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Sam Northshield science forum beginner
Joined: 05 Jun 2006
Posts: 2

Posted: Mon Jun 05, 2006 3:00 pm Post subject:
projective geometry



Suppose I have a smooth curve in R^2 so that any line hits it at at
most two points and so that it obeys the following "hexagon rule": if
six points P(1), P(2),... P(6) on the curve define a hexagon so that
two of the three pairs of opposite sides are parallel (i.e., the line
through P(1) and P(2) is parallel to the line through P(4) and P(5),
etc.), then the third pair of sides are parallel. Is such a curve
necessarily a conic? These properties indeed hold for conics, but I
believe only for conics. I figure this must be known or easy (but not
to me!). Thanks in advance for any help. 

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jpdahl@lycos.com science forum beginner
Joined: 18 Jun 2006
Posts: 1

Posted: Sun Jun 18, 2006 7:15 pm Post subject:
Re: projective geometry



Sam Northshield skrev:
Quote:  Suppose I have a smooth curve in R^2 so that any line hits it at at
most two points and so that it obeys the following "hexagon rule": if
six points P(1), P(2),... P(6) on the curve define a hexagon so that
two of the three pairs of opposite sides are parallel (i.e., the line
through P(1) and P(2) is parallel to the line through P(4) and P(5),
etc.), then the third pair of sides are parallel. Is such a curve
necessarily a conic? These properties indeed hold for conics, but I
believe only for conics. I figure this must be known or easy (but not
to me!). Thanks in advance for any help.
There is a reversal of the Pascal theorem saying that if the points 
of intersection of the three pairs of opposite sides of a simple
hexagon are collinear, the vertices of the hexagon are on a conic,
cf Ayres: PG (Schaum's), Th 10.2. Parallel lines in PG is another
way of saying that they intersect on the line at infinity and are
hence collinear. Thus, your smooth curve has the same locus as
a conic. 

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Sam Northshield science forum beginner
Joined: 05 Jun 2006
Posts: 2

Posted: Wed Jun 28, 2006 5:56 pm Post subject:
Re: projective geometry



Quote:  There is a reversal of the Pascal theorem saying that if the points
of intersection of the three pairs of opposite sides of a simple
hexagon are collinear, the vertices of the hexagon are on a conic,
cf Ayres: PG (Schaum's), Th 10.2. Parallel lines in PG is another
way of saying that they intersect on the line at infinity and are
hence collinear. Thus, your smooth curve has the same locus as
a conic.

Thanks for the reply. There still seems a problem though. Although
its true that every 'parallel hexagon' on the curve lies on a conic,
it's not clear that its always the *same* conic for every choice of
'parallel hexagon' inscribed in the curve. That is, it may be
possible that for any parallel hexagon inscribed in the curve, the
conic determined by those six points intersects the curve only at
those six points.
[Reformatted by moderator. Please keep lines shorter than 80 characters.] 

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