Chord partition of regular polygon: same fraction of area and perimeter? 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?
 A: 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. 
