How many pentagons can you make with four corners of $120^\circ$ and five sides whose lengths are consecutive integers (but not necessarily in order)? I’m trying to prepare for a national math competition, but I’m stuck at this and can’t figured anything out. The challenge is to prove your answer.

How many pentagons can you make with four corners of $120^\circ$ and five sides whose lengths are consecutive integers (but not necessarily in order)?

I’ve only come as far as to realize that the shape will be very similar to a hexagon, and that you will probably need to use the Pythagorean theorem to solve it. 
Thanks in advance!
 A: Let the side-lengths be $a$, $b$, $c$, $d$, $e$, as shown in the figure:

We see that we must have
$$\begin{align}
a+b &= d+e \tag{1} \\
a+e &= b+2c+d \tag{2}
\end{align}$$
Let $m$ be the minimum side-length, and define $a^\prime := a-m$, etc. Then $\left(a^\prime, b^\prime, c^\prime, d^\prime, e^\prime\right)$ is some permutation of $\left(0, 1, 2, 3, 4\right)$, so that $a'+b'+c'+d'+e'=10$. We can re-write $(1)$ and $(2)$ as
$$\begin{align}
a^\prime + b^\prime &= d^\prime + e^\prime \tag{1'} \\
2m &= 3a'+b'+d'+3e'-20\tag{2'}
\end{align}$$
Now, $(1')$ has limited solutions, arising from permuting the terms and sides of $0+3=1+2$, $0+4=1+3$, and $1+4=2+3$; very few of these give rise to feasible values of $m$. There aren't unreasonably-many cases to check, but a simple observation can save some work: If neither $a'$ nor $e'$ is $4$, then $3a'+b'+d'+3e'$ is at most $3\cdot 3+3\cdot 2+4+1=20$, and even this value is unattainable in light of $(1')$; but that sum must be at least $20$ for a valid $m$ by $(2')$, so we must have that either $a'$ or $e'$ is $4$. This leaves the following solutions:
$$\begin{align}(a',b',d',e')\;\text{or}\;(e',d',b',a') 
&= (4,0,1,3) \quad\to\quad m = \phantom{-}1 \quad\checkmark \\
&= (4,0,3,1) \quad\to\quad m = -1 \\
&= (4,1,2,3) \quad\to\quad m = \phantom{-}2 \quad\checkmark \\
&= (4,1,3,2) \quad\to\quad m = \phantom{-}1 \quad\checkmark
\end{align} \tag{3}$$
Consequently, there are six pentagons with sides $(a,b,c,d,e)$.

$$\begin{align}
(5,1,3,2,4) \quad (6,3,2,4,5) \quad (5,2,1,4,3) \\
(4,2,3,1,5) \quad (5,4,2,3,6) \quad (3,4,1,2,5)
\end{align} \tag{$\star$}$$

If reflections are ignored, there are only three (one from each column of $(\star)$). $\square$
A: Let $a$, $b$, $c$ be consecutive sides not adjacent to the pentagon's $60^\circ$ angle, as shown:

By appending a side-$b$ equilateral triangle, the pentagon becomes a parallelogram, and we see that the pentagon's remaining sides must be $a+b$ and $b+c$. As in my previous answer, we take $m$ to be the smallest side (necessarily, the smallest of $a$, $b$, $c$) and define $a^\prime := a-m$, $b^\prime := b-m$, $c^\prime := c-m$. Then the five sides, reduced by $m$, are
$$( a^\prime, \;b^\prime, \;c^\prime, \;a^\prime+b^\prime+m, \;b^\prime+c^\prime+m) \tag{1} $$
which must be a permutation of $(0,1,2,3,4)$. Note that $0$ must be one of the first three terms; and, since two values are larger than $m$ (which itself is non-zero), we must have that $m=1$ or $m=2$. Moreover, if $m=2$, then $b^\prime=0$, as there is no other way for the last two values to be less than $5$.
Thus, we have the following possibilities:
$$\begin{array}{cc:ccccc:l}
m=1 & a^\prime = 0 & 0 & b^\prime & c^\prime & b^\prime+1 & b^\prime+c^\prime+1
& (0,2,1,3,4) \\
    & b^\prime = 0 & a^\prime & 0 & c^\prime & a^\prime+1 & c^\prime+1 
& (1,0,3,2,4)\;(3,0,1,4,2) \\
    & c^\prime = 0 & a^\prime & b^\prime & 0 & a^\prime+b^\prime+1 & b^\prime+1 
& (1,2,0,4,3)\\
\hline
m=2 & b^\prime = 0 & a^\prime & 0 & c^\prime & a^\prime+2 & c^\prime+2 
& (1,0,2,3,4)\;(2,0,1,4,3)
\end{array}\tag{2}$$
In addition to the obvious $a^\prime \leftrightarrow c^\prime$ symmetry, I suspect there may be ways to simplify the consideration of cases. In any event, we arrive at the same six results as in my previous answer (with sides cyclically permuted):

$$\begin{align}
(1,3,2,4,5) \quad (3,2,4,5,6) \quad (2,1,4,3,5) \\
(2,3,1,5,4) \quad (4,2,3,6,5) \quad (4,1,2,5,3)
\end{align} \tag{$\star$}$$

As before, if reflections are ignored, then there are three solutions (one from each column of $(\star)$). $\square$ 
