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∑ (a^4 + b^2c^2) x (c^2 y^2  b^2 z^2) = 0 ∏ (b^2 x  c^2 y) + ∏ (b^2 z  c^2 x) = 0 

X(1), X(3), X(4), X(32), X(39), X(76), X(83), X(194), X(384), X(695), X(2896), X(3491) up to X(3503), X(14370) excenters, vertices of cevian triangle of X(384) vertices A1, B1, C1 and A3, B3, C3 of the first and third Brocard triangles reflections A1', B1', C1' of A1, B1, C1 in BC, CA, AB and their isogonal conjugates. These points lie on the cevians of X(83), X(39) respectively. A1, B1, C1 and A1', B1', C1' are the vertices of the Kiepert triangles with base angle Ω and Ω respectively. See Table 32 and also K797. 

K020 is the isogonal pK with pivot X(384) hence a member of the Euler pencil. See Table 27. Locus properties : Let A1B1C1 be the first Brocard triangle and M a variable point. The lines MA1, MB1, MC1 meet BC, CA, AB at Ma, Mb, Mc respectively. These three points are collinear if and only if M lies on K017. They form a triangle perspective to ABC if and only if M lies on K020. This is also true with the third Brocard triangle. In the case of the first Brocard triangle, the locus of the perspector is K322 and for the third Brocard triangle, the locus of the perspector is K532. Further details and figure below. See also a generalization at CL041. The isotomic transform of K020 is K743. The symbolic substitution SS{a > √a} transforms K020 into K132. 



K020 is a pK with pivot X(32) and isopivot X(3493) with respect to the first Brocard triangle. X(3493) is the tangential of X(32) in K020 and also the perspector of the first Brocard triangle and the anticevian triangle of X(694). 
K020 is a pK with pivot X(76) with respect to the third Brocard triangle. Its isopivot is X(8871) = X(76)', the tangential of X(76) in K020. X(8871) is also the perspector of the third Brocard triangle and the anticevian triangle of X(694). 



Let Q be a variable point on K020. Denote by BR1 = A1B1C1 and BR3 = A3B3C3 the 1st and 3rd Brocard triangles. • If P1a = QA1 /\ BC, P1b = QB1 /\ CA, P1c = QC1 /\ AB then P1aP1bP1c is the cevian triangle (with respect to ABC) of a point P1which lies on K322 and P1aP1bP1c, BR1 are perspective at Q. Recall that ABC and BR1 are perspective et X(76). • Similarly with BR3 instead of BR1, we find P3 on K532 such that the cevian triangle P3aP3bP3c of P3 is perspective at Q to BR3. Recall that ABC and BR3 are perspective at X(32). • Let P4 be the P3Ceva conjugate of Q and P2 the X(385)Ceva conjugate of P4. These points lie on K128 and they are collinear with X(694), the isopivot of K128, hence they are X(385)Ceva conjugate points since X(385) is the pivot of K128. The anticevian triangle of P2 (resp. P4) is perspective at Q to BR1 (resp. BR3). Note the following collinearities : Q, P1, P3, X(384) – Q, P2, X(32) – Q, P4, X(76) and X(385), P1, P2 – X(385), P3, P4. Also X(694), P2, P4 as already said. The following table gives the correspondences between Q and the four points P1, P2, P3, P4. 


Notes 1. when Q is the pivot X(384) of K020, the points P1, P2, P3, P4 are the pivots of their corresponding cubic (pink cells). These pivots X(385), X(694) lie on the three cubics. 2. two points Q on K020 collinear with X(695) – the isopivot of K020 – hence X(384)Ceva conjugates correspond to swapped X(385)Ceva conjugate points P2, P4 on K128. See for example the yellow and orange lines. 3. two isogonal conjugate points Q, Q* on K020 hence collinear with X(384) correspond to : • the pairs {P1, P1'} on K322 and {P3, P3'} on K532 also collinear with X(384). • P2 and P2' on K128 on a line passing through the isogonal conjugate of X(3505), a point obviously on K128. • P4 and P4' on K128 on a line passing through a fixed point on K128 with first barycentric coordinate : (a^4+b^2 c^2)/(a^6 b^6a^4 b^4 c^4+a^6 c^6b^6 c^6) and SEARCH = 3.67358153099718. This point is the isogonal conjugate of the X(385)Ceva conjugate of X(3505). 
