infinite configuration of lines I was looking at some random problems and questions I liked when I was in high school and I found this one which I still cannot prove.
Does there exist a configuration of a countable number of straight lines in the plane such that:
1) no two are parallel
2) no three are concurrent
3) any bounded subset of the plane is intersected by a finite number of lines
4) the area of every minimal polygon is equal, where a minimal polygon is a polygon formed by a finite subset of the set of lines such that no lines pass through the inside of the polygon.
The answer is certainly no, but it is not that easy to prove. Any ideas?
 A: I can sketch a proof based on assuming this "finite" result:
A). For any pentagonal star one of the 5 triangles will have area strictly
smaller than that of the central pentagon.
(I think a brute force attack should yield a proof here.)
 
The proof of the original problem would then go as follows.
b). A) generalizes to n-agons by considering the pentagon
spanned by any 5 vertices.

c). b) implies that a tiling with polygons of EQUAL AREA
is not possible unless all polygons are either triangles or
quadrilaterals.
d). Take one 4-tile and continue tiling next to it inside
the cone enclosed by the converging lines of 2 opposite edges;
we have a sequence of quadrilaterals which must end with a
triangle were the the 2 lines meet. This shows that the tiling
must contain a 3-tile somewhere.

e). By d) take a 3-tile and continue tiling outwards, inside each of the
3 beams generated by the lines of each pair of edges; the original 3-tile
will be the first tile in each beam, but every other tile after it must be
a 4-tile (build them one at a time and keep using c)).
We can ignore what happens in the 3 leftover cones radiating from the 3 vertices.

f). In one of the 3 beams (which now look like ladders) take any one of the new rungs
from step e) and extend it - that line will then collide with one of the other
2 beams (but cannot overlap with any of its rungs). That will cut
one of the 4-tiles, creating a 5-tile.

Apologies for bumping up the question repeatedly while trying to edit my answer.
