## 19649Re: A point related to a quadrilateral

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• Jan 2, 2011
Dear Jean-Pierre,

I am quite thrilled.
As always when I find something beautiful.
You described a point M depending on the 4 points A1, A2, A3, A4 of a quadrilateral.
Here is the structure behind it.

M = the inverse in the circumcircle of AiAjAk of the isogonal conjugate of Al wrt AiAjAk,
where (i,j,k,l) is any permutation of (1,2,3,4).

Strangely enough this is true for each permutation.
There should be more properties/references relating to this point.
Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...> wrote:
>
>
> Dear Chris
> [JP]
> > > Consider a quadrilateral A_1,A_2,A_3,A_4
> > > For k =1,2,3,4, T_k is the triangle with vertices the A_i except A_k; O_k is the circumcenter of T_k, (O_k) the circumcircle and B_k the isogonal conjugate of A_k wrt T_k
> > > Then the inverse of B_k in (O_k) doesn't depend on k and this point M is the center of the homothecy mapping O_1,O_2,O_3,O_4 to B_1,B_2,B_3,B_4.
> > > Is there a special name for this point M? Do you know some references?
> [Chris]
> > I do not know a special name for point M.
> > However I noticed this. Maybe you know it already.
> > Let T(A_1,A_2,A_3,A_4) = Transform A_1,A_2,A_3,A_4 --> O_1,O_2,O_3,O_4.
> > Then T^2(A_1,A_2,A_3,A_4) produces a quadrilateral homethetic with A_1,A_2,A_3,A_4 only rotated 180 degrees. Again Center of Homothecy = M.
> > T^4(A_1,A_2,A_3,A_4) produces a quadrilateral homothetic and with same orientation as A_1,A_2,A_3,A_4.
>
> Thank you for your nice remark.
> In fact, if O_1' is the circumcenter of O_2O_3O_4,...,
> the same homothecy maps A_i to O_i' and B_i to O_i
> The point M is characterized by the angular relations
> <A_iMA_j = <A_iA_kA_j +<A_iA_lA_j (oriented angles modulo Pi) where (i,j,k,l) is any permutation of (1,2,3,4)