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A simple construction with a simple perspector

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  • César Lozada
    Dear geometers: This construction leads to a very simple perspector, which mades me think that it’s surely already known. Sorry if this is the case. Through
    Message 1 of 2 , Jan 23, 2013
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      Dear geometers:



      This construction leads to a very simple perspector, which mades me think
      that it’s surely already known. Sorry if this is the case.



      Through G, the centroid of DABC, draw the lines (a) // (BC), (b) // (AC) and
      (c) // (AB). Lines (b) and (c) cut (BC) in B1 and C1 , respectively, making
      the triangle DTA =DGB1C1.

      Build triangles DTB=DGC2A2 and DTC=DGA3B3 similarly.



      Let P be any center and A’ = center P of DTA, B’ = center P of DTB and C’ =
      center P of DTC.



      Then:

      a) DA’B’C’ is perspective with DABC.

      b) If P =( u : v : w ) then the perspector in (a) is Q = ( (2*a*u +
      b*v + c*w)/a : (a*u + 2*b*v + c*w)/b : (a*u + b*v + 2*c*w)/c ) (trilinears)

      c) If the perspector Q=( U : V : W ) then P = ( (3*a*U - b*V - c*W)/a
      : (-a*U + 3*b*V - c*W)/b : (-a*U - b*V + 3*c*W)/c )

      d) If P” = center P of DABC then G, Q and P” are aligned



      P=X(I) and Q=X(J) for the next (I,J): (1, 1125) (2, 2) (3, 140) (4, 5)
      (5, 3628) (6, 3589) (7, 142) (8, 10) (10, 3634)

      (145, 1) (20, 3) (3146, 4) (193, 6) (3621, 8) (144, 9) (149,11)
      (5059, 20)



      Regards

      Cesar Lozada







      [Non-text portions of this message have been removed]
    • Nikolaos Dergiades
      Dear Cesar, If I understand you mean the following: The parallells from the centroid G to the sides of ABC give 3 similar triangles to ABC the triangles GAbAc,
      Message 2 of 2 , Jan 24, 2013
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        Dear Cesar,
        If I understand you mean the following:
        The parallells from the centroid G to the sides
        of ABC give 3 similar triangles to ABC the triangles
        GAbAc, GBcBa, GCaCb.
        If P = (u:v:w) in barycentrics is a triangle center of ABC
        then since triangle GAbAc is homothetic to ABC
        in the homothecy (Ga, 1/3) Ga is the mid point of BC
        the P point of GAbAc is A' = (u : u+2v+w : u+v+2w).
        Similarly define B', C' and the triangles ABC, A'B'C'
        are perspective at Q = (2u+v+w : u+2v+w: u+v+2w) = (P + 3G)/4
        or Q divides GP in ratio 1/3.

        Best regards
        Nikos Dergiades


        > Dear geometers:
        >
        >
        >
        > This construction leads to a very simple perspector, which
        > mades me think
        > that it’s surely already known. Sorry if this is the
        > case.
        >
        >
        >
        > Through G, the centroid of DABC, draw the lines (a) // (BC),
        > (b) // (AC) and
        > (c) // (AB). Lines (b) and (c) cut (BC) in B1 and C1 ,
        > respectively, making
        > the triangle DTA =DGB1C1.
        >
        > Build triangles DTB=DGC2A2 and DTC=DGA3B3 similarly.
        >
        >
        >
        > Let P be any center and A’ = center P of DTA, B’ =
        > center P of DTB and C’ =
        > center P of DTC.
        >
        >
        >
        > Then:
        >
        > a)       DA’B’C’ is
        > perspective with DABC.
        >
        > b)       If P =( u : v : w ) then
        > the perspector in (a) is Q = ( (2*a*u +
        > b*v + c*w)/a : (a*u + 2*b*v + c*w)/b : (a*u + b*v + 2*c*w)/c
        > ) (trilinears)
        >
        > c)       If the perspector Q=( U :
        > V : W ) then P = ( (3*a*U - b*V - c*W)/a
        > : (-a*U + 3*b*V - c*W)/b : (-a*U - b*V + 3*c*W)/c )
        >
        > d)       If P” = center P of
        > DABC  then G, Q and P” are aligned
        >
        >
        >
        > P=X(I)  and Q=X(J) for the next (I,J): (1, 1125) 
        > (2, 2)  (3, 140)  (4, 5)
        > (5, 3628)  (6, 3589)  (7, 142)  (8, 10) (10,
        > 3634) 
        >
        > (145, 1)  (20, 3)  (3146, 4)  (193, 6) 
        > (3621, 8)  (144, 9)  (149,11)
        > (5059, 20)
        >
        >
        >
        > Regards
        >
        > Cesar Lozada
        >
        >
        >
        >
        >
        >
        >
        > [Non-text portions of this message have been removed]
        >
        >
        >
        > ------------------------------------
        >
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        >
        >
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        >
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