Place 9 balls ("o") in the intersections of the grid below to achieve 10 straight lines, each line containing 3 and exactly 3 balls. You may assume, if you need, that each cell is a perfect square.
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
To Avalon: I believe that DejMar's Comment #8 is a solution which is not just a rotation of the 4x6 (in my last post, I should have written 4x6 rather than 3x5, since I advocated sizing by the matrix of points, not of the cells) which Leming gave earlier (Comment #3). Its general orientation is a 45º R rotation, but the center is skewed: I'm not a topologist so I'm not certain this qualifies as a different solution (i.e. not derivable from the first). He says there are 80+ different exact solutions, but I'm not sure how many are topologically nonequivalent. We could use a proposer ex machina to explain what was intended.
Round and round we go...
