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

Altimedial Triangles

Expand Messages
  • Antreas P. Hatzipolakis
    Definition: Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle of a triangle ABC is named after, namely: M_i : the midpoints of sides
    Message 1 of 4 , Jan 30, 2000
    • 0 Attachment
      Definition:
      Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle
      of a triangle ABC is named after, namely:

      M_i : the midpoints of sides BC,CA,AB, resp.
      H_i : the feet of the altitudes from A,B,C, resp.
      E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)
      (i = 1,2,3)

      I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3
      The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three
      altimedials are congruent (and similar to ABC).

      Theorem:
      Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles
      H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are
      concurrent.

      Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

      Proof:

      Lemma:
      Let ABA'C be a quadrilateral, with sides AB = c, BA' = c', A'C = b', CA = b,
      and angles CAA' = A_1, BAA' = A_2, ABC = B, ACB = C, A'BC = w (omega),
      A'CB = q (theta).Then

      sinA_1 b' sin(C+q)
      ------ = ---- * ---------
      sinA_2 c' sin(B+w)


      A
      /\
      / \
      c b
      / \
      / \
      B---------C
      \ w q /
      \ /
      c' b'
      \ /
      \/
      A'

      b' AA'
      Triangle ACA': -------- = ------- (1)
      sinA_1 sin(C+q)

      c' AA'
      Triangle ABA': -------- = ------- (2)
      sinA_2 sin(B+w)


      sinA_1 b' sin(C+q)
      (1) & (2) ===> ------ = ---- * ---------
      sinA_2 c' sin(B+w)


      A
      /\
      / \
      / \
      / \
      / \
      / \
      / \
      M_3 M_2
      / \
      / \
      / I_1 \
      / \
      / \
      / \
      B-------------------H_1------C


      In the quadrilateral AM_3I_1M_2 we have:
      angles AM_2M_3 = C = H_1M_2M_3, AM_3M_2 = B = H_1M_3M_2
      angles I_1M_3M_2 = B/2, I_1M_2M_3 = C/2
      AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'sin(B/2), M_2I_1 = r'sin(C/2)
      [r' = inradius of H_1M_2M_3 = r/2]
      Denote angles I_1AM_2 := A_1, I_1AM_3 := A_2.

      By ap[lyimg the lemma above we get:

      sinA_1 I_1M_2 sin(C+ C/2) sin(B/2)sin(3C/2)
      ------ = -------- * ------------ = -------------------
      sinA_2 I_1M_3 sin(B+ B/2) sin(C/2)sin(3B/2)

      Similarly, for the incenters I_2, I_3 we get:

      sinB_1 sin(C/2)sin(3A/2)
      ------ = -------------------
      sinB_2 sin(A/2)sin(3C/2)

      sinC_1 sin(A/2)sin(3B/2)
      ------ = -------------------
      sinC_2 sin(B/2)sin(3A/2)

      Therefore the lines AI_1, BI_2, CI_3 concur. (Trig. version of Ceva Theorem).

      Note:
      From the above Ceva ratia we get the trilinears of the point of concurrence:
      (x:y:z) = (sin(A/2)sin(3B/2)sin(3C/2) ::)

      Locus Problem:
      Let ABC be a triangle, A'B'C' the pedal triangle of a point P, and
      M_1M_2M_3 the medial triangle of ABC, and I_1, I_2, I_3 the incenters of
      the triangles A'M_2M_3, M_1B'M_3, M_1M_2C', respectively.
      Which is the locus of P such that the lines AI_1, BI_2, CI_3 concur?
      (In the thoerem above P = H).
      The same for the circumcenters instead of the incenters
      (the three circumcenters above coincide with N, the the center of the 9pc)


      Antreas
    • Antreas P. Hatzipolakis
      ... Read instead: AM_3 = c/2, AM_2 = b/2, M_3I_1 = r /sin(B/2), M_2I_1 = r /sin(C/2) Sorry for the typos. Antreas
      Message 2 of 4 , Jan 30, 2000
      • 0 Attachment
        I wrote:

        >AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'sin(B/2), M_2I_1 = r'sin(C/2)

        Read instead:

        AM_3 = c/2, AM_2 = b/2, M_3I_1 = r'/sin(B/2), M_2I_1 = r'/sin(C/2)

        Sorry for the typos.


        Antreas
      • Antreas P. Hatzipolakis
        ... We have: sin(A/2)sin(3B/2)sin(3C/2) :: = sin(A/2)*[3sin(B/2) - 4sin^2(B/2)]*[3sin(C/2) - 4sin^2(C/2)] :: = (3 - 4sin(B/2))*(3 - 4sin(C/2)) :: Can it be
        Message 3 of 4 , Feb 1, 2000
        • 0 Attachment
          I wrote:

          >Definition:
          >Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle
          >of a triangle ABC is named after, namely:
          >
          >M_i : the midpoints of sides BC,CA,AB, resp.
          >H_i : the feet of the altitudes from A,B,C, resp.
          >E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)
          >(i = 1,2,3)
          >
          >I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3
          >The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three
          >altimedials are congruent (and similar to ABC).
          >
          >Theorem:
          >Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles
          >H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are
          >concurrent.
          >
          >Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

          We have:

          sin(A/2)sin(3B/2)sin(3C/2) :: =

          sin(A/2)*[3sin(B/2) - 4sin^2(B/2)]*[3sin(C/2) - 4sin^2(C/2)] :: =

          (3 - 4sin(B/2))*(3 - 4sin(C/2)) ::

          Can it be further reduced ?


          Antreas
        • Paul Yiu
          Dear Anteas, This perspector of I_1I_2I_3 and ABC is the point X_{79} in Kimberling s list. As such, its TRILINEAR coordinates can be written as 1/(1+2cos A) :
          Message 4 of 4 , Feb 1, 2000
          • 0 Attachment
            Dear Anteas,

            This perspector of I_1I_2I_3 and ABC is the point X_{79} in
            Kimberling's list. As such, its TRILINEAR coordinates can be
            written as

            1/(1+2cos A) : 1/(1+2cos B) : 1/(1+2cos C).

            Here is the derivation of the homogeneous BARYCENTRIC
            coordinates.

            *************
            The incenter I_1 of triangle H_1M_2M_3 is the reflection of the Spieker point

            b+c:c+a:a+b

            in the perpendicular bisector of the segment M_2M_3. This line has equation

            (b^2-c^2)x + (2a^2-b^2+c^2)y - (2a^2+b^2-c^2)z = 0.

            From this The incenter I_1 has coordinates

            a(b+c) : a^2+ab+b^2-c^2 : a^2-b^2+c^2+ac
            = *** : 1/(c^2+ca+a^2-b^2) : 1/(a^2+ab+b^2-c^2).

            Similarly, we write down the coordinates of I_2
            and I_3. It is evident then that I_1I_2I_3 and
            ABC are perspective at

            1/(b^2+bc+c^2-a^2) : 1/(c^2+ca+a^2-b^2) :
            1/(a^2+ab+b^2-c^2).

            As pointed above, this is X_{79} in Kimberling's list.

            Best regards.
            Sincerely,
            Paul

            ----------
            From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]
            Reply To: Hyacinthos@onelist.com
            Sent: Tuesday, February 01, 2000 2:37 PM
            To: Hyacinthos@onelist.com
            Subject: Re: [EMHL] Altimedial Triangles

            From: xpolakis@... (Antreas P. Hatzipolakis)

            I wrote:

            >Definition:
            >Let M_i, H_i, E_i (i=1,2,3) be the nine points that the nine-point circle
            >of a triangle ABC is named after, namely:
            >
            >M_i : the midpoints of sides BC,CA,AB, resp.
            >H_i : the feet of the altitudes from A,B,C, resp.
            >E_i : the midpoints of AH,BH,CH, resp. (H: orthocenter)
            >(i = 1,2,3)
            >
            >I call "altimedial triangles" the triangles H_1M_2M_3, M_1H_2M_3, M_1M_2H_3
            >The medial triangle (M_1M_2M_3), the Euler triangle (E_1E_2E_3) and the three
            >altimedials are congruent (and similar to ABC).
            >
            >Theorem:
            >Let I_1, I_2, I_3 be the Incenters of the three altimedial triangles
            >H_1M_2M_3, M_1H_2M_3, M_1M_2H_3, resp. Then the lines AI_1, BI_2, CI_3 are
            >concurrent.
            >
            >Trilinears of the point of concurrence: (sin(A/2)sin(3B/2)sin(3C/2) ::)

            We have:

            sin(A/2)sin(3B/2)sin(3C/2) :: =

            sin(A/2)*[3sin(B/2) - 4sin^2(B/2)]*[3sin(C/2) - 4sin^2(C/2)] :: =

            (3 - 4sin(B/2))*(3 - 4sin(C/2)) ::

            Can it be further reduced ?


            Antreas





            --------------------------- ONElist Sponsor ----------------------------

            ONElist: your connection to online communities.

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