Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x, y,$ and $z$ are each chosen from the set $\{ 0, 1, 2\}$. How many equilateral triangles have all their vertices in $S$? $\mathrm {(A) } 72 \qquad$ $ \mathrm {(B) } 76 \qquad$ $ \mathrm {(C) } 80 \qquad$ $ \mathrm {(D) } 84 \qquad$ $ \mathrm {(E) } 88$
Solution
Each coordinate difference between two points of $S$ is $0,\pm1$, or $\pm2$, so the squared side length is $s=a^2+b^2+c^2$ with $a,b,c\in\{0,1,2\}$ in absolute value. For an equilateral triangle with side vectors $u,v$ from one vertex, $|u-v|^2=s$ gives $u\cdot v=\frac s2$. Since $u\cdot v$ is an integer, $s$ is even, so $s\in\{2,4,6,8,12\}$.
\emph{$s=4$ and $s=12$ are impossible.} If $s=4$ the side vectors are $(\pm2,0,0)$ and its permutations; two such vectors that are not opposite differ by a vector with two entries $\pm2$, of squared length $8\ne4$. If $s=12$ the side vectors are $(\pm2,\pm2,\pm2)$; two different ones differ by $\pm4$ in some coordinate, which is impossible in $S$.
\emph{$s=2$ (sides $\sqrt2$).} The side vectors are $(\pm1,\pm1,0)$ and permutations, so in each coordinate the three vertices differ by at most $1$: all three lie at corners of one unit cube $[i,i+1]\times[j,j+1]\times[k,k+1]$, and the cube is determined by the triangle. A unit cube has $8$ corners, forming two regular tetrahedra (the corners of the same parity) whose faces are the equilateral triangles with side $\sqrt2$: $2\cdot4=8$ triangles per unit cube. There are $2\cdot2\cdot2=8$ unit cubes, so $8\cdot8=64$ triangles.
\emph{$s=8$ (sides $2\sqrt2$).} The side vectors are $(\pm2,\pm2,0)$ and permutations. Then in each coordinate the three vertices differ pairwise by $0$ or $\pm2$. No coordinate can be the same for all three vertices: the triangle would lie in a plane, where three corners of a $2\times2$ square always include two at distance $2$. So each coordinate takes the two values $a$ and $a+2$ with $a=0$, i.e. the vertices have coordinates in $\{0,2\}$. They are corners of the big cube $[0,2]^3$, and exactly as in (iii) there are $2\cdot4=8$ such triangles.
\emph{$s=6$.} Here $6=4+1+1$, so each side vector is a permutation of $(\pm2,\pm1,\pm1)$: each side has exactly one coordinate in which its endpoints differ by $2$, namely $0$ and $2$. These coordinates are different for the three sides (if two sides had the $2$ in the same coordinate, the third side would have $0$ there, which is not allowed). Name the vertices $U,V,W$ so that the side $VW$ has its $2$ in $x$, $WU$ in $y$, $UV$ in $z$. Then $U_x=1$ (it differs by $1$ from both $0$ and $2$), $V_y=1$, $W_z=1$. The remaining choices are free: $\{V_x,W_x\}=\{0,2\}$, $\{W_y,U_y\}=\{0,2\}$, $\{U_z,V_z\}=\{0,2\}$, each in $2$ ways, and all three sides then have squared length $4+1+1=6$. This gives $2\cdot2\cdot2=8$ triangles.
$\mathrm {(A) } 72 \qquad$ $ \mathrm {(B) } 76 \qquad$ $ \mathrm {(C) } 80 \qquad$ $ \mathrm {(D) } 84 \qquad$ $ \mathrm {(E) } 88$ 