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.

Wherever two tiles are adjacent, one of the adjacent edges must

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.