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

A geometrical property of points X(1300) and X(254)

Expand Messages
  • Angel
    In: Eric Danneels and Nikolaos Dergiades.- A Theorem on Orthology Centers. Forum Geometricorum Volume 4 (2004) 135–141.
    Message 1 of 1 , Dec 5, 2011
    • 0 Attachment
      In:

      Eric Danneels and Nikolaos Dergiades.- A Theorem on Orthology Centers. Forum Geometricorum Volume 4 (2004) 135–141.

      (http://forumgeom.fau.edu/FG2004volume4/FG200417index.html)

      we read:

      (p. 135) "Let A'B'C' be the cevian triangle of P with respect to a
      given triangle ABC. Denote by Oa, Ob, Oc the circumcenters of
      triangles PB'C', PC'A', PA'B' respectively. Since ObOc, OcOa, and
      OaOb are perpendicular to AP, BP, CP, the triangles OaObOc and ABC
      are orthologic at P".

      (p. 139) "If P=X(1300), OaObOc is perspective to ABC at X(254)".

      (p. 140) "In each of the cases P = X(7) and X(80), the triangle
      OaObOc is also perspective to ABC at the incenter".

      ------
      It also has:


      If ABC is an acute triangle, X(1300) is the "ONLY POINT"
      (other than A, B and C) on the circumcircle, for which the
      triangles ABC and OaObOc are perspective.



      The locus of points P=(x:y:z) (barycentric coordinates with
      respect to ABC), such that triangles ABC and OaObOc are
      perspective are the quintic (c5) and septic (c7) of equations:

      (c5): y*z(x^2(2SA*x + c^2y) + a^2y*z(y+z)) + cyclic =0

      (c7): y^2z(SA(SA(c^2+5SB)-S^2)x^4 +
      (b^2-a^2)(5SA*SB-S^2)x^3y - SB((c^2 + 5SA)SB-S^2)x^2y^2 +
      6(c^2-b^2)SA^2x^3z + a^4SB*y^2z^2 - a^4SC*y*z^3) + cyclic =0

      The vertices of the triangle ABC are triple points of (c7), the
      orthocenter H is a double point of (c7), and the lines AH, BH and
      CH are tangent in the traces of H to the curve (c7). The septic
      (c7) also contains the points X(7), X(13), X(14), X(80), X(1300),...

      The perspectors of the points X(4), X(7), X(13), X(14), X(80),
      X(1300) are X(4), X(1), X(17), X(18), X(1), X(254), respectively.

      --------------

      For a generic point P on the circumcircle with coordinates
      (a^2/p:b^2/q:c^2/r)=(a^2qr:b^2rp:c^2pq) for an infinite point
      (p:q:r), the triangles ABC and OaObOc are perspective iif:

      (c^2q^2 + 2qrSA + b^2r^2)^3=0 or

      (c^4q^2 + (b^4 + c^4 - a^4)q r +b^4r^2)=0 or

      c^2q SC((b^2-c^2)S^2-SA(S^2-3SB SC)) +
      b^2r SB((b^2-c^2)S^2+SA(S^2-3SB SC))=0.



      --The first condition gives no real solutions for p, q and r.

      --The second condition gives real solutions if ABC is an acute
      triangle:-SA*SB*SC> 0. Obtaining the common points of the
      circumcircle and the De Longchamps line and quintic (c5).

      --The third condition implies that

      q =-r(b^2SB((b^2 - c^2)S^2)+SA(S^2 - 3SB SC))/(c^2SC((b^2-c^2)S^2-SA(S^2-3SB SC))),
      p + q + r = 0

      So (a^2/p: b^2/q: c^2/r) is

      (1/((b^2+c^2-a^2)(a^4(b^2 + c^2)- 2a^2(b^4 - b^2c^2 + c^4) +
      (b^2 - c^2)^2(b^2 + c^2) )): ... : ... ).

      This is the point X(1300) in ETC. The fourth intersection of circum-hyperbola tangent at H to the Euler line. In this case,
      the perspector of ABC and OaObOc is X(254).

      http://webpages.ull.es/users/amontes/pdf/ejtr2509.pdf


      Angel Montesdeoca
    Your message has been successfully submitted and would be delivered to recipients shortly.