Total area for a natural nested set of convex polygons. Suppose we have a convex polygon $P_0$ with $n$ given vertices, and we want to "nest" polygons $P_j$ for $j > 0$ by taking the midpoints between edges of $P_{j-1}$ as the vertices. For a regular polygon the total area of all the polygons will form a geometric series that is pretty easy to solve for (at least in terms of trig functions) in terms of the area $A$ of the original polygon $P_0$. However, what if $P_0$ is not regular? If we know the area of $P_0$, is there a formula for the sum of areas of all the nested polygons, in terms of $n$ and the original area? Or do we need more information, like the vertices of $P_0$? If we need to know the vertices, are there formulas in terms of the vertices at least for some small values of $n$, like $n=3,4$?
 A: Let the vertices of the polygon be $\mathbf p_1, \mathbf p_2, \ldots, \mathbf p_n$ in counterclockwise order, and for convenience define $\mathbf p_{n+1}=\mathbf p_1, \mathbf p_{n+2}=\mathbf p_2,$ and so on. Then the (signed) area of the polygon is given by
$$A=\frac12\sum_{i=1}^n(\mathbf p_i\wedge\mathbf p_{i+1}),$$
where $\mathbf p\wedge\mathbf q=\begin{vmatrix}p_x & q_x\\p_y & q_y\end{vmatrix}=p_xq_y - p_yq_x$. Note that $\mathbf p\wedge\mathbf p=0$.
The nested polygon has vertices $\frac12(\mathbf p_1+\mathbf p_2), \frac12(\mathbf p_2+\mathbf p_3), \ldots, \frac12(\mathbf p_n+\mathbf p_{n+1})$, and so its area is
$$\begin{align}
A' &= \frac12\sum_{i=1}^n\left(\frac{\mathbf p_i+\mathbf p_{i+1}}2\wedge\frac{\mathbf p_{i+1}+\mathbf p_{i+2}}2\right) \\
&= \underbrace{\frac18\sum_{i=1}^n(\mathbf p_i\wedge\mathbf p_{i+1})}_{A/4} + \frac18\sum_{i=1}^n(\mathbf p_i\wedge\mathbf p_{i+2}) + \frac18\sum_{i=1}^n\underbrace{(\mathbf p_{i+1}\wedge\mathbf p_{i+1})}_0 + \underbrace{\frac18\sum_{i=1}^n(\mathbf p_{i+1}\wedge\mathbf p_{i+2})}_{A/4} \\
&= \frac12A+\frac14\left(\frac12\sum_{i=1}^n\mathbf p_i\wedge\mathbf p_{i+2}\right).
\end{align}$$
So here's the main result:

The area of the nested polygon is $\frac12$ times the area of the original polygon, and plus $\frac14$ times the total "area" of the polygon(s) obtained by connecting each vertex to the next-to-next one.

Geometrically, connecting vertices by twos gives two cases. When $n$ is even, you get two polygons $\mathbf p_1\mathbf p_3\cdots\mathbf p_{n-1}$ and $\mathbf p_2\mathbf p_4\cdots\mathbf p_n$. When $n$ is odd, you get a single "doubly-wrapped" polygon $\mathbf p_1\mathbf p_3\cdots\mathbf p_n\mathbf p_2\mathbf p_4\cdots\mathbf p_{n-1}$, such as a pentagram. In this case, the formula double-counts the area of the central region, which is why I've put "area" in quotes above.
For small $n$, life is pretty simple, though:
When $n=3$, the polygon $\mathbf p_1\mathbf p_3\mathbf p_2$ is just the same triangle with its vertices connected in backwards order, so it has the opposite (clockwise) orientation and negative signed area. Then you get $A' = \frac12A + \frac14(-A) = \frac14A$.
When $n=4$, the "polygons" $\mathbf p_1\mathbf p_3$ and $\mathbf p_2\mathbf p_4$ are the diagonals of the quadrilateral and have zero area. So $A' = \frac12A$.
In both these cases, the areas of the sequence of nested polygons form a geometric series no matter whether the original polygon was regular or not, and so you can easily compute the total area of the entire series.
When $n\ge5$, it turns out that the area of the alternating polygon(s) also depends on the arrangement of the vertices, not just the area of the original polygon. So one cannot determine $A'$ from $A$ alone.
A: I can't answer all of your questions but there is a formula for the Area of a nested square with unit length edges:
$\sum\limits_{k=0}^{\infty} \frac{(-1)^k}{2^k} = \frac{2}{3}$.
Here is a picture:

Note: The formula calculates the Area of the black region.
