This is a variation of a question posed by James Tanton on Twitter.
Let $P$ be a regular $n$-gon, $n \ge 3$. A chord $c$ of $P$ is a segment connecting two distinct points of the boundary of $P$, on two distinct edges (so not two points on one edge). Such a chord partitions both the perimeter and area of $P$ into two non-zero parts. For a given chord $c$, let $a(c)$ and $p(c)$ be the smaller area fractions and smaller perimeter fractions of the parts. In other words, $a(c)$ is the area to the smaller side of $c$ divided by the area of $P$. And similarly for $p(c)$.
Q. For which $n$-gons do there exist chords $c$ and fractions $a(c) = p(c)$ not equal to $\frac{1}{2}$?
Tanton's question asks for $n=4$ (a square), can $\frac{1}{3}$ be achieved? "Simultaneously divide off one-third of the area of a square and one-third of the perimeter of the square?" And the answer is No.
(Image from Tanton.)
The generalization asks for all those non-half fractions achievable for regular $n$-gons. Perhaps the answer to Q is: For none?
Partial answer: no good chord for the square (with any ratio different from $\frac{1}{2}$).
Proof: Assume that we connect two points on opposite sides. Then we obtain a trapezoid. Denote the bases by $x,y$. Then $a(c)= \frac{x+y}{2}$ and $p(c)=\frac{x+y+1}{4}$, which yields $x+y=1$, and thus $a(c)=p(c)=\frac{1}{2}$.
If the two points are on adjacent sides, then we obtain a right triangle. Let $x,y$ denote the legs this time. Then $a(c)=\frac{xy}{2}$ and $p(c)=\frac{x+y}{4}$, which yields $(x-\frac{1}{2})(y-\frac{1}{2})=\frac{1}{4}$. As $0\leq x,y\leq 1$, we have $|x-\frac{1}{2}|\leq \frac{1}{2}$ and $|y-\frac{1}{2}|\leq \frac{1}{2}$, thus $(x-\frac{1}{2})(y-\frac{1}{2})=\frac{1}{4}$ is possible only if $|x-\frac{1}{2}|=|y-\frac{1}{2}|=\frac{1}{2}$. Hence, either the triangle is degenerate (a point), or $x=y=1$, in which case $a(c)=p(c)=\frac{1}{2}$ again.
The fact that we needed very different arguments in the two cases suggests that the general question might be hard to handle.