 3627), (15690, 3860), (15691, 3850), (15704, 549), (19710, 2)
= anticomplement of X(15690)
= complement of X(15685)
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (4, 15640, 3534), (3830, 15640, 5066), (15640, 15682, 15684)
= [ -16.0591046651956000, -16.8862602644993600, 22.7430468181896000 ]
César Lozada
[APH]:
Antireflection triangle: I do not know if CL has already named it, but the triangle I mean is that one whose the vertices are the reflections of the feet of the altitudes in the respective vertices of the triangle, that is:
Let ABC be a triangle and A'B'C' the orthic triangle.
Let A", B", C" be the reflections of A', B', C' in A, B, C resp,.
A"B"C" is the antireflection triangle of ABC.
[César Lozada]:
I have arbitrarily adopted the following rules for naming triangles:
1) I use the prefix anti- when the triangles are mutually “inverse”. Example, T’ is the anti-triangle of T if T-of-T’ = T’-of-T = ABC. I have quoted the word “inverse”, but the coordinates matrices of such a pair of triangles in fact behave like inverse matrices.
2) I have reserved the word reflection for triangles made by reflecting some elements in some lines (most of times the sidelines of the reference triangle ABC).
3) When a triangle is built by reflecting some points in A, B, C, I thing a good term is symmetric, and when it is built by reflecting A,B,C in some points maybe a good name is counter-symmetric, again avoiding to use the prefix anti-, for which I haven`t found other synonym. Precisely, I am currently writing an article where I use these terms.
4) According to these “rules”, I would rename your “anti-reflection triangle” to “symmetric of orthic triangle” and your triangle N1N2N3 as “symmetric of NaNbNc”
Note: The anti-reflection triangle is an already taken name and a very old and not yet solved problem: To find the triangle whose reflection triangle is ABC.
Just some ideas attempting to order the lexicon.
César Lozada
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου