20181002, 13:28  #1 
Aug 2002
7×23×53 Posts 
October 2018

20181003, 03:16  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}×5^{2}×23 Posts 
I wonder if there has ever been a challenge that has gone unsolved for an entire month.
Or at least a couple of weeks. No solvers listed on 2nd day of the month. The challenge has been listed for 6 days so far. 
20181003, 04:10  #3 
Jun 2003
3^{4}·67 Posts 
Not listed does not mean no solvers at all. When they get around to updating the page, we'll see the actual dates. Almost sure someone would've solved it within hours.

20181003, 05:00  #4  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·5^{2}·23 Posts 
Quote:
Anyhow it's quite rare for no solvers to be listed on the 1st day of the month. 

20181003, 06:28  #5 
Jun 2003
3^{4}·67 Posts 

20181003, 16:24  #6 
Jul 2015
2·7 Posts 
I think that using Bruteforce method is not intended at all because the search space is too large even if an efficient algorithm is applied to remove the overlapped cases.
However, I still have not figured out any approaches or ideas to reach the solution. There should be 165 different areas of triangles. Any of three points cannot be on the same line. Any of two segments cannot be parallel. Is there any other constraints? Maybe dividing the space (with grid of dimension 600) into small rectangular pieces with a point inside can be a good approach. ...needs more pondering... 
20181003, 17:06  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4374_{8} Posts 
I don't think being parallel is relevant. What is, is that no 3 points should be on a straight line.
None of the constraints are much of challenge other than having 165 triangles (formed by 11 vertexes) which none have the same area. Plenty of triangles will end up with equal areas even with differing side lengths. I think that perhaps a manual arrangement is the path of least resistance. 
20181003, 18:58  #8  
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 
Quote:


20181004, 07:08  #9 
Jul 2015
1110_{2} Posts 
I tried to figure out more simple cases with smaller numbers of points.
The results are shown below with minimum area. Any ideas or intuitions can be derived from the simple cases? 3 points: (0,0) (0,1) (1,0) with area 1 (1*1) 4 points: (0,0) (0,1) (1,2) (2,0) with area 4 (2*2) 5 points: (0,0) (0,1) (2,3) (4,2) (5,0) with area 15 (5*3) 6 points: (0,0) (0,1) (1,4) (3,5) (4,0) (6,2) with area 30 (6*5) 7 points: (0,0) (1,0) (2,4) (5,6) (10,5) (11,1) (11,3) with area 66 (11*6) 8 points: (0,1) (0,3) (1,6) (4,6) (7,7) (12,0) (14,8) (15,2) with area 120 (15*8) 
20181005, 04:38  #10 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4374_{8} Posts 
I have been on the subject for a few days now. I don't see any feasible approach other than dedicating 10s of cores to the job.
The best I have found was for a 39x36 = 1404 matrix, which I failed to reduce without duplicating areas. 
20181005, 16:49  #11 
"Ben"
Feb 2007
3·17·73 Posts 
Still no correct answers listed on the page...
I've written a solver that finds a solution to N=10 (on the maxsized grid) in a couple minutes, but the best it's done so far on N=11 is a 161triangle solution (4 duplicates). I should probably take that one and try to tweak it by hand. 
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