Saturday, July 27, 2013

TWO CENTERS CIRCLE – CEVIANS

Let P and Q be two points in the plane of ∆ABC.

The circle with diameter [PQ] cuts the line (AP) in P and AP and the line (AQ) in Q and AQ.

Build BP, BQ, CP, CQ cyclically.

RESULTS:

1) If Q=H (the orthocenter of 
ABC) then lines (APAQ), (BPBQ), (CPCQ) are concurrent.      

·        If P = u : v : w (trilinears) then the point of concurrence is:

ZH(P)= u*( v*cos(B) + w*cos(C))  :  :

·         If P is on the circumcircle of ∆ABC then ZH(P) lies on the line at infinity.
·         { P, ZH(P) } = { X(I), X(J) } for these (I,J):

(1,65), (2,6), (3,185), (4,4), (5,3574), (6,51), (7,1836), (8,1837), (9,1864), (10,1834), (13,5318), (14,5321), (17,397), (18,398), (19,1824), (21,1858), (24,52), (25,1843), (27,1839), (28,1829), (33,1827), (34,1828), (54,389), (57,2262), (69,1899), (75,3914), (76,5254), (90,1898), (92,1826), (93,1594), (98,1503), (99,3566), (105,3827), (106,2390), (107,523), (108,513), (110,924), (111,2393), (112,512), (158,225), (186,1986), (225,407), (226,1901), (250,1112), (253,1853), (254,3), (262,5480), (264,427), (270,1831), (277,2082), (278,19), (280,1854),  (281,33), (308,1194), (330,3959), (393,25), (403,113), (451,2906), (459,393),  (468,5095), (477,2777), (485,3070), (486,3071), (514,1146), (523,125), (525,1562), (648,2501), (650,3270), (653,3064), (733,3852), (842,2781), (847,5), (915,517), (917,516), (933,1510), (935,690), (953,2818), (1061,1905), (1068,46), (1093,235), (1105,1885), (1113,2574), (1114,2575), (1119,1851), (1123,2362), (1138,74), (1166,973), (1172,1859), (1217,1593), (1289,525), (1300,30), (1301,520), (1304,526), (1309,900), (1785,1846), (1826,430), (1870,1845), (1989,1495), (2006,2182), (2052,53), (2165,184), (2184,1903), (2190,1825), (2374,524), (2383,1154), (2687,2778), (2998,3981), (3144,2907), (3296,4295), (3316,1587), (3317,1588), (3346,64), (3459,54), (3542,155), (3563,511), (4213,2905)  

2) For others Q<>H, it seems that if P lies on a hyperbola depending on Q then lines (APAQ), (BPBQ), (CPCQ) concur. So, considering the first 10 ETC-centers, they concur if:

·         If Q is the incenter X(1),  the Gergonne point X(7), the Nagel point X(8) or the Mittenpunkt X(9) and P lies on the Feuerbach hyperbola; 
·        If Q is the centroid X(2) or the Spieker center and P lies on Kiepert hyperbola;
·        If Q is circumcenter X(3) or the Symmedian point X(6) and P lies on Jerabek hyperbola.
·        If Q is the NPC=N (X(5))  concurrence occurs if P lies on the unnamed hyperbola:
(SB-SC)*(S^2+SB*SC)/(a*u) = 0
which has center X(137) and passes through X(4), X(5), X(53), X(311), X(327), X(1141), X(1263), X(1487), X(2165), X(2980),  X(3459), X(3613).
If P lies on this last conic then ZN(P) lies on line (5,51).
{ P, ZN(P) } = { X(I), X(J) } for these (I,J): (4,3574), (53,51), (1263,1154)
ZN( X(311) ) = (b^2+c^2)*(a^2*(b^2+c^2)-(b^2-c^2)^2)/a : : = (2,98)∩(5,51) = complement of X(5012)

1 comment:

  1. César, congratulations and best wishes for your blog!

    Given a point Q, the locus of P such that ApAq, BpBq and CpCq are concurrent is the rectangular hyperbola through Q.

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