Loading ...
Sorry, an error occurred while loading the content.

Two Apollonius configurations

Expand Messages
  • pldx1
    Dear Friends, First configuration. Let (XYZ) be the circumcircle of triangle XYZ. The Apollonius circles relative to (ABH), (BCH), (CAH), i.e. the circles
    Message 1 of 3 , Oct 1, 2009
    • 0 Attachment
      Dear Friends,

      First configuration.
      Let (XYZ) be the circumcircle of triangle XYZ. The Apollonius circles relative to (ABH), (BCH), (CAH), i.e. the circles tangent to all of the three given circles, are (H,0) four times, (H,2R) once and three other circles, Ta,Tb,Tc.

      Circles Ta,Tb,Tc are ever external to each other, and their common orthogonal circle To is real. Condition of (external) tangency is : (a^2-b^2)^2-(a^2+b^2)c^2=0 (etc) or ABC rectangular.

      Second configuration.
      The Apollonius circles of Ta, Tb, Tc are (ABH) etc, their inverses in To and two others. Center X=x:y:z and radius omega of the first one are :

      x=(b^2+c^2-a^2)(a^8-2(b^2+c^2)a^6+2(b^4-b^2 c^2+c^4)a^4 -2(b^2+c^2)(b^2-c^2)^2 a^2+(b^2-c^2)^4)

      omega=2*(a^2 b^2 c^2 R)/(a^6+b^6+c^6-a^2 b^4-a^4 b^2-c^2 b^4-a^4 c^2-b^2 c^4-c^4 a^2+4 a^2 b^2 c^2)

      while the second is less simple.

      Moreover, X lies on X(3)X(68) and on X(4)X(94)
    • Quang Tuan Bui
      Dear Friend, X lies on line X(i)X(j) for following (i,j): 2,49 3,68 4,94 66,1351 70,1993 125,1147 155,2072 182,1209 539,1092 631,2888
      Message 2 of 3 , Oct 2, 2009
      • 0 Attachment
        Dear Friend,

        X lies on line X(i)X(j) for following (i,j):

        2,49 3,68 4,94 66,1351 70,1993 125,1147 155,2072 182,1209 539,1092 631,2888 1352,1656

        omega = 2*R^3/(R^2 + OH^2)

        Best regards,
        Bui Quang Tuan

        --- On Fri, 10/2/09, pldx1 <pldx1@...> wrote:

        > From: pldx1 <pldx1@...>
        > Subject: [EMHL] Two Apollonius configurations
        > To: Hyacinthos@yahoogroups.com
        > Date: Friday, October 2, 2009, 1:14 AM
        > Dear Friends,
        >
        > Center X=x:y:z and radius
        > omega of the first one are :
        >
        > x=(b^2+c^2-a^2)(a^8-2(b^2+c^2)a^6+2(b^4-b^2 c^2+c^4)a^4
        > -2(b^2+c^2)(b^2-c^2)^2 a^2+(b^2-c^2)^4)
        >
        > omega=2*(a^2 b^2 c^2 R)/(a^6+b^6+c^6-a^2 b^4-a^4 b^2-c^2
        > b^4-a^4 c^2-b^2 c^4-c^4 a^2+4 a^2 b^2 c^2)
        >
        > while the second is less simple.
        >
        > Moreover, X lies on X(3)X(68) and on X(4)X(94)
        >
      • ndergiades
        Dear Friend, your first configuration can be generalized as The Apollonius circles of three concurrent circles . Let three circles (Oa), (Ob), (Oc) be
        Message 3 of 3 , Oct 3, 2009
        • 0 Attachment
          Dear Friend,
          your first configuration can be generalized as
          "The Apollonius circles of three concurrent circles".
          Let three circles (Oa), (Ob), (Oc) be concurrent at P,
          let A be the second intersection of (Ob), (Oc)
          let B be the second intersection of (Oc), (Oa)
          let C be the second intersection of (Oa), (Ob).
          Let A'B'C' be the circumcevian triangle of P
          relative to ABC and k be the power of P relative to
          the circumcircle (ABC).
          In the inversion (P, k) with pole P and power k
          the inverse of the circle (BPC)=(Oa) is the line B'C',
          the inverse of the circle (CPA)=(Ob) is the line C'A',
          the inverse of the circle (APB)=(Oc) is the line A'B'.
          Hence if AoBoCo is the intouch triangle of A'B'C'
          and the line PAo meets the circle (AA'Ao) at A'o
          the inverse of Ao in the inversion (P, k)
          this point lies also on the circle (Oa)
          and from this point pass also the circles
          (BB'Ao), (CC'Ao).
          Similarly define the points B'o, C'o on the circles
          (Ob), (Oc) and they are the inverse points of Bo, Co.
          Hence the inverse (To) of the incircle of A'B'C'
          is the circle passing through A'o, B'o, C'o
          and tangent to (Oa), (Ob), (Oc) at the above
          points. Similarly working with the excircles of A'B'C'
          and their tangency points with the sides of triangle
          A'B'C' the points (Aa, Ba, Ca), (Ab, Bb, Cb), (Ac, Bc, Cc)
          we construct other three circles (Ta), (Tb), (Tc)
          tangent to the given circles (Oa), (Ob), (Oc).
          It is obvious that the other 4 Apollonius solutions
          coincide with the point P.

          Sinse the NPC of A'B'C' is tangent to incircle and
          excircles of A'B'C' the inverse of NPC of A'B'C'
          is a circle that is tangent to (To), (Ta), (Tb), (Tc).
          Best regards
          Nikos Dergiades

          >
          > Dear Friends,
          >
          > First configuration.
          > Let (XYZ) be the circumcircle of triangle XYZ. The Apollonius circles relative to (ABH), (BCH), (CAH), i.e. the circles tangent to all of the three given circles, are (H,0) four times, (H,2R) once and three other circles, Ta,Tb,Tc.
          >
          > Circles Ta,Tb,Tc are ever external to each other, and their common orthogonal circle To is real. Condition of (external) tangency is : (a^2-b^2)^2-(a^2+b^2)c^2=0 (etc) or ABC rectangular.
          >
          > Second configuration.
          > The Apollonius circles of Ta, Tb, Tc are (ABH) etc, their inverses in To and two others. Center X=x:y:z and radius omega of the first one are :
          >
          > x=(b^2+c^2-a^2)(a^8-2(b^2+c^2)a^6+2(b^4-b^2 c^2+c^4)a^4 -2(b^2+c^2)(b^2-c^2)^2 a^2+(b^2-c^2)^4)
          >
          > omega=2*(a^2 b^2 c^2 R)/(a^6+b^6+c^6-a^2 b^4-a^4 b^2-c^2 b^4-a^4 c^2-b^2 c^4-c^4 a^2+4 a^2 b^2 c^2)
          >
          > while the second is less simple.
          >
          > Moreover, X lies on X(3)X(68) and on X(4)X(94)
          >
        Your message has been successfully submitted and would be delivered to recipients shortly.