More Rows of Checkers
A B C
D E F G
H I J K L
M N O P
Q R S
Consider a hexagonal arrangement of 19 circles, like the configuration shown.

Can you place 10 checkers on the hexagon such that there are
2 checkers in each row in each of the three primary directions?

Can you place 9 black and 9 white checkers on the hexagon (leaving only
one location empty) such that no more than 2 checkers of the same color
are in any row in each of the three primary directions? Or can you
prove it cannot be done? (If it can't be done, what is the largest
number of checkers you can place under these conditions?)
Source: Original, though I wouldn't be surprised to hear of another source.
Solutions were received from:
Philippe Fondanaiche,
Al Zimmermann,
Wackychad@aol.com,
Erich Friedman.
Al Zimmermann found all possible solutions to the first problem, showing
that there are only three solutions when you disregard rotation:
Group 1:
A B D G J L M P Q R
B C D G H J M P R S
Group 2:
A C D E K L M P Q R
A B E G H L N P Q R
A B D G K L M N Q S
B C D G H I O P Q S
B C D F H L M O R S
A C F G H I M P R S
Group 3:
B C D E H L O P Q R
A C D G I K M P Q S
A B F G H L M N R S

A B .
D . . G
. . J . L
M . . P
Q R .

A . C
D E . .
. . . K L
M . . P
Q R .

. B C
D E . .
H . . . L
. . O P
Q R .


For the second problem, several people sent proofs that a solution cannot be
found with 9 black and 9 white checkers. The following solution is from
Philippe Fondanaiche:
Preliminary comment: it is easy to check that there is no possible solution
with a checker placed at the center J. [KD: if a checker is at J, there
are three rows of 5 through J which would need to have a blank space to allow
only four checkers in the row, but this wouldn't allow all 18 checkers to
be placed.]
Let consider the triangle DGR.There are two possible configurations:

There are 3 identical checkers at the vertices D, G, R. So there are
identical checkers of opposite color at E, F, I, K, N, O and it is impossible
to place other checkers at the other points.

There are 2 black (or white) checkers and 1 white (or black) one, for
example black at D and G, white at R. So white at E and F.
There are again two possible configurations at B:

if white at B, black at I, K, M, P, then white at N and O, then black
at A, S, C, Q, then white at H and L ==> 10 black and 8 white checkers.

if black at B, there again two possibilities:

if black at M, then white at I, then black at N, then white at O
and P, then black at K, A, S, then white at H, then black at L, then white at
C, at last black at Q ==> 10 black and 8 white checkers again.

if white at M, black at I, then white at N, then black at O and P,
then white at K, then black at C and Q, then white at A, then black at S, then
white at L, at last black at H ==> 10 black and 8 white checkers again.
In conclusion 17 checkers can be placed at a maximum.
[KD: Actually, what
I'd intended was the proof (as above) that you could place 18 checkers, but
only if you have 10 of one color and 8 of the other. These can be found
in the two configurations above where J is left empty (Groups 2 and 3).]
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