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

De Longchamps point of the orthic triangle

Expand Messages
  • Eric Danneels
    Dear Hyacinthians, message 7504 led to some results on orthopoles. In message 7550 I made the following statement: (rephrased) Consider a triangle ABC with H =
    Message 1 of 1 , Dec 3, 2003
    • 0 Attachment
      Dear Hyacinthians,

      message 7504 led to some results on orthopoles.

      In message 7550 I made the following statement:

      (rephrased)

      Consider a triangle ABC with
      H = orthocenter;
      E = center of the ninepointcircle;
      A' the foot of the altitude on BC.
      Let At be the orthopole of the tangent to the circumcircle in A.
      At is het reflection of A' wrt the midpoint of AH.
      Let Ad be the orthopole of the diameter of the circumcenter parallel
      to the tangent to the circumcircle in A.
      Ad is the second intersection of the line bisector of BC with the
      ninepointcircle.
      Define Bt, Ct, Bd and Cd cyclically.

      ==> The triangles AtBtCt and AdBdCd are perspective

      Using synthetical methods it is easy to prove that AtAd is
      perpendicular to the A-tangent and intersects BC at the isotomic
      conjugate of A' wrt BC.
      Using that I found the barycentrics of the perspector

      P = (a.cos²A.cos(B-C) : b.cos²B.cos(C-A) : c.cos²C.cos(A-B))

      Replacing cos²A by 1-sin²A ....

      it is immediately clear that P lies on the line E - X(51) = the
      Euler line of the orthic triangle.

      With some trigonometrical manipulations I proved that P is the
      reflection of X(52) wrt to E. Since X(52) is the orthocenter of the
      orthic triangle the perspector is the De Longchamps point of the
      orthic triangle.

      This result can also be proved synthetically: since the sides of the
      orthic triangle are parallel to the sides of tangential triangle it
      follows that P is the orthocenter of triangle AdBdCd.
      Ad, Bd and Cd are the antipodes of A', B' and C' in the
      ninepointcircle so the orthocenters of the orthic triangle and the
      triangle AdBdCd are symmetrical wrt to E.

      Expanding the x-coordinate gives
      a.cos²A.cosB.cosC + a.cos²A.sinB.sinC
      The first term can be reduced to a.cosA or sinA.cosA or sin(2A)
      => circumcenter of ABC
      The second term becomes cos²A => X(394) in the ETC
      So the perspector also lies on the line X(3) - X(394)
      (= line 3,49 in the ETC)

      Remark:

      According to the ETC X(185) also lies on this line.
      However, my constructions showed that this is not the case if you
      construct it as the Nagel point of the orthic triangle as its
      definition says.
      The coordinates in the ETC correspond with the search-value and this
      point seems to be on the line 3,49.
      It think that the definition of this point is incorrect


      Greetings from Bruges


      Eric Danneels
    Your message has been successfully submitted and would be delivered to recipients shortly.