Re: Families of Straight Lines
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 14 Aug 2005 18:10:33 -0700
On 14 Aug 2005 13:01:12 -0700, cbrown@xxxxxxxxxxxxxxxxx wrote:
>Ulysse Keller wrote:
>> On Sat, 13 Aug 2005 07:15:27 EDT, Maury Barbato
>> <mauriziobarbato@xxxxxxxx> wrote:
>>
>> >Hello,
>> >I'm thinking about some particular families of straight lines, which I called "closed under intersection".
>> >I say that a family F of straight lines in the plane is "closed under intersection" if for every four elements
>> >a_1, a_2, b_1, b_2 of F such that a_1 and a_2 intersect in a point A, and b_1 and b_2 intesect in
>> >a point B, then the stright line passing through A and B belongs to F.
>> >Dou you know if someone explored the properties of these particular objects?
>> >(...)
>> >
>> No, but do you have a non trivial example of such a family ?
>
>I can think of several ways to interpret "trivial".
>
>First, consider a set of any number of parallel lines. Since there are
>no intersections, the axioms hold. Call this class of families "truly
>trivial"
>
>Next consider any truly trivial family, with the addition of a single
>line not parallel to the others. Since every point of intersection
>falls on this line, it meets the axioms. Call this class of solutions
>"simply trivial".
>
>Then there's the set of families where all lines in the family meet at
>a single point; I suppose we could call this "radially trivial".
>
>The family of any three lines which forms a triangle is a sort of
>trivial non-trivial case.
>
>For a non-trivial finite example with more than 3 lines, consider four
>lines whose intersections form a rectangle. Add the two diagonals; the
>resulting family obeys the axioms.
>
>Are there other finite non-trivial families?
>
>Cheers - Chas
Any parallelogram together with the 2 diagonals can be used instead of
just a rectangle.
Also, the radially trivial case can be relaxed to include one
additional line not through the common point. This case then obsoletes
the triangle case.
Other than these corrections, I believe you have identified all finite
configurations. I'll sketch a proof ...
Firstly if no intersection points, then the configuration must be a
single line or a finite set of parallel lines. These are trivial
configurations.
If all intersection points are collinear then the configuration is
either a pair of intersecting lines or a line intersected by a finite
set of parallel lines, so also trivial.
Hence we can assume at least 3 intersections points, not all
collinear.
Let P be the polygonal boundary of the convex hull of the intersection
points, and let n=number of vertices of P. Thus P is a convex n-gon.
By the above discussion, we must have n>2.
If n>=5 then some pair of sides would necessarily intersect outside
the convex hull, but that contradicts the concept of the convex hull,
hence we must have n<5.
If n=4 then the convex hull is a quadrilateral and again there would
be outside intersection points unless the quadrilateral is a
parallelogram. Starting with the 4 vertices of a parallelogram, the
diagonals are forced, and also the intersection point of the
diagonals. Suppose there is an additional point besides the vertices
of the parallelogram and the intersection point of the diagonals. Then
connect the additional point with the intersection point of the
diagonals. If the new line is not parallel to one of the pairs of
opposite sides of the parallelogram, then we can extend to get an
intersection point outside the convex hull, contradiction. Hence all
new additional points must be on the lines through the midpoints of
the opposite sides. But any such point when connected to the 4
vertices gives new intersection points on the sides of the
parallelogram which are not on these 2 new lines. Therefore if n=4,
there can only be the trivial parallelogram configuration with just
the sides and the diagonals.
Next suppose n=3. If all additional points are on one side of the
triangle then the configuration would be radial with an additional
line as mentioned above. Thus there must be 2 points in the interior
of adjacent sides. The line between them must be parallel to the 3rd
side otherwise there would be an intersection point outside the
triangle. So a trapezoid is formed inside the triangle and we can
connect the 2 diagonals to create an interior intersection point.
Connecting this new intersection point with the remaining vertex
results in an intersection point on the 3rd side, and so we can now
form an inscribed triangle inside the trapezoid. All sides of the
inscribed triangle must be parallel to the opposite sides of the big
triangle so the 2 vertices of the inscribed triangle must be the
midpoints of the respective sides of the big triangle. All we need for
a contradiction is one new point on the boundary of the big triangle
besides the vertices and the midpoints. To get this, draw any median
of the big triangle and take the intersection point created with the
inscribed triangle. Connect this new intersection point to either of
the 2 other vertices of the big triangle and extend to get a new
intersection point on the boundary of the big triangle. Since this new
intersection point is not a vertex or a midpoint, connect it to one of
the midpoints on an adjacent side and extend to an intersection point
outside the big triangle, contradiction. Thus n=3 is resolved (whew,
painful).
Therefore the only finite configurations are the ones discussed.
quasi
.
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