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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 circum-cubic, 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 KHO-cubics namely K1192, K1193, K1207, K1218, K1219, K1221a, K1221b, K1222, respectively.

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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 KHO-curve and then, the condition on x, y, z will be called KHO-equation of the curve. In the same manner, KHO-points 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 KHO-cubic with KHO-equation :

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^4-a^2 b^2-2 b^4-a^2 c^2+4 b^2 c^2-2 c^4) : : , SEARCH = -3.902587776484656

P18 = (5,-6,3) = 10 a^2 S -3 Sqrt[3] (3 a^4-a^2 b^2-2 b^4-a^2 c^2+4 b^2 c^2-2 c^4) : : , SEARCH = 11.02164993880089

Q1 = (-45,4,23) = 19 a^4-23 a^2 b^2+4 b^4-23 a^2 c^2-8 b^2 c^2+4 c^4+30 Sqrt[3] a^2 S : : , SEARCH = -1.074693164798429

Q2 = (45,4,23) = 19 a^4-23 a^2 b^2+4 b^4-23 a^2 c^2-8 b^2 c^2+4 c^4-30 Sqrt[3] a^2 S : : , SEARCH = 2.002201263625273

Q3 = (9,-4,19) = 23 a^4-19 a^2 b^2-4 b^4-19 a^2 c^2+8 b^2 c^2-4 c^4-6 Sqrt[3] a^2 S : : , SEARCH = 5.444982415362372

Q4 = (9,4,-19) = 23 a^4-19 a^2 b^2-4 b^4-19 a^2 c^2+8 b^2 c^2-4 c^4+6 Sqrt[3] a^2 S : : , SEARCH = -194.3189305674863

Q5 = (3,2,-8) = Sqrt[3] a^2 S + (5 a^4-4 a^2 b^2-b^4-4 a^2 c^2+2 b^2 c^2-c^4) : : , SEARCH = 78.79635321375187

Q6 = (-12,1,5) = 4 a^4-5 a^2 b^2+b^4-5 a^2 c^2-2 b^2 c^2+c^4+8 Sqrt[3] a^2 S : : , SEARCH = -0.4989323744221265

Q7 = (12,1,5) = 4 a^4-5 a^2 b^2+b^4-5 a^2 c^2-2 b^2 c^2+c^4-8 Sqrt[3] a^2 S : : , SEARCH = 1.820060958660224

Q8 = (3,-2,8) = Sqrt[3] a^2 S - (5 a^4-4 a^2 b^2-b^4-4 a^2 c^2+2 b^2 c^2-c^4) : : , SEARCH = 6.303447823194458

Q9 = (2,-3,2) = 4 a^2 S -Sqrt[3] (5 a^4-2 a^2 b^2-3 b^4-2 a^2 c^2+6 b^2 c^2-3 c^4) : : , SEARCH = 13.68454696008417

Q10 = (-4,3,5) = 8 a^2 S +Sqrt[3] (2 a^4-5 a^2 b^2+3 b^4-5 a^2 c^2-6 b^2 c^2+3 c^4) : : , SEARCH = 9.736488348800315

Q11 = (4,3,5) = 8 a^2 S -Sqrt[3] (2 a^4-5 a^2 b^2+3 b^4-5 a^2 c^2-6 b^2 c^2+3 c^4) : : , SEARCH = 1.589708866177522

Q12 = (2,3,-2) = 4 a^2 S +Sqrt[3] (5 a^4-2 a^2 b^2-3 b^4-2 a^2 c^2+6 b^2 c^2-3 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