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.
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
| | | | | | | | |
+---+---+---+---+---+---+---+---+
I see from brianjn's posting that ambiguities of this sort have beset previous problems. I am glad to read I am not the only one who is bothered by these. In the linked problem (about tree planting in "rows") I encounted "7 lines intersect, but only 2 lines intersect at a point" which almost tempted me to look up the definition of an intersection). It seems that (without spec in the that problem statement) perhaps "rows" is to be taken as horizontal lines, not as all lines. Perhaps likewise there is a restriction on "when is a line not a line" in the current problem.
DejMar's diagram at first seemed as though he may have solved the problem as I interpret it (10 and only 10 lines), but then it seems he has disregarded several lines which would be determined by pairs of his points. If one labels the grid (of points, not cells) from (1,1) as the upper left corner, what about:
Line thru (1,5) and (3,1)
Line thru (1,5) and (5,6)
Line thru (5,4) and (7,5)
These lines have only two balls each.
Interpretation redivivus
