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.

           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+
           |   |   |   |   |   |   |   |   |
           +---+---+---+---+---+---+---+---+

In my recent post I listed three cases where DejMar's proposed solution had lines with only two balls each.

The last of these was a typo: instead of (5,4) and (7,5) the last should have been (5,9) and (7,5).

Also, two other pairs had only two balls, viz. a line thru (3,4) and (5,9); and a line thru (3,4) and (7,5).

In comment 8, DejMar states he has found 88 "possible" solutions to this problem. (Does this mean "possible" depending upon how one interprets the specs?) Do any of them use the nine balls in such a way that all lines are considered and there are exactly ten lines of exactly three balls each?  If so, I think we would all agree that would be a solution.

Posted by ed bottemiller on 2008-10-14 14:15:55