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catalogue

name

R

circular

nodal

K+

K60

K016

Tucker+

X(2)

 

 

YES

 

K093

isotomic equilateral nK0

2a^2(b^2+c^2)-b^2c^2 : :

 

 

 

YES

K149

 

X(6)

 

 

 

 

K185

Hirst-Kiepert

X(523)

 

YES

 

 

K197

isotomic circular nK0

a^4+b^4+c^4-3b^2c^2 : :

YES

 

 

 

K296

 

X(514)

 

YES

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Given a point R = u : v : w (the root of the cubic), the isotomic nK0(X2, R) has barycentric equation :

u x (y^2 + z^2) = 0.

It always passes through the intercepts U, V, W of the trilinear polar of R on the sidelines of ABC. The points A and U, B and V, C and W share the same tangentials A', B', C' respectively which are three collinear points on the cubic.

 

Circular nK0(X2, R)

nK0(X2, R) is circular if and only if R = a^4 + b^4 + c^4 - 3b^2c^2 : : , intersection of the lines X(2)X(76) and X(115)X(316). This is the cubic K197.

 

Equilateral nK0(X2, R)

nK0(X2, R) is equilateral if and only if R = 2a^2(b^2 + c^2) - b^2c^2 : : , reflection of X(76) in X(2). This is the cubic K198.

 

nK0+(X2, R)

nK0(X2, R) has concurring asymptotes if and only if R lies on the cubic which is the reflection of K015 (Tucker nodal cubic) in G. In particular, with R = G, we obtain nK0+(X2, X2) = K016 = Tucker+ cubic.

 

Nodal nK0(X2, R)

nK0(X2, R) is a nodal cubic if and only if it contains one and only one of the points G, Ga, Gb, Gc (vertices of the antimedial triangle) which is therefore the node. If we take the centroid G = X(2), R must lie on the line at infinity.

When R = X(523), we obtain K185 the only such cubic with perpendicular nodal tangents.