TITLE: Can all convex $3n$-iamonds be tiled by $3$-iamonds?
QUESTION [9 upvotes]: Background
A polyiamond is a plane figure constructed by joining together equilateral triangles of the same size along their edges.
The number of convex polyiamonds is given by A096004.
Based on enumerating examples up to $n=8$, it looks plausible that all $A096004(3n)$ convex $3n$-iamonds can be tiled by the $3$-iamond (i.e. the triamond).


Question
If the conjecture is true, I'm curious to see a proof (or even some heuristics). If the conjecture isn't true, I'd be curious to see a counterexample.
(It's perhaps worth noting that not all convex $2n$-iamonds can be tiled by the $2$-iamond.)

REPLY [3 votes]: Here is another argument to show that indeed, a convex polyiamond with $3n$ cells is tileable by 3-bars.
Any convex polyiamond is really a "hexagon" with side lengths $a$, $b$, $c$, $d$, $e$ and $f$, some of which may be 0 (so it may in fact be a pentagon, quadrangle, or triangle). This is because any convex polygon is the intersection of half-planes, and there are only 6 orientations of half-planes in a triangular grid.

Here is an example with $b = 0$:


If both sides of any pair of opposite sides have length 3 or more, we can cut off a tileable strip (where each row has $6$ triangles, and is tileable by two bars), and get a smaller convex "hexagon".



If both sides of any pair of sides with another side in between (for example $a$ and $c$) have length 3 or more, we can cut of a tileable "trapezium" (possibly with one side 0 to give a triangle), and get a smaller convex "hexagon".

A trapezium with 3 rows is tileable as shown in this figure (Peter Taylor also uses this idea in his analysis):


Suppose we have a "hexagon" that does not satisfy one of the two conditions above. Then at last 4 adjacent sides must have length 2 or shorter. But then, the figure must lie within a hexagon with all sides 2, and so the maximum area of such a figure is $24 = 3\times 8$, and all such figures is tileable by enumeration (according to OP), and so any convex polyiamond can be partitioned into a set of strips, trapeziums, and a polyiamond with area 24 or less, all of which are tileable, and hence, so is the original polyiamond.