Search   Memberlist   Usergroups
 Page 1 of 1 [3 Posts]
Author Message
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.]
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.
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.

 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
 Page 1 of 1 [3 Posts]
 The time now is Mon Feb 18, 2019 5:02 pm | All times are GMT
 Jump to: Select a forum-------------------Forum index|___Science and Technology    |___Math    |   |___Research    |   |___num-analysis    |   |___Symbolic    |   |___Combinatorics    |   |___Probability    |   |   |___Prediction    |   |       |   |___Undergraduate    |   |___Recreational    |       |___Physics    |   |___Research    |   |___New Theories    |   |___Acoustics    |   |___Electromagnetics    |   |___Strings    |   |___Particle    |   |___Fusion    |   |___Relativity    |       |___Chem    |   |___Analytical    |   |___Electrochem    |   |   |___Battery    |   |       |   |___Coatings    |       |___Engineering        |___Control        |___Mechanics        |___Chemical

 Topic Author Forum Replies Last Post Similar Topics Wavelengths, Frequency and Geometry harmonic Electromagnetics 0 Sun Sep 24, 2006 5:30 pm Question about modular function related to the projective... David Sevilla Math 4 Wed Jul 12, 2006 9:57 pm Geometry Problem Bractals Math 7 Wed Jul 05, 2006 5:38 am geometry & vector analysis tamiry Math 1 Wed Jun 28, 2006 7:01 pm Spherical Geometry Problem jecottrell@comcast.net Math 2 Sun Jun 25, 2006 2:20 am