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X(1), X(105), X(243), X(296), X(518), X(1155), X(1156), X(2651), X(2652), X(5205), X(57061), X(9432), X(12008) foci of the Mandart ellipse (see Generalized Mandart Conics in the Downloads page and also K351) foci of the Kellipse (inellipse with center K when the triangle ABC is acute angle) See at the bottom for many other centers on the curve and also Table 29 : QIxanticevian points. 

The Pelletier point X(650) is the perspector of the Feuerbach hyperbola. The Pelletier strophoid K040 is the isogonal circumstrophoid with root X(650). Its singular focus is X(105). It is a member of the class CL003 of cubics. The polar conic of X(105) is the circle passing through X(1), X(105), X(919), tangent at X(1) to the orthic line X(1)X(6). Its node is X(1) with tangents parallel to the asymptotes of the rectangular circumhyperbola centered at X(116), isogonal transform of the parallel at O to the line X(1)X(7). K040 also the Hirst transform of the Feuerbach hyperbola under the Hirst transformation with pole X(1) and conic the circumconic with perspector X(1), with center X(9). See CL030 for a generalization. K040 is the pedal curve with respect to X(1) of the parabola (P) with directrix X(1)X(6) and focus X(5540), the reflection of X(1) in X(105). (P) is the pivotal conic of K040. It is inscribed in the excentral triangle and it is tangent to the nodal tangents of K040. See figure below. K040 is spK(X518, X1) in CL055. K040 is the locus : 1. of foci of inscribed conics whose center lies on the line IK = X(1)X(6). See also K086, K165, K351 and Z+(O) = CL025. 2. of M such that M and its isogonal conjugate M* are conjugated with respect to the Feuerbach hyperbola. See Table 4. *** Let f : M = u:v:w > M' = a(a^2vw  bcu^2) / ((bc)(b+ca)) : : . f maps any point – except X(1) – on K040 to a point on the circumcircle. The reciprocal transformation g of f is defined by g : M = u:v:w > M' = [b^2c^2(bc)^2(b+ca)^2u^2  a^4(ca)(ab)(c+ab)(a+bc)vw]/a : : . g maps any point on the circumcircle to a point on K040. You may want to download a text file containing more than a hundred of centers on K040. 



Let m be a variable point on the line through X(1) and X(6). The perpendicular bisector (B) of {m, X(5540)} is tangent at M to (P). M lies on the perpendicular (m) at m to the line X(1)X(6). The perpendicular (B') at X(1) to (B) meets (B) at P which lies on K040 and (m) at N which lies on the rectangular hyperbola (H). The line PP* envelopes (P). (H) is the Apollonius hyperbola of X(1) with respect to (P) hence (P) and (H) meet at E such that the line X(1)E is normal to (P) at E. The center of (H) is the reflection of X(5540) about the line X(1)X(6). The tangent at E to (P) is also tangent to K040 and must pass through E* which is a point of inflexion on K040. The contact conic (C) also contains E and its isogonal transform (L) is the line passing through E*, X(5540), X(2820). 



K040 is an isogonal cubic in infinitely many triangles 

The results below are easily generalized for any (isogonal or not) circumstrophoid. 

The inverse of K040 in the (orange) circle with center X(1) passing through X(105) is the rectangular hyperbola (H) passing through X(1), X(105), the inverse of A, B, C. It is tangent at X(1) to the line X(1)X(6). (H) meets this circle at X(105) and three other points which are the vertices of a (light blue) equilateral triangle inscribed in K040. Let Q1 be a point on (H) different of X(1). The circle (C) with center M on the line X(1)X(6), passing through X(105) and Q1 meets the line X(1)Q1 again at T1. The perpendicular bisector of X(1)T1 meets (C) at Q2, Q3. The inverses R1, R2, R3 of Q1, Q2, Q3 are three points on K040 and K040 is an isogonal cubic cK with respect to the triangle R1R2R3. In particular, K040 is isogonal with respect to the equilateral triangle above. 

The inverses S1, S2, S3 of the antipodes of Q1, Q2, Q3 in (H) are three collinear points on the line (L) and on the sidelines of the triangle R1R2R3. When Q1 traverses (H), the line (L) passes through a fixed point S which is the intersection of the perpendicular at X(1) to the line X(1)X(6) and the perpendicular bisector of X(1)X(105). S is now X(14205) in ETC (20170830). Note that the position of the isogonal conjugate of a point on K040 is independent of the choice of the triangle R1R2R3.

