Hexagonal rooks Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$.  A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.
What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?
More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?
 A: A visualization aid, $n=10$:
          

And now here are Will Swain's rook placements:

A: For n=6 you can fit 5 rooks
(0,2,4)
(4,0,2)
(1,4,1)
(3,3,0)
(2,1,3)
For n=9 you can fit 7 rooks
(0,3,6)
(6,0,3)
(2,6,1)
(4,5,0)
(3,1,5)
(5,2,2)
(1,4,4)
A: This has a connection with additive permutations, asked about here:
Are there enough additive permutations? .
http://oeis.org/A002047 has references to a similar problem involving a hexagonal board and counting the number of placements. (The Bennett and Potts reference calls the piece a 'brook'.) This number also counts the permutations.  A recent ArXiv paper http://arxiv.org/abs/1510.05987 appears to provide an asymptotic estimate.  Note that placement on a hexagonal board gives one for the triangle, but there may be placements in a triangle corner that are not realizable in the hexagon.
Gerhard "Cutting Corners For More Results" Paseman, 2017.03.13
A: Nice question!
For the maximum number of pairwise non-defending rooks,
Will Sawin proved an upper bound of $(2n/3) + 1$
in his comment to the original question.  This bound is attained,
at least to within $O(1)$, by two rows of $n/3 - O(1)$ rooks each,
starting from around $(2n/3,n/3,0)$ and $(n/3,2n/3,1)$
and proceeding by steps of $(-1,-1,2)$ until reaching the
$y=0$ or $x=0$ edge of the triangle.  This construction
generalizes Sawin's five-Rook placement for $n=6$.
On further thought, it seems we actually achieve
$\lfloor (2n/3) + 1 \rfloor$ exactly for all $n$.
Here's how it works for $n=12$ and $n=15$, 
with $(2n/3)+1 = 9$ and $11$ respectively:
                                            .   
                                           . . 
                                          . . . 
            .                            . . . . 
           . .                          . . . . . 
          . . .                        R . . . . . 
         . . . .                      . . . . . . R 
        R . . . .                    . R . . . . . . 
       . . . . . R                  . . . . . . . R . 
      . R . . . . .                . . R . . . . . . . 
     . . . . . . R .              . . . . . . . . R . . 
    . . R . . . . . .            . . . R . . . . . . . . 
   . . . . . . . R . .          . . . . . . . . . R . . . 
  . . . R . . . . . . .        . . . . R . . . . . . . . . 
 . . . . . . . . R . . .      . . . . . . . . . . R . . . . 
. . . . R . . . . . . . .    . . . . . R . . . . . . . . . . 

Starting from such a solution with $n=3m$, we can add an empty row 
to get an optimal solution for $n=3m+1$, and remove an edge
(and the Rook it contains) to get an optimal solution for $n=3m-1$.
So this should solve the problem for all $n$.
Jeremy Martin also asks:

More generally, is anything known about the graph whose vertices
  are these ordered triples and whose edges are rook moves?

I don't remember reading about this graph before.
Experimentally (for $3 \leq n \leq 16$) its adjacency matrix
has all eigenvalues integral, the smallest being $-3$ with huge multiplicity
$n-1\choose 2$; more precisely:

Conjecture. For $n \geq 3$ the eigenvalues of the adjacency matrix are:
  a simple eigenvalue at the graph degree $2n$; a $n-1\choose 2$-fold
  eigenvalue at $-3$; and a triple eigenvalue at each integer
  $\lambda \in [-2,n-2]$, except that $\mu := \lfloor n/2 \rfloor - 2$
  is omitted, and $\mu - (-1)^n$ has multiplicity only $2$.

This is probably not too hard to show.  For example, the $\lambda = -3$
eigenvectors constitute the codimension-$3n$ space of functions
whose sum over each of the $3(n+1)$ Rook lines vanishes.
[Added later: in the comment Jeremy Martin reports that
he and Jennifer Wagner already made and proved the same conjecture.]
Given that the minimal eigenvalue is $-3$,
it follows by a standard argument in "spectral graph theory"
that the maximal cocliques have size at most $3(n+1)(n+2)/(4n+6) = 3n/4 + O(1)$.
But that's asymptotically worse than $2n/3 + O(1)$, though it's still
good enough to prove the optimality of Will Sawin's cocliques of size
$5$ for $n=6$ and of size $7$ for $n=9$.
Here's some gp code to play with this graph and its spectrum:
{
R(n)=
  l = [];
  for(a=0,n,for(b=0,n-a,l=concat(l,[[a,b,n-a-b]])));
  matrix(#l,#l,i,j,vecmin(abs(l[i]-l[j]))==0) - 1
}

running "R($n$)" puts a list of the vertices in "l" and returns
the adjacency matrix with the corresponding labeling.  So for instance
matkerint(R(7)-2)~
matkerint(R(8)-1)~

returns matrices whose rows are nice generators of the
$2$-dimensional eigenspaces of the $n=7$ and $n=8$ graphs.
A: Here is a paper about this problem: "Non-attacking queens on a triangle".
And here's another one "Putting Dots in Triangles"
