Anyway, I described to you the basis for a parity problem with the pieces
from the Triominoes game.  Here is a summary of the rules for these pieces:

56 triangular tiles, with a number from 0 to 5 at each point.
All 6 possible triples and 30 possible doubles exist.  For the tiles
with 3 different numbers, only one of the two mirror images exists for
each of the 20 possible combinations of numbers; except for 134 and 234,
these always exist in the form where the numbers increase in a clockwise
direction.  These pieces are always played in a triangular grid, with
the numbers at all adjacent points matching.

What convex shapes can be made using all 56 Triominoes, following the
usual matching rule?

My first parity observation dealt with the edges.
An initial count of the edges which increase in a clockwise or
counterclockwise direction around the tile notes a discrepancy of
16 edges between the clockwise and counterclockwise edges.
be clockwise and one must be counterclockwise, or both must be
"double" edges.

A more detailed analysis considers the parity for each type of edge,
since an 01 edge cannot be adjacent to an 05 edge regardless of the
parity.  Here, we consider only the edges on the 20 pieces with
3 different numbers, because each double has two edges with the same
pair of numbers in both orders, and thus they always balance.
In the table, the numbers in the middle two columns are the other
which appear on a piece having the specified edge in the specified
orientation (clockwise [cw] or counterclockwise [ccw]).

edge cw    ccw    difference
01   2345         4
02   345   1      2
03   45    12     0
04   5     123    2
05         1234   4
12   0345         4
13   05    24     0
14   035   2      2
15   0     234    2
23   015   4      2
24   0135         4
25   01    34     0
34   05    12     0
35   012   4      2
45   0123         4

total difference: 30

As a result, any figure made using all 56 Triominoes must have a perimeter
of at least 30, since there are that many "extra" edges that can't be
matched up with other edges.  This eliminates all possible convex figures
except the 2-unit-wide parallelogram and trapezoid, and the 1-unit-wide
parallelogram, and the 2-wide figures are very close to the limit
(perimeter 32).

After this, though, I noticed another issue that, all at once, eliminates
all possible convex figures.  Each number appears 28 times in the set
of Triominoes.  Note that at each vertex in a completed figure, the
pieces which meet must all have the same number at that vertex.  As a
result, in a convex figure, most vertices have either 6 like points
(interior vertices) or 3 like points (vertices on straight portions of
edges).  Only the vertices at the corners of the entire figure can have
non-multiple-of-3 numbers of like vertices.  There are at most 6 of these,
but each normally has 2 like points, so you need two of these corners to
make the points for a single number to come out to 28 (= 1 mod 3).
You can have some 1-point corners if your figure has less than 6 sides
(and thus less than 6 corners) but only one for each side less than 6.
Thus, at most you can balance 3 of the 6 numbers.  If you removed one
piece with 3 different numbers on it, you might be able to make a
convex figure with the other 55 pieces, with all of the numbers at
the figure's corners being the 3 numbers not on the piece removed.
(The edge analysis tells us such a shape would still need to be
at most two units wide, since even the best choice of piece to remove
leaves us needing a perimeter of at least 27.)

You could also use these results to find other interesting shapes you might
be able to make from the Triominoes.  A hexagonal shell one unit thick (not
a regular one, though) could work; a two-unit-wide one doesn't have enough
perimeter (including the inside perimeter).  Two separate hexagons
could work, if they have enough perimeter; a regular 2-unit hexagon plus
another similar hexagon stretched one unit in one direction don't have
enough combined perimeter, but some skinnier ones could work.