- [Antreas P. Hatzipolakis]:Let ABC be a triangle, A'B'C' the orthic triangle

and L a line passing through H.

Let A",B",C" be the reflections of A',B',C' in L, resp.

The circumcircles of the triangles:

1. A"BC, B"CA, C"AB

2. AB"C", BC"A", CA"B"

are concurrent.Reference: Anopolis #816 etc

[Seiichi Kirikami]:As for our communication of Anopolis #816 and others, I obtained the following results.[1] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of D’BC, E’CA, F’AB concur in X(115).[2] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of AE’F’, BF’D’, CD’E’ concur in X(935).[3] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of A’EF, B’FD, C’DE concur in X(112).[4] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of DB’C’, EC’A’, FA’B’ concur in a point P.Bricentrics of P: {(b^2-c^2)^2(b^2+c^2-2a^2)(b^4+c^4-a^4-b^2c^2); ; }.I replaced Euler line by a general line {l,m,n}. Mathematica produced the factor of (l(a^4-(b^2-c^2)^2)+m(b^4-(c^2-a^2)^2)+n(c^4-(a^2-b^2)^2) for the difference between the radical center and the radius of the circumcircle. So if the line passes the orthocenter of ABC, 3 circles concur in a point.Reference: email

- [Antreas P. Hatzipolakis]:Let ABC be a triangle, A'B'C' the orthic triangle

and L a line passing through H.

Let A",B",C" be the reflections of A',B',C' in L, resp.

The circumcircles of the triangles:

1. A"BC, B"CA, C"AB

2. AB"C", BC"A", CA"B"

are concurrent.Reference: Anopolis #816 etc

[Seiichi Kirikami]:As for our communication of Anopolis #816 and others, I obtained the following results.[1] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of D’BC, E’CA, F’AB concur in X(115).[2] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let D’, E’, F’ be the reflections of D, E, F in L respectively, the circumcircles of AE’F’, BF’D’, CD’E’ concur in X(935).[3] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of A’EF, B’FD, C’DE concur in X(112).[4] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of DB’C’, EC’A’, FA’B’ concur in a point P.Bricentrics of P: {(b^2-c^2)^2(b^2+c^2-2a^2)(b^4+c^4-a^4-b^2c^2); ; }.I replaced Euler line by a general line {l,m,n}. Mathematica produced the factor of (l(a^4-(b^2-c^2)^2)+m(b^4-(c^2-a^2)^2)+n(c^4-(a^2-b^2)^2) for the difference between the radical center and the radius of the circumcircle. So if the line passes the orthocenter of ABC, 3 circles concur in a point.Reference: email

[Peter Moses]:

[SK]That point is X(5099)[4] Given a triangle ABC, its Euler line L and its orthic triangle DEF, let A’, B’, C’ be the reflections of A, B, C in L respectively, the circumcircles of DB’C’, EC’A’, FA’B’ concur in a point P.Bricentrics of P: {(b^2-c^2)^2(b^2+c^2-2a^2)(b^ 4+c^4-a^4-b^2c^2); ; }.