Involutions from Concurrency of Three Algebraic Curves

Cosymmedian Involution on Miquelian? Benz Planes

Given a pair of cosymmedian triangles ABC and A'B'C' and a point P on a Miquelian Benz plane, the three cycles PAA' and PBB' and PCC' concur in a point P'. The transformation P↦P' is the cosymmedian involution. If P is the point at infinity on the classical Möbius plane, then P' is the symmedian point of both ABC and A'B'C'. The isodynamic points of both ABC and A'B'C' on the classical Möbius plane are the same. They are the images of the involution to each other. On the extended complex plane, P'=(A(B-C)((A-P)(B-C))*+B(C-A)((B-P)(C-A))*+C(A-B)((C-P)(A-B))*)/((B-C)((A-P)(B-C))*+(C-A)((B-P)(C-A))*+(A-B)((C-P)(A-B))*), where * is the complex conjugation.

Gibert-Burek-Moses(Incenter) Inversion on the Extended Complex Plane?

Given a quadrangle ABCD on the extended complex plane, we can get the common harmonic conjugation triple A₊A_- and B₊B_- and C₊C_-, as follows:

Common Harmonic Conjugation Triple A₊A_- and B₊B_- and C₊C_- of ABCD

Harmonic Conjugates	Formula
A+A- AD	A+=(AD-BC+sqrt((A-B)(D-C)(A-C)(D-B)))/((A+D)-(B+C))
A+A- BC	A-=(AD-BC-sqrt((A-B)(D-C)(A-C)(D-B)))/((A+D)-(B+C))
B+B- BD	B+=(BD-CA+sqrt((B-C)(D-A)(B-A)(D-C)))/((B+D)-(C+A))
B+B- CA	B-=(BD-CA-sqrt((B-C)(D-A)(B-A)(D-C)))/((B+D)-(C+A))
C+C- CD	C+=(CD-AB+sqrt((C-A)(D-B)(C-B)(D-A)))/((C+D)-(A+B))
C+C- AB	C-=(CD-AB-sqrt((C-A)(D-B)(C-B)(D-A)))/((C+D)-(A+B))

A₊A_- and B₊B_- and C₊C_- are a harmonic conjugation triple. The Gibert-Burek-Moses inversion is the harmonic conjugation triple involution. If D is the point at infinity, then the Gibert-Burek-Moses involution image D' of D is the incenter of ABC.

Harmonic Conjugation Triple Involution on Miquelian? Benz Planes

Given a harmonic conjugation triple A₊A_- and B₊B_- and C₊C_- on a Miquelian Benz plane, the radical involution from the three cycles B₊B_-C₊C_- and C₊C_-A₊A_- and A₊A_-B₊B_- is the harmonic conjugation triple involution.

Inversion with Respect to the Circumcircle of a Triangle on the Extended Complex Plane?

Given a circle c on a seperable Miquelian Möbius plane and a point P, there is the inverse P' of P with respect to c. Given a triangle ABC on the extended complex plane, we have the two harmonic conjugation triples AA₊,BB₊,CC₊ and AA_-,BB_-,CC_- of ABC such that AA₊,BB₊,CC₊ are harmonic conjugates of each other. So are AA_-,BB_-,CC_-.

Harmonic Conjugation Triples AA₊,BB₊,CC₊ AA_-,BB_-,CC_- of ABC

Harmonic Conjugates	Formula
BB+ CC+	A+=(Ai(B-C)+B(A-C)+C(A-B))/(i(B-C)+(A-C)+(A-B))
BB- CC-	A-=(A(-i(B-C))+B(A-C)+C(A-B))/(-i(B-C)+(A-C)+(A-B))
CC+ AA+	B+=(A(B-C)+Bi(C-A)+C(B-A))/((B-C)+i(C-A)+(B-A))
CC- AA-	B-=(A(B-C)+B(-i(C-A))+C(B-A))/((B-C)-i(C-A)+(B-A))
AA+ BB+	C+=(A(C-B)+B(C-A)+Ci(A-B))/((C-B)+(C-A)+i(A-B))
AA- BB-	C-=(A(C-B)+B(C-A)+C(-i(A-B)))/((C-B)+(C-A)-i(A-B))

Moreover, A₊B₊C₊ and A_-B_-C_- are inverses with respect to the circumcircle of ABC and quadrangles B₊B_-C₊C_- and C₊C_-A₊A_- and A₊A_-B₊B_- are concyclic. Given the circumcircle c of ABC, the inverse P' of P with respect to c is the radical involution image of P with respect to the three circles. That is, the three circles PA₊A_- and PB₊B_- and PC₊C_- concur in P'=(A(B-C)(AP+BC)*+B(C-A)(BP+CA)*+C(A-B)(CP+AB)*)/((B-C)(AP+BC)*+(C-A)(BP+CA)*+(A-B)(CP+AB)*), where * is the complex conjugation.

Isogonal Conjugation on Miquelian? Benz Planes

Given a quadrangle ABCD and a point P on a Miquelian Benz plane, we have three involutions I_A(X)=XAD∩XBC and I_B(X)=XBD∩XCA and I_C(X)=XCD∩XAB and can construct the isogonal conjugate P' of P with respect to ABCD by applying the three involutions in any order to P. That is, P'=I_A(I_B(I_C(P)))=I_C(I_A(I_B(P)))=... etc. If D is the point at infinity on the classical Möbius plane, then P' is the isogonal conjugate of P with respect to triangle ABC. If P is the point at infinity on the classical Möbius plane, P' is the isogonal center QA-P4 of quadrangle ABCD. Given a quadrangle ABCD and any point from the eight points A₊B₊C₊D₊ and A_-B_-C_-D_- on a Miquelian Benz plane, we can construct the other 7 points with the following table:

Isogonal Conjugates A₊B₊C₊D₊ and A_-B_-C_-D_- with Respect to ABCD Respectively

Consctrction of A-B-C-D- from A+B+C+D+ and ABCD					as Cyclologic Center
A-		=IC(B+)	=IB(C+)	=IA(D+)	=B+CD∩C+DB∩D+BC
B-	=IC(A+)		=IA(C+)	=IB(D+)	=C+DA∩D+AC∩A+CD
C-	=IB(A+)	=IA(B+)		=IC(D+)	=D+AB∩A+BD∩B+DA
D-	=IA(A+)	=IB(B+)	=IC(C+)		=A+BC∩B+CA∩C+AB
Consctrction of A+B+C+D+ from A-B-C-D- and ABCD					as Cyclologic Center
A+		=IC(B-)	=IB(C-)	=IA(D-)	=B-CD∩C-DB∩D-BC
B+	=IC(A-)		=IA(C-)	=IB(D-)	=C-DA∩D-AC∩A-CD
C+	=IB(A-)	=IA(B-)		=IC(D-)	=D-AB∩A-BD∩B-DA
D+	=IA(A-)	=IB(B-)	=IC(C-)		=A-BC∩B-CA∩C-AB

Moreover, A_-B_-C_-D_- and A₊B₊C₊D₊ are the mutual cyclologic centers of each other with respect to ABCD respectively. In fact, A₊B₊C₊D₊ are the Miquel conjugates of each other with respect to ABCD. So are A_-B_-C_-D_-.

Radical Involution on Ovoidal Benz Planes

Given three cycles abc and a point D₊ on an ovoidal Benz plane, we have the three pairs of points of intersection A₊A_-=b∩c and B₊B_-=c∩a and C₊C_-=a∩b. According to the bundle theorem, the three cycles D₊A₊A_- and D₊B₊B_- and D₊C₊C_- concur in a point D_-. The transformation D₊↦D_- is the radical involution. If D₊ is the point at infinity on the classical Möbius plane, D_- is the radical center of abc. Moreover, A₊A_- and B₊B_- and C₊C_- are image pairs of radical involutions.

Radical Involution Image Pairs A₊A_- and B₊B_- and C₊C_- and D₊D_-

Radical Involution Image Pair	Cycles
A+A-	(B+B-C+C-)(C+C-D+D-)(D+D-B+B-)
B+B-	(C+C-D+D-)(D+D-A+A-)(A+A-C+C-)
C+C-	(D+D-A+A-)(A+A-B+B-)(B+B-D+D-)
D+D-	(A+A-B+B-)(B+B-C+C-)(C+C-A+A-)

If A₁A₂ and B₁B₂ and C₁C₂ are pairs of points of intersection of cycle pairs (B₊B_-C₊C_-),(A₊A_-D₊D_-) and (C₊C_-A₊A_-),(B₊B_-D₊D_-) and (A₊A_-B₊B_-),(C₊C_-D₊D_-), respectively, then A₊A_- and B₊B_- and C₊C_- and D₊D_- are radical involution image pairs of cycles (B₁B₂C₁C₂)(C₁C₂A₁A₂)(A₁A₂B₁B₂)