Making a polygon using equilateral triangles and squares.

You are given some equilateral triangles and squares, all with side length 1. You have to make a convex polygon with side $n$, if both square and equilateral triangle must be use, what is all the possible value for $n$? Assume there is sufficient supply of the pieces.

For above problem, my idea is to use some Diophantine equation, when an equilateral triangle and square meet, they will formed a 150 degree angle, also if 2 equilateral triangles meet, they will formed a 120 degree angle, so we will get 2 information, assume there are $x$ 120 degree angle, and $y$ 150 degree angle, so the informations are : $60x+30y=360$ (exterior angle sum) $180|120x+150y$ (interior angle sum) By that I successfully created the polygon for $(x,y)=(0,12),(2,8),(3,6),(4,4)$, but I don’t manage to create polygon for $(x,y)=(1,10),(5,2),(6,0)$, and I also can’t prove that we can’t make a polygon using that value of $(x,y)$. Can anyone help me ?

Denote the convex polygone by $A_1A_2...A_n$ and the angle $\alpha_i$ between two vectors $\overrightarrow{A_{i-1}A_{i}}$ and $\overrightarrow{A_iA_{i+1}}$ for $i=1,..,n$ (we assume that $A_{-1}=A_{n}$ and $A_{n+1} = A_1$), then $$\sum_{i=1}^n \alpha_i =360°$$

The possible value of $\alpha_i$ is $30°$(1 square, 1 triangle), $60°$ (2 triangles), $90°$ (1 square) or $120°$ (1 triangle), denote $x,y,z,t$ the numbers of vertex with these 4 angles, then: $$30x+60y+90z+120t = 360\iff x+2y+3z+4t=12\tag{1}$$ From $(1)$, it's easy to deduce that $4\le n = x+y+z+t \le 12$.

Really by chance, I have some magnetic blocks, here are somes convex polygons :). All values of $n$ from $5$ to $12$, except $n =4, 11$ are in the photo.



For $n = 4$, we have $(x,y,z,t) = (0,2,0,2)$ then no square is used. So, $n$ cannot be $4$.

For $n = 11$, we show that it is not possible to contruct the convex polygon if all edges having length exactly equal to $1$ (solutions do exist if some sides are allowed to be two units, see the answer of @OscarLanzi). Suppose the contrary, then $(x,y,z,t) =(10,1,0,0)$.

The vertex corresponds to the angle $\alpha = 60°$ is formed by two triangles, the other vertex are formed by $1$ triangle and $1$ square. Therefore, there is one possibility to contruct the edges polygone as follows: $$...-\color{blue}{S}-\color{red}{T}-\color{blue}{S_3}-\color{red}{\mathbf{T_1}}-\color{red}{\mathbf{T_2}}-\color{blue}{S_1}-\color{red}{T_3}-\color{blue}{S_2}-...$$



Let's study the inner-polygon.

The outer-convex-polygon is well defined based on the inner-polygon. But we can only obtain a decagon (as in the left of the first photo). We conclude then it is not possible to have a hendecagon of all edges having length exactly equal to $1$.

We conclude that:

If all sides are of unit length, then $n$ can receive values in $\{5,6,7,8,9,10,12\}$.

If sides of length two or more areallowed, then $n$ may also be $11$.

If we construct sides of two or more units (there are different interpretations about whether this is allowed), we actually can render a hendecagon:



By placing pairs of squares together along the outer boundary we make the perimeter 13 but two sides having two units instead of one, removing the apparent contradiction.

If $n$ is even (greater than 4), we can create a line with $\frac{n-4}{2}$ squares, and add one triangle at each end of this line.

And if $n$ is odd, we can create a line with $\frac{n-3}{2}$ squares, and add one triangle at one end of this line.