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 reply to re: Generic answer (Spoiler) by Daniel)

Daniel: I have the same question about your solution, as that given by Leming: you have ten lines of three each, but you ALSO have other lines which have only two balls, e.g. along the left side from upper left to lower left.  This "works" only on the interpretation that you may have 10 OR MORE lines in your diagram.  Your interpretation (and Leming's) may or may not be what the propounder intended.

If we label your diagram as

ABCDE

FGHIJ

KLMNO

Then lines through AK, EO, AI, EG, KI, and GO are lines with exactly TWO balls, not three.  Regarding interpretation, generally I would suppose that a good puzzle (not in the "jokes" category) should have only a single solution (perhaps in a case like this allowing for rotational symmetries as the same solution). Only pcb can say what he intended, I guess.

 

Posted by ed bottemiller on 2008-10-13 18:17:39