In the plane are given 100 lines such that no 2 are parallel and no 3 meet in a point. The intersection points are marked. Then all the lines and k of the marked points are erased. Given the remained points of intersection for what max k one can reconstruct the lines? Proposed by A. Golovanov
Problem
Source: Tuymaada 2004
Tags: combinatorics proposed, combinatorics
ciprian
29.06.2005 23:54
Can anyone hepl me with a link at wich i can find the pb form tuymaada (yakutsk) contest 2003 and 2002?
ciprian
30.06.2005 09:07
No solution? No one likes this pb?
valeriano
01.07.2005 06:16
ciprian wrote: No solution? No one likes this pb? Puedes encontrar los problemas de Tuymaada en: (Review this URL:) http://www.guas.info/competit/tuyme.htm I found these problems in PDF format but I don't remember where!!! Enrique.
manuel
01.07.2005 07:02
Que es tuymaada?
perfect_radio
16.01.2006 19:25
Anyone solved this problem?
da-kid
01.01.2008 19:02
I'm not sure if this is right but in this case three points define a line. Every point denotes an intersection between two lines so if three lines are collinear, there must be a line passing through it. Since two lines share a point each point "contributes" one point to where the line passes through. Since there are 100 lines we need 150 points (since each point contributes points to two lines and each line needs 3) in order to define 100 lines.