most recent 30 from http://mathoverflow.net 2013-05-19T20:15:08Z http://mathoverflow.net/feeds/question/66508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66508/infinite-configuration-of-lines Manuel Rivera 2011-05-31T00:53:09Z 2011-06-01T09:33:50Z

<p>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.</p> <p>Does there exist a configuration of a countable number of straight lines in the plane such that:</p> <p>1) no two are parallel</p> <p>2) no three are concurrent</p> <p>3) any bounded subset of the plane is intersected by a finite number of lines</p> <p>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.</p> <p>The answer is certainly no, but it is not that easy to prove. Any ideas?</p>

http://mathoverflow.net/questions/66508/infinite-configuration-of-lines/66618#66618 Yaakov Baruch 2011-06-01T04:39:12Z 2011-06-01T09:33:50Z

<p>I can sketch a proof based on assuming this "finite" result:</p> <p>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.) <img src="http://img268.imageshack.us/img268/6733/proofay.png" alt="star"></p> <p>The proof of the original problem would then go as follows.</p> <p>b). A) generalizes to n-agons by considering the pentagon spanned by any 5 vertices. <img src="http://img263.imageshack.us/img263/1743/proofbe.png" alt="hexagon"></p> <p>c). b) implies that a tiling with polygons of EQUAL AREA is not possible unless all polygons are either triangles or quadrilaterals.</p> <p>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.</p> <p><img src="http://img202.imageshack.us/img202/7531/proofdm.png" alt="4-tile"></p> <p>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.</p> <p><img src="http://img191.imageshack.us/img191/4290/proofen.png" alt="3-tile"></p> <p>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.</p> <p><img src="http://img585.imageshack.us/img585/4230/prooffh.png" alt="contradiction!"></p> <p>Apologies for bumping up the question repeatedly while trying to edit my answer.</p>