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TCS: P22

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  • Antreas Hatzipolakis
    [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,
    Message 1 of 2 , Sep 25 4:37 AM
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      [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

    • xpolakis
      [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,
      Message 2 of 2 , Sep 25 3:21 PM
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        [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]
        [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); ; }.
         
        That point is X(5099)
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