Dlanod is building huts

The best that can be done is $\boxed{6}$ huts.

Put points $A$ and $E$ into corners of the plot, $3$ m away from the plot edges. Then place point $C$ a distance of $18$ m from both $A$ and $E$ . Extend the lines $AC$ and $EC$ to place huts $B$ and $F$ so that $ABFE$ is a rectangle with $C$ at its centre. Then create triangle $BDF$ congruent to $ACE$ . Note that $AB = CD = EF = 6\sqrt{11} > 18$ , while $9\sqrt{11} = 29.8496 < 30$ , and so there is just room to build huts with centres at $A,B,C,D,E,F$ .

This set-up has some huts exactly $12$ m from each other. However, Dlanod has some $16$ cm of slack in the the placement of point $D$ , and so he could ease this construction slightly, perhaps constructing point $C$ at distance of $18.02$ m from $A$ and $E$ , and satisfy the building regs. Or else see below!

For integral distances to the boundary, keep $A$ and $E$ where they are. Move $B$ and $F$ a little to the right, so that $AB = EF = 20$ m (the previous value of $AB$ was $6\sqrt{11} = 19.8997$ m). Place $D$ midway down the right-hand side, a distance $3$ m from the edge, and have $C$ on the same level, with $CD = 20$ m. Then $A,B,D,E,F$ are all $3$ m from the boundary, while $C$ is $13$ m from the boundary. Also $AC = BC = EC = FC = 5\sqrt{13} = 18.0278$ m

Mark Hennings has written a careful solution showing how we can build six huts on the plot. To summarize: We can place the centers of the huts at the points $(13,3),(33,3), (3,18), (23,18), (13,33), (33,33)$ , with the origin in the lower left corner. Any two of these points will be at least $\sqrt{325}>18$ meters apart, which implies that the distance between the huts will be more than 12 meters as required.

But there is more work to be done: We need to convince Mr Dlanod that he will not be able to build seven huts (or more) to these specifications. Rather than proving this fact from first principles, which is quite tedious, we will relate this problem to another, better understood question that turns out to be equivalent: Can we fit seven non-overlapping circles of radius 9 into a $48\times 48$ meter square (we are moving each side of the plot out by 6 meters). Indeed, if we have a solution to the original problem, then we can solve the new problem by enlarging the radius of all the huts from 3 to 9, and vice versa.

The problem of whether one can fit $n$ non-overlapping circles of radius $r$ into a square with side-length $a$ has long been studied and is well understood, at least for small values of $n$ . In particular, it turns out that seven circles can be packed into the square if $\frac{a}{r}\geq 4+\sqrt{3}$ . In our case we have $\frac {a}{r}=\frac{48}{9}< 4+\sqrt{3}$ , so that it is impossible to pack 7 circles of radius 9 into a square with side-length 48.

Thus the largest number of huts that can be built is $\boxed{6}$ .