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X(2), X(6), X(13), X(14), X(15), X(16), X(17), X(18), X(30), X(550) X(42625) → X(42648) other points detailed below 

Geometric properties : 

Let f be the isoconjugation in the improper triangle T = X(2)X(6)X(30) having fixed points X(13), X(14), X(15), X(16). See Table 62 for other analogous isoconjugations. If P is a point different from the seven points already mentioned, the locus K(P) of M such that P, M, F(M) are collinear is a pK with respect to T which must pass through these seven points and also P and f(P). With P = X(74), we obtain the only circumcubic, namely K505. With P = X(111), we obtain the only circular cubic, namely K881. With P = X(550), we obtain K1191, the only cubic passing through X(17) and X(18), hence meeting the Evans conic at the six very familiar points X(13), X(14), X(15), X(16), X(17), X(18). See CL034 for related topics. P17 and P18 are f(X17) and f(X18) respectively. X(6) is a point of inflexion on the curve with inflexional tangent passing through X(631). The harmonic polar line of X(6) is the Euler line hence the tangents to K1191 at X(2), X(30), X(550) concur at X(6). It follows that one asymptote is the line X(6)X(30). The remaining infinite points are complicated. The tangents to K1191 at X(13), X(14), X(15), X(16) meet at X(550) since it is the pivot and the tangent at X(550) passes through the isopivot X(6), the tangential of X(550). With P = X(20), X(140), X(4), X(35018), X(376), X(397), X(398), X(5056), we obtain other pivotal KHOcubics namely K1192, K1193, K1207, K1218, K1219, K1221a, K1221b, K1222, respectively. *** Consider a triple (x,y,z) of real numbers and define the point F(x,y,z) = S / √3 x X(6) + 2y X(4) + z X(3) where S = 2 area(ABC). If x^2  2 y^2  (y  z)^2 = 0, then F(x,y,z) lies on the Evans conic. If x^2  3z (2y  z) = 0, then F(x,y,z) lies on the Kiepert hyperbola. If x^2 (2y + z)  3 (2y  z) (y + z)^2 = 0, then F(x,y,z) lies on K369. If x^2 (2y  3 z)  3 (y  z)^2 (2y  z) = 0, then F(x,y,z) lies on K458. If x^2 (4y  z)  (2y  z) (y + z) (3y + z) = 0, then F(x,y,z) lies on K1191. Remark : each 1st degree factor represents a line. For instance, x = 0 is the Euler line, 2y  z = 0 is the line GK, etc. Obviously, a missing x, y, z in a factor means the line passes through K, H, O respectively. More generally, a curve whose barycentric equation can be transformed into a condition independent of a, b, c (and S) will be said to be a KHOcurve and then, the condition on x, y, z will be called KHOequation of the curve. In the same manner, KHOpoints are those associated to a triple of real numbers. On some occasions, complex numbers are also to tbe considered : for instance, (i √3, 0, ±1) represent the imaginary foci of the Brocard ellipse. Further properties in CL075. When P is on the Euler line, the homothetic of H under h(O, T), then K(P) is always a KHOcubic with KHOequation : T (x + y + z) (x + y + z) (2y  z) + y (x^2  3y^2 + z^2  y z) = 0. The first two factors represent the lines X(14)X(16) and X(13)X(15). The last factor represents a conic passing through X(2), X(13), X(14), X(30), X(543). The equation above can be rewritten under the form : x^2 (y + 2 T y  T z)  (2 y  z) (y + z) (y + T y + T z) = 0. In this case, y + 2 T y  T z = 0 represents the inflexional tangent at X(6) which meets the Euler line at Q, the barycenter of (O, 1), (G, 3T). The last three factors represent three lines passing through X(6) and X(2), X(30), P respectively. These lines are the tangents to K(P) at these three points. Let Z be the complement of Q. The line X(6)Z is the polar line of P in the Evans conic and their two common points E1, E2 lie on K(P). They are their remaining common points. This generalizes the role of X(17), X(18) for the cubic K1191. For example with P = X(20) and P = X(140), these cubics are K1192 and K1193 respectively. In this latter cubic, E1 and E2 are X(3070) and X(3071). See also CL075 for other analogous cubics. *** More centers on K1191 with corresponding triples (x,y,z) : these are the third points on lines through two of the centers above. P17 = (5,6,3) = 10 a^2 S +3 Sqrt[3] (3 a^4a^2 b^22 b^4a^2 c^2+4 b^2 c^22 c^4) : : , SEARCH = 3.902587776484656 P18 = (5,6,3) = 10 a^2 S 3 Sqrt[3] (3 a^4a^2 b^22 b^4a^2 c^2+4 b^2 c^22 c^4) : : , SEARCH = 11.02164993880089 Q1 = (45,4,23) = 19 a^423 a^2 b^2+4 b^423 a^2 c^28 b^2 c^2+4 c^4+30 Sqrt[3] a^2 S : : , SEARCH = 1.074693164798429 Q2 = (45,4,23) = 19 a^423 a^2 b^2+4 b^423 a^2 c^28 b^2 c^2+4 c^430 Sqrt[3] a^2 S : : , SEARCH = 2.002201263625273 Q3 = (9,4,19) = 23 a^419 a^2 b^24 b^419 a^2 c^2+8 b^2 c^24 c^46 Sqrt[3] a^2 S : : , SEARCH = 5.444982415362372 Q4 = (9,4,19) = 23 a^419 a^2 b^24 b^419 a^2 c^2+8 b^2 c^24 c^4+6 Sqrt[3] a^2 S : : , SEARCH = 194.3189305674863 Q5 = (3,2,8) = Sqrt[3] a^2 S + (5 a^44 a^2 b^2b^44 a^2 c^2+2 b^2 c^2c^4) : : , SEARCH = 78.79635321375187 Q6 = (12,1,5) = 4 a^45 a^2 b^2+b^45 a^2 c^22 b^2 c^2+c^4+8 Sqrt[3] a^2 S : : , SEARCH = 0.4989323744221265 Q7 = (12,1,5) = 4 a^45 a^2 b^2+b^45 a^2 c^22 b^2 c^2+c^48 Sqrt[3] a^2 S : : , SEARCH = 1.820060958660224 Q8 = (3,2,8) = Sqrt[3] a^2 S  (5 a^44 a^2 b^2b^44 a^2 c^2+2 b^2 c^2c^4) : : , SEARCH = 6.303447823194458 Q9 = (2,3,2) = 4 a^2 S Sqrt[3] (5 a^42 a^2 b^23 b^42 a^2 c^2+6 b^2 c^23 c^4) : : , SEARCH = 13.68454696008417 Q10 = (4,3,5) = 8 a^2 S +Sqrt[3] (2 a^45 a^2 b^2+3 b^45 a^2 c^26 b^2 c^2+3 c^4) : : , SEARCH = 9.736488348800315 Q11 = (4,3,5) = 8 a^2 S Sqrt[3] (2 a^45 a^2 b^2+3 b^45 a^2 c^26 b^2 c^2+3 c^4) : : , SEARCH = 1.589708866177522 Q12 = (2,3,2) = 4 a^2 S +Sqrt[3] (5 a^42 a^2 b^23 b^42 a^2 c^2+6 b^2 c^23 c^4) : : , SEARCH = 6.043981456203585
The following list gives triples of collinear points on K1191. Naturally, the 3 x 3 determinant formed by the three associated triples must be 0.

X2 – X13 – X16 X2 – X14 – X15 X2 – X17 – Q1 X2 – X18 – Q2 X2 – X30 – X550 X2 – P17 – Q4 X2 – P18 – Q3 X2 – Q5 – Q12 X2 – Q6 – Q11 X2 – Q7 – Q10 X2 – Q8 – Q9 X6 – X13 – X14 X6 – X15 – X16 X6 – X17 – X18 X6 – P17 – P18 
X6 – Q1 – Q2 X6 – Q3 – Q4 X6 – Q5 – Q8 X6 – Q6 – Q7 X6 – Q9 – Q12 X6 – Q10 – Q11 X13 – X15 – X30 X13 – X17 – Q5 X13 – X18 – Q6 X13 – P17 – Q8 X13 – P18 – Q7 X13 – Q1 – Q10 X13 – Q2 – Q9 X13 – Q3 – Q12 X13 – Q4 – Q11 
X14 – X16 – X30 X14 – X17 – Q7 X14 – X18 – Q8 X14 – P17 – Q6 X14 – P18 – Q5 X14 – Q1 – Q12 X14 – Q2 – Q11 X14 – Q3 – Q10 X14 – Q4 – Q9 X15 – X17 – Q9 X15 – X18 – Q10 X15 – P17 – Q12 X15 – P18 – Q11 X15 – Q1 – Q6 X15 – Q2 – Q5 
X15 – Q3 – Q8 X15 – Q4 – Q7 X16 – X17 – Q11 X16 – X18 – Q12 X16 – P17 – Q10 X16 – P18 – Q9 X16 – Q1 – Q8 X16 – Q2 – Q7 X16 – Q3 – Q6 X16 – Q4 – Q5 X17 – X30 – Q3 X17 – X550 – P17 X18 – X30 – Q4 X18 – X550 – P18 X30 – P17 – Q2 
X30 – P18 – Q1 X30 – Q5 – Q10 X30 – Q6 – Q9 X30 – Q7 – Q12 X30 – Q8 – Q11 X550 – Q1 – Q4 X550 – Q2 – Q3 X550 – Q5 – Q6 X550 – Q7 – Q8 X550 – Q9 – Q10 X550 – Q11 – Q12 


