Math Forum / Mathematics / Recreational Math / June 2009

Kumar wrote :

Hi,

Consider 9 lines intersecting each other, giving
2 Choose 9 = 36 intersection points, too much as we want only
18 trees. We can reduce the number of trees by :

1) discarding some intersections
2) collapsing intersection points so that more than two lines
  concur
3) combination of both methods
4) if too much reduced, adding trees outside intersection points

The second method in its simplest way, that is more than 2 lines
at some trees, and a tree at every intersection, gives after a
combinatoric only study :

 9 trees where three lines meet (3-trees),
 and 9 trees where 2 lines meet (2-trees),
 on each line there are three 3-trees and two 2-trees

This configuration can be obtained, starting from three pencils
of three lines each and adjusting the rays of the pencils to
satisfy the other conditions :
6 more 3-trees in total, three 3-trees per line.
Then complete the 2-trees on the other intersections.

Best Regards.

Start with 5 points having the coordinates (approximately)
p_1: (10,25)
p_2: (15,10)
p_3: (15,33)
p_4: (10,15)
p_5: (30,25)

L[i,j] - the line through points p_i and p_j

p_6 = L[4,5] x L[1,2] - the two lines cross at p_6
p_7 = L[1,2] x L[3,4]
p_8 = L[3,2] x L[4,5]
L[3,6]
L[1,8]
L[5,7]
p_9 = L[5,7] x L[3,2]
p_10 = L[3,4]xL[1,8]
L[1,9]
L[5,10]

Now we have 9 lines and 10 points. The remaining 8 points are easily identified as intersections.

The lines L[3,6], L[5,10] and L[1,9] going through one point needs to be proved.