------ Mensagem original ------
Assunto : Enc: caminho mínimo (suite 2)
De : Luís Lopes
Data : 09/08/2018 15:30:04
Para : "[email protected]" , "[email protected]"
Sauda,c~oes,
Outra tentativa.
________________________________
De: Luís Lopes
Enviado: quinta-feira, 9 de agosto de 2018 16:39
Para: [email protected]
Assunto: caminho mínimo (suite 2)
Sauda,c~oes,
Este problema reapareceu recentemente em uma outra lista
da qual faço parte.
Resposta definitiva:
Dear Luis and Ralf
If Cr is the circle O(R),
A, B are arbitrary points inside Cr
P is a point on Cr and Es is the ellipse with foci A, B that passes through P then
PA + PB is the shortest path if Es is inside Cr and
PA + PB is the longest path if Cr is inside Es.
This means in both cases that Cr is tangent to Es at P or
that the reflection of the line AP in OP is the line BP.
This is the Alhazen's circular billiard problem see FG 16 2012 pag. 193-196.
The points P are constructed as intersections of the Cr and a rectangular
hyperbola Hp as follows:
Let A', B' be the inverses of A, B in Cr,
H be the orthocenter of triangle OA'B',
O' be the symmetric of O relative to the mid point of A'B' then
the hyperbola Hp is the conic that passes through O, A', B', H, O'.
Best regards
Nikolaos Dergiades
Referência:
http://forumgeom.fau.edu/FG2012volume12/FG201216.pdf
Alhazen’ s Circular Billiard Problem - Forum Geometricorum
forumgeom.fau.edu
Alhazen’s circular billiard problem 195 3. Alhazen’s problem 2 Given two points A and B inside a circle (O), to construct a point P on the circle such that the reflection of AP at P passes through B.
Sds,
Luís
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