For a convex $n$-gon $P_1P_2\cdots P_n$, let $M_i$ be the mid-point of the line segment $P_iP_{i+1}\ (i=1,2,\cdots,n)$ where $P_{n+1}=P_1$. Also, let $Q_1Q_2\cdots Q_n$ be an inner $n$-gon made by $n$ line segments $P_iM_{i+1}\ (i=1,2,\cdots,n)$ where $M_{n+1}=M_1$.
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Letting $S,S^{\prime}$ be the area of $P_1P_2\cdots P_n,Q_1Q_2\cdots Q_n$ respectively, then here is my question.
Question : What are the min and the max of $\frac{S^{\prime}}{S}$ ?
Motivation : I've just been able to solve the $n=4$ case. The answer for $n=4$ case is $$\begin{align}\frac 16\lt \frac{S^{\prime}}{S}\le \frac 15\qquad(\star)\end{align}$$ However, I'm facing difficulty for $n$ in general. Can anyone help?
Proof for $(\star)$ : Since $M_1M_2M_3M_4$ is a parallelogram, we may suppose that $M_1M_2M_3M_4$ is a square by the affine transformation for $P_1P_2P_3P_4$. Clearly, we know that two diagonals $P_1P_3$ and $P_2P_4$ are orthogonal to each other and that they have the same length. Then, letting $\overline {P_1P_3}=\overline {P_2P_4}=4$, let us take the center of $M_1M_2M_3M_4$ as the origin. Suppose that $\overline {AB}$ represents the Euclidean length of the line segment from $A$ to $B$. We may suppose the followings : $$\overline{M_1M_2}=2, M_1(1,1),M_2(-1,1),M_3(-1,-1),M_4(1,-1).$$ Letting $P(x,y)$ be the intersection of $P_1P_3$ and $P_2P_4$, we get $$P_1(2-x,y),P_2(x,2-y),P_3(-2-x,y),P_4(x,-2-y).$$ Note that $-1\lt x\lt 1,-1\lt y\lt 1.$
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Then, letting $Q_1,Q_2,Q_3,Q_4$ be the intersection point of $P_1M_2, P_4M_1$ and $P_2M_3, P_1M_2$ and $P_3M_4,P_2M_3$ and $P_4M_1,P_3M_4$ respectively, $$S^{\prime}=|Q_1Q_2Q_3Q_4|=|M_1M_2M_3M_4|-|Q_1M_1M_2|-|Q_2M_2M_3|-|Q_3M_3M_4|-|Q_4M_4M_1|$$$$=4-\left\{\frac{2(15-4xy-3y^2)}{25-(2x-y)^2}+\frac{2(15+4xy-3x^2)}{25-(x+2y)^2}\right\}$$ because $$|Q_1M_1M_2|=\frac{3-2y-y^2}{5-2x+y},|Q_3M_3M_4|=\frac{3+2y-y^2}{5+2x-y},$$$$|Q_2M_2M_3|=\frac{3+2x-x^2}{5-2y-x},|Q_4M_4M_1|=\frac{3-2x-x^2}{5+2y+x}.$$ Suppose that $|F|$ represents the area of a figure $F$. Here, using $u,v$ such that $x=2u+v,y=2v-u$, we can get $$\frac{S^{\prime}}{S}=\frac 15\left\{1-\frac{(u^2-v^2)^2}{(1-u^2)(1-v^2)}\right\}$$ where $|2u+v|\lt 1,|2v-u|\lt 1$.
Since we can easily prove that if $|2u+v|\lt 1,|2v-u|\lt 1$, then $$0\le \frac{(u^2-v^2)^2}{(1-u^2)(1-v^2)}\lt\frac 16,$$ we now know that the proof is completed.