Home page | Catalogue | Classes | Tables | Glossary | Notations | Links | Bibliography | Thanks | Downloads | Related Curves

Written from an idea by Wilson Stothers and with his help

The Ix-anticevian points

Let P be a point and PaPbPc its anticevian triangle. We say that P is a Ix-anticevian point if and only if P is an in/excenter of PaPbPc. This is generalized in : Table 28 : cevian and anticevian points. See the related paper Two Related Transformations and Associated Cubics.

The isogonal conjugates of these Ix-anticevian points are also interesting since they lie on many remarkable curves. See below.

An Ix-anticevian point P must lie on a bisector of the lines PaPb, PaPc (i.e. the reflection of B in AP must lie on PaPb) hence P lies on the nodal cubic K(A) with equation : x(c^2 y^2 - b^2 z^2) + 2(SC z - SB y)yz = 0.

the nodal tangents at A are the bisectors of ABC.

the tangents at B and C to K(A) meet at the point Xa = 2SB SC : b^2 SB : c^2 SC, on the cubic and on the line AO.

K(A) contains :

• Ha foot of the A-altitude,
• Ga midpoint of the A-median,
• E1, E2 intersections of the circumcircle with the perpendicular at Ga to BC,
• Ka = -2SA : b^2 : c^2 on the symmedian AK.

K(A) is the locus of P such that A, P and the isogonal conjugate P' of P with respect to the anticevian triangle of P are collinear.

With P = x : y : z, P' is the point with 1st barycentric coordinate : x (- SA x + SB y + SC z) - a^2 y z.

Two other cubics K(B), K(C) are defined similarly, thus P is a Ix-anticevian point if and only if it lies on the three cubics and consequently on any cubic of the net of circumcubics generated by them. If Q = u:v:w is a point, we write the net under the form K(Q) = u K(A) + v K(B) + w K(C). K(Q) is always a K0 (without term in x y z) and always contains Q.

K(Q) is the locus of P such that Q, P and the isogonal conjugate P' of P with respect to the anticevian triangle of P are collinear. See at the bottom of this page for other cubics related to K(Q).

Since any point on K(A) and K(B) must lie on K(C) and since K(A) and K(B) have already five common points, we see that there are four Ix-anticevian points although not neccessarily all real.

The figure below shows a triangle ABC with two real points.

Here the three conics C(A), C(B), C(C) meet at two real points (and two other which are imaginary).

Each point P is represented with its anticevian triangle and the corresponding in/excircle with center P.

Another figure with four real points is given in the section concerning isogonal conjugates of these points.

(H) is the rectangular hyperbola described below.

The net contains three decomposed cubics obtained when Q is a vertex of the reflection triangle.

For example, if A' is the reflection of A in BC, the cubic K(A') is the union of the line BC and the conic C(A) with equation :

(4SB^2 – a^2c^2) y^2 – (4SC^2 – a^2b^2) z^2 = [(a^2 – c^2)^2 – b^2(a^2 + c^2)] xz – [(a^2 – b^2)^2 – c^2(a^2 + b^2)] xy

C(A) contains the five following points :

A with a tangent through X(5),

A' (reflection of A in BC) with a tangent through X(382), the reflection of O in H,

A3 = AX(1141) /\ A'X(30) where X(1141) is the second intersection of the line X(5)X(110) with the circumcircle,

B' = BA3 /\ AC,

C' = CA3 /\ AB.

Note that the tangents at B', C' pass through the foot of the trilinear polar of X(5) on BC.

C(A), C(B), C(C) belong to a same pencil of conics passing through the Ix-anticevian points. This gives the construction of these points.

This pencil contains one rectangular hyperbola passing through X(5), X(6), X(52), X(195), X(265), X(382), X(2574), X(2575). Its asymptotes are parallel to those of the Jerabek hyperbola. See the red curve above.

Recall that K(Q) is a circum-cubic passing through Q and the Ix-anticevian points.

K(Q) meets the line at infinity at the same points as pK(X6, cQ) where cQ is the complement of Q.

K(Q) meets the circumcircle (O) at the same points as pK(X6, S) where S is the point with barycentrics :

a^2 (2 b^2 c^2 u+a^2 c^2 v-c^4 v+a^2 b^2 w-b^4 w) : : .

S lies on the line passing through X(6) and the complement of the barycentric quotient Q ÷ X(251), also on the line passing through Q parallel to the trilinear polar of X(110) ÷ Q. S is the homothetic of the centroid of the pedal triangle of Q under h(ccQ, 3).

These three cubics coincide in K005 if and only if Q = X(3).

When Q ≠ X(3), denote by Z the infinite point of the perpendiculars to the line QX(3).

K(Q) and pK(X6, S) meet at three other points lying on the line (L) passing through X(6), with trilinear pole the isogonal conjugate of Z.

K(Q) and pK(X6, cQ) meet at six other points lying on the circum-conic (C) passing through X(2), with perspector Z.

Special cubics of the net

This net of circumcubics through the Ix-anticevian points contains a large number of remarkable cubics.

Circular cubics

K(Q) is circular if and only if Q lies at infinity.

In this case, Q lies on K(Q) and the singular focus is a point of the circumcircle of the antimedial triangle i.e. the circle centered at H with radius 2R.

The real asymptote envelopes a deltoid homothetic of the Steiner deltoid H3.

Obviously, these circular cubics form a pencil of cubics passing through the Ix-anticevian points.

The most interesting is obtained when Q is X(30), the infinite point of the Euler line, the cubic is K060 = Kn.

The table below shows a selection of such cubics.

Equilateral cubic

K(Q) is equilateral if and only if Q = H. The cubic is K049, the McCay orthic cubic.

K+ cubics

K(Q) is a K+ (i.e. has three concurring asymptotes) if and only if Q lies on a non-circumcubic K60+ passing through H, X(550), X(3146).

The asymptotes concur at a point X lying on the line HX(51) but not on the cubic. They are parallel to those of the McCay cubic.

When Q = H, we find K049, the McCay orthic cubic, and when Q = X(550), the cubic is K123 = D(-1/2).

The third point on the Euler line and on the cubic is X(3146), the homothetic of H under h(O,3). This gives another K+ which is K127 = D(3).

See below for more on these cubics D(k).

When Q is any point on the Euler line, K(Q) passes through H and X(5), the nine point center. Hence all these cubics form a pencil of cubics generated by K(O) = K005 (Napoleon cubic) and K(X30) = K060 (Kn cubic). This is precisely the pencil of D(k) cubics we met in "Two Remarkable Pencils..." to be found in the Downloads page. Here is a selection of these cubics given Q or k, the abscissa of Q in (O,H).

K(Q) meets the sidelines of ABC at three points U, V, W. Since K(Q) is a K0, it is a pK if and only if the triangles ABC and UVW are perspective and a nK0 if and only if the points U, V, W are collinear. This gives the two following results.

K(Q) is a pK if and only if Q lies on the Neuberg cubic K001. In this case

• the pole W lies on the cubic Co = K095,
• the pivot P lies on Kn = K060,
• the isopivot P* lies on the Napoleon cubic K005.

Here is a selection of these cubics (a SEARCH number is given when the point is not in ETC).

K(Q) is a nK0 if and only if Q lies on K216, a nK(X6, X5, ?) without any known center on it.

K(Q) passing through a given point Z

Let Z be a point distinct of A, B, C which is not a Ix-anticevian point. Any K(Q) passing through Z contains eight known points and must pass through a nineth point Z' which only depends of Z.

For example, when Z = H we have Z' = X(5) and when Z = X(61) we have Z' = X(62). This gives a conjugation studied below.

A related conjugation

Wilson Stothers comments about the conjugation Z -> Z' seen above. Suppose we write the equation of K(A) as KA = 0, and that of C(A) as CA = 0 then define KB, KC, CB, CC by symmetry.

The conjugation is then given by x:y:z maps to CA/KA : CB/KB : CC/KC.

Some conjugate pairs are :{X1,X3336}, {X2,X3629}, {X3,X3470}, {X4,X5}, {X6,X251}, {X54,X195}, {X61,X62}, {X79,X79}.

The set of fixed points is actually the sextic S(Ix).

The sextic S(Ix) contains :

the double points A, B, C - the tangents at A are the bisectors, as for K(A),

the four Ix-anticevian points

X(79) and its extraversions.

the six traces of the points fixed by X(1989)-isoconjugation - those A1, A2 on BC are also on C(A).

***

Wilson notes a similarity in Table 11 and Table 23 which are reversing the roles of H and Ix. The analogous sextic is S(H).

Higher degree curves through the Ix-anticevian points

The table gives several curves passing through these points.

Isogonal conjugates of the Ix-anticevian points

Let P be an Ix-anticevian point and Q = P* its isogonal conjugate.

Q must lie on the isogonal transform of the nodal cubic K(A) which is a conic K(A)* passing through A (with tangent through O).

Two other conics K(B)*, K(C)* are defined likewise.

These conics K(A)*, K(B)*, K(C)* are in a same pencil of conics that contains one rectangular hyperbola (H) which is the complement of the Jerabek hyperbola.

(H) is also the bicevian conic C(X2, X110) hence the G-Ceva conjugate of the Brocard axis.

(H) contains X(3), X(5), X(6), X(113), X(141), X(206), X(942), X(960), X(1147), X(1209), X(1493), X(1511), X(2574), X(2575), X(2883).

All these conics must contain the isogonal conjugates Qi of the Ix-anticevian points. Note that these are not always real.

Let S = u : v : w be any point of the plane and let K(S) = u x K(A)* + v y K(B)* + w z K(C)*.

K(S) is obviously a circum-cubic passing through the points Qi and all these cubics are in a same net of cubics and the isogonal transform K(S)* of K(S) is a circum-cubic passing through the Ix-anticevian points Pi.

Naturally, the Napoleon cubic K005 contains all points Pi and Qi.

K(S) is a pivotal cubic if and only if S lies on K276, the isotomic transform of the Neuberg cubic K001.

The following table gathers together a good selection of cubics passing through these points Pi and Qi.

Recall that K(S)* is a pK for any S on the Neuberg cubic K001. Consequently, K(S) is a pK for any S on K276, the isotomic transform of K001.

a figure with four real Ix-anticevian points Pi and their isogonal conjugates Qi

All these points lie on the Napoleon cubic K005 and X(5), Pi, Qi are collinear.

The two hyperbolas (H) are those mentioned above, green through the Ix-anticevian points and blue through their isogonal conjugates.

The pedal triangles of Q2, Q3 are also represented and their nine point centers are Q2, Q3. This is naturally true for the other points Q1, Q4.

In other words, if Pi is an Ix-anticevian point then its isogonal conjugate Qi is a X(5)-pedal point. See Table 45.

Higher degree curves through the isogonal conjugates of the Ix-anticevian points

The table gives several curves passing through these points.

Other cubics related to K(Q)

Recall that, for a given point Q = u : v : w, K(Q) is the locus of P = x : y : z such that Q, P, and P' are collinear.

P' is the isogonal conjugate of P with respect to the anticevian triangle of P with barycentric coordinates :

x (- SA x + SB y + SC z) - a^2 y z : y (SA x - SB y + SC z) - b^2 z x : z (SA x + SB y - SC z) - c^2 x y.

Denote by :

• P* = a^2 y z : : , the isogonal conjugate of P,

• ocP = x (- SA x + SB y + SC z) + a^2 y z : : , the orthocorrespondent of P,

• H/P = x (- SA x + SB y + SC z) : : , the H-Ceva conjugate of P.

The four points P', P*, ocP, H/P are obviously collinear and (P*, H/P, ocP, P') = -1. Each point X corresponds to a cubic, the locus of P such that Q, P and X are collinear. More precisely :

• when X = P*, the cubic is pK1 = pK(X6, Q),

• when X = H/P, the cubic is pK2 = pK(H x Q, H),

• when X = ocP, the cubic is the orthopivotal cubic O(Q),

• when X = P', the cubic is K(Q).

These four cubics belong to a same pencil and all pass through A, B, C, Q. Hence they must meet at five other points Qi depending of Q.

Note that this pencil contains a third (rather complicated) pK3 and three (very complicated) nK0s, one of them being always real. Recall that O(Q) is the circular cubic of the pencil which also contains an equilateral cubic when Q lies on the Euler line. Naturally, two of the mentioned cubics might coincide.

These points Qi lie on Q003. Indeed, the triads Q, Qi, Qi* and Q, Qi, H/Qi are made of collinear points hence Qi, Qi*, H/Qi are also collinear. This is property 5 in Q003.

When Q lies on the Euler line, one of the five points Qi is H, hence pK1 and pK2 meet again at four points on Q003. This is discussed in the page Q003.