## 21056Re: [EMHL] Re: Simson lines

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• Jun 7, 2012
Dear Daniel, Nikos, Cosmin and Barry!
Yes, I also found these general facts. Thank you for the references. Also the envelop of Simson lines is the astroid and its center coincides with the center of NPC. And what is the locus of point P such that the circumcevian triangle of P is S-triangle? It is clear that this is some cubic, but which?

Sincerely Alexey

----- Original Message -----
From: Barry Wolk
To: Hyacinthos@yahoogroups.com
Sent: Thursday, June 07, 2012 11:33 PM
Subject: [EMHL] Re: Simson lines

> From: Daniel <dvacaretu@...>
>
> Yes, it is a particular case of S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle are
> S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and the Simson
> lines of A,B,C, with respect to A'B'C' are concurrent in the
> midpoint of [HH'] where H and H' are the orthocenters of ABC and
> A'B'C'.

A routine calculation using complex numbers as coordinates gives
the following: If A,B,C,D,E are on the unit circle, then the Simson
lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2

This formula is simpler than expected, even though it was known
in advance that it had to have some symmetry and scaling properties.
And the S-triangle result follows easily from this.
--
Barry Wolk

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