## A simple construction with a simple perspector

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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
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

[Non-text portions of this message have been removed]
• 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

> Dear geometers:
>
>
>
> This construction leads to a very simple perspector, which
> 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
>
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>