Hi Jon and Benoit,
My formula can be seen as either a change of coordinates (into a system with
CA, CB as the unit vectors), or as barycentric coordinates: this shall be the
same if I’m not mistaken.
The class variables E/D and F/D shall represent these coordinates and vary
between 0 and 1 inside of the unit parallelogram. I just found it more elegant
to have E and F vary between 0 and D, this avoids a division (and its corner
cases), and results in a shorter formula.
E and F are indeed built using cross products (or matrix determinents, again,
this shall be the same).
Of course Benoit is right, my formula would match the full parallelogram, it
misses something like ( E + F ) e. ( 0 [,] D ) to cut it in half.
I chose this representation because, as Benoit says, I suspect it is rather
compact.
Another way may be to see the triangle as the intersection of 3 half planes and
write S = { <. x , y >. | ( ph /\ ps /\ ch ) } where ph, ps and ch are the
formula for each half plane.
Jon, do you plan your final statement formula to still require to have the
three points ordered with abscisses in ascending order? I’m still interested to
see it. In full, like my attempt.
BR,
_
Thierry
--
You received this message because you are subscribed to the Google Groups
"Metamath" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion on the web visit
https://groups.google.com/d/msgid/metamath/192A12C4-9415-4DB9-B7A0-38AC4219F84B%40gmx.net.
