## Re: [EMHL] Poncelet points on the Euler line

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• Dear Randy Thanks!! I think it would be interesting to study this special case: Let p(pN) be the point where concur the NPCs of N1N2N3, pNN1N2, pNN2N3, pNN3N1
Message 1 of 3 , Feb 11, 2013
Dear Randy

Thanks!!

I think it would be interesting to study this special case:

Let p(pN) be the point where concur the NPCs of N1N2N3, pNN1N2,
pNN2N3, pNN3N1 (ie we replace N with pN).

Is it interesting ? ie ls it lying on some interesting curves, lines?

In my figure it lies on the circle centered at N with radius NppN

APH

[Randy Hutson]
> pP is the center of the rectangular circumhyperbola through P.
> ppP does not, in general, lie on the Euler line.
>
> Some results:
>
> P=X(1), the NPCs are concurrent, with center = non-ETC 1.121590125545969 (on lines 1,5 3,962).
> P=X(2), ppP=non-ETC 1.690358502447462
> P=X(3), ppP=X(140)
> P=X(4), ppP=undefined
> P=X(5), ppP=X(3628)
> P=X(6), ppP=non ETC 0.780037257060191
> P=X(7), ppP=non ETC 0.750876768572663
> P=X(8), ppP=non ETC 2.966801160450799
> P=X(9), ppP=non-ETC 0.972023454564163
> P=X(10), ppP=non-ETC 2.238481946743318
> P=X(20), ppP=non-ETC 6.363850996796102
> P=X(21), ppP=non-ETC -1.717011738240629
> P=X(22), ppP=non-ETC -4.036288926987237
>
> Of these, only X(140) and X(3628) lie on the Euler line.Â
> The ppP for points P on the Euler line do not even lie on the
> same conic. Locus?
[APH]
> >Let ABC be a triangle, (N),(N1),(N2),(N3) the NPCs of
> >ABC,NBC,NCA,NAB, resp. [concurrent at pN]. The NPCs of the
> >triangles N1N2N3, NN1N2, NN2N3, NN3N1 concur at point ppN
> >on the Euler Line of ABC.
> >
> >Coordinates of ppN?
> >
> >Generalization:
> >
> >Let ABC be a triangle, P a point, (N),(N1),(N2),(N3) the NPCs
> >of ABC, PBC, PCA, PAB resp. [concurrent at pP]. If P is on
> >the Euler line of ABC, then the NPCs of N1N2N3, PN1N2, PN2N3,
> >PN3N1 concur at point ppP on the Euler Line of ABC.
> >
> >True??
> >
> >APH
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