## 9884Re: [EMHL] The Brocard points of a quadrilateral (Geoff Millin)

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• Jun 9, 2004
>[AZ]
>> I think that this two conditions are equivalent. If the point L
>exists then
>> the projective transformation T conserving the circumcircle and
>such that
>> T(L)=O transforms our polygon to regular. In the circle this
>transformation
>> coincide with some inversion.
>
>Suppose that A_1,...,A_n has Brocard points P,Q; let M be a Poncelet
>point of the pencil generated by the circumcircle (O) of the polygon
>and the circle OPQ. the line MA_k intersects again (O) at B_k.
>Then B_1,...,B_n is regular.
>Of course, it is necessary to prove that such a point M is real, ie
>that (O) and the circle OPQ cannot intersect.
>It is necessary too to prove the reciprocal.
>I think that it could be possible to name these two points M the
>isodynamic points of the Brocardian polygon because
>MA_(i+k)/M_A(i-k) = A_iA_(i+k)/A_iA_(i-k) for all i and k.
>
Yes, this is the same transformation. Let we considere a sphere with
diameter circle (O) and the pole N. The perpendicular to the plan (O) from
my point L intersect the sphere in point L' (the plan (O) divide L' and N).
Then the line NL' intersect (O) in your point M. (That is the corresponding
between Klein and Poincare models of Lobachevsky geometry). So the inversion
with center M is equivalent to projective transformation with center L.

Sincerely Alexey
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