If we assume there is such a function $P(a,b,c,d)$, how could it look like?
$P$ should be Linear
The perimeter should be $0$ if every length is $0$, and the perimeter should be twice as long if every side is twice as long. It means that $P$ is linear, and can be written:
$$P(a,b,c,d) = ma + nb + oc + pd$$
$P$ should be symmetrical
The perimeter shouldn't change if we consider the mirror image of the hexagon. It means that :
$$P(a,b,c,d) = P(d,c,b,a)\\
\implies \begin{cases}
p = m\\
o = n
\end{cases}\\
\implies P(a,b,c,d) = m(a + d) + n(b + c)$$
$P$ should be valid for any equiangular hexagon
In particular, $P$ should be valid for a regular hexagon:
$$P(a,a,a,a) = 6a\\
\implies 2ma + 2na = 6a\\
\implies m + n = 3\\
\implies P(a,b,c,d) = (3-n)(a + d) + n(b+c)$$
$P$ should be valid for very small $a$ and $d$

If we let $a$ and $d$ tend to $0$ and keep $b=c$, the hexagon will tend to a rhombus. It means that
$$P(0,b,b,0) = 4b\\
\implies n(b+b) = 4b\\
\implies n = 2\\
\implies m = 1\\
\implies P(a,b,c,d) = a + 2(b+c) + d \\
$$
3(a+d)
? Never been good at writing proofs, but my (rusty) intuition makes me suspect that the sum of opposite sides (a+d, b+e, c+f) is always the same, and sincea,b,c,d
omitse
andf
,a+d
is the only pair of opposite sides that we can make. $\endgroup$