Expert Assistance for Polygons: Comprehensive Resources and Practice Problems

Inequality for convex polygons
Let P be a convex n-gon on the plane. For $k=\stackrel{¯<!--\; ¯\; -->}{1,n}$ define $a_k$ as the length of k-th side of P and $d_k$ as the length of projection of P onto the line containing k-th side of the polygon P. Prove that
$2<\frac{a_1}{d_1}+\frac{a_2}{d_2}+\dots <!--\; \dots \; -->+\frac{a_n}{d_n}\le <!--\; \le \; -->4.$
Firstly, let us prove the first inequality. Indeed, if p is the perimeter of the polygon P, then it's clear that $2d_k<p$ for all $k\in <!--\; \in \; -->\{1,2,\dots <!--\; \dots \; -->,n\}$. Hence, we obtain
$\frac{a_1}{d_1}+\frac{a_2}{d_2}+\dots <!--\; \dots \; -->+\frac{a_n}{d_n}>\frac{2(a_1+a_2+\dots <!--\; \dots \; -->+a_n)}{p}=\frac{2p}{p}=2,$
as desired.
Now, for the second part note that equality holds if, for example, P is a rectangle, so the second inequality is sharp. For polygon P denote $f(P):=\frac{a_1}{d_1}+\frac{a_2}{d_2}+\dots <!--\; \dots \; -->+\frac{a_n}{d_n}.$
Then, it can be shown that if P′ is polygon which is centrally symmetric to P, then the Minkowski sum $Q=P+P^{\prime }$ satisfy the following equality $f(Q)=f(P).$
Thus, it's sufficient to prove the inequality for Q, i. e. for centrally symmetric polygons (it's well-known that $P+P^{\prime }$ is a centrally symmetric polygon). However, it's quite unclear how to continue this approach.
So, is there any way to end this solution?

What is the Cartesian product of n polygons?
How do we calculate the Cartesian product between n polygons? Is it always a polytope? If so, can we say anything about its faces?
For example, n squares: $\{S_1:[a_1,a_2]\times <!--\; \times \; -->[b_1,b_2],\dots <!--\; \dots \; -->,S_n:[a_k,a_{k+1}]\times <!--\; \times \; -->[b_k,b_{k+1}]\}$ so, $S_1\times <!--\; \times \; -->\dots <!--\; \dots \; -->\times <!--\; \times \; -->S_n\in <!--\; \in \; -->{\mathcal{R}}^n$ forms a $n-<!--\; -\; -->$ dimensional box.

Sum of the interior angles of a polygon is $${3060}^{\circ <!--\; \circ \; -->}$$ . How many sides does it have?

The sum of the measures of the interior angles of a polygon is ${720}^{\circ <!--\; \circ \; -->}$ . What type of polygon is it?

One exterior angle of a regular polygon measures 40. What is the sum of the polygon's interior angle measures? Please show working.

You need to construct a regular polygon. When you draw two sides, the interior angle created between them is ${120}^{\circ <!--\; \circ \; -->}$ . What will be the sum, in degrees, of the measures of the interior angles of this polygon when it is completed?

A polygon has 6 sides. How would I find the measure of each exterior and interior angle?

How many sides does a regular polygon have if one exterior angle is 72?

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 polygon has n sides. Three of its exterior angels are 70 degrees, 80 degrees and 90 degrees. The remaining ${\displaystyle (n-3)}$ exterior angels are each 15 degrees. How many sides does this polygon have?

What is the measure of an interior angle of a regular polygon with 20 sides?

A regular polygon has 21 sides. Find the size of each interior angle?

What is the sum of the measures, in degrees, of the interior angles of a 20-sided polygon?

Calculate the measure of each exterior angle of a 90 sided regular polygon? Explain.

Minkowski sum of convex sets in the plane which are not polygons
Can the Minkowski sum of two convex sets in the plane which are not polygons be a polygon? Explicitly my convex set is of the form $C=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:x,y\ge <!--\; \ge \; -->0\text{ , }\sqrt{y}+\sqrt{x}\ge <!--\; \ge \; -->1\text{ and }y\le <!--\; \le \; -->1-<!--\; -\; -->x\}$
I am interested in knowing if there is a convex set C' such that the Minkowski sum
$C+C^{\prime }=P$
where $P=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:x,y\ge <!--\; \ge \; -->0\text{ and }1/2\le <!--\; \le \; -->x+y\le <!--\; \le \; -->1\}.$

Comparing between regular polygons
In this experiment, I am adding the inradius (let's call it A) and circumradius (let's call it B) of different polygons with equal sides each equal 1 (starting with a square and adding one side each time). The result is $A+B=C$ when side of polygon $=1$.
When comparing the C of one polygon with the C of a polygon with one side more, the difference seems to go smaller, as if approaching a version of $\mathrm{\pi <!--\; \pi \; -->}$ number with 0. before (possibly such as 0.314159265359...).
Can anyone confirm it or elaborate on it?
I can not go over a polygon with 1000 sides in my computation power, and would like to know what to expect while going towards a polygon with infinity sides.
Here are some examples:
4 sided polygon: $0.5+0.707106781=1.207106781$
5 sided polygon: $0.68819096+0.850650808=1.5388417680000002$ (Difference of 0.33173498700000015 from previous result)
6 sided polygon: $0.866025404+1=1.866025404$ (Difference of 0.3271836359999998 from previous result)
7 sided polygon: $1.0382607+1.15238244=2.1906431399999997$ (Difference of 0.32461773599999977 from previous result)
8 sided polygon: $1.20710678+1.30656296=2.51366974$ (Difference of 0.3230266000000004 from previous result)
9 sided polygon: $1.37373871+1.4619022=2.83564091$ (Difference of 0.32197116999999986 from previous result)
10 sided polygon: $1.53884177+1.61803399=3.15687576$ (Difference of 0.3212348500000002 from previous result)
11 sided polygon: $1.70284362+1.77473277=3.47757639$ (Difference of 0.3207006299999997 from previous result)
12 sided polygon: $1.8660254+1.93185165=3.79787705$ (Difference of 0.32030066 from previous result)
13 sided polygon: $2.02857974+2.08929073=4.11787047$ (Difference of 0.31999341999999986 from previous result)
14 sided polygon: $2.19064313+2.2469796=4.43762273$ (Difference of 0.31975226000000045 from previous result)
15 sided polygon: $2.35231505+2.40486717=4.757182220000001$ (Difference of 0.3195594899999996 from previous result)
...
999 sided polygon: $158.995264+158.99605=317.991314$
1000 sided polygon: $159.154419+159.155205=318.309624$ (Difference of 0.31830999999999676 from previous result)

Number of r-polygons in an n-polygon with no side coinciding
Here is the full question:
r-sided polygons are formed by joining the vertices of an n - sided polygon. Find the number of polygons that can be formed, none of whose sides coincide with those of the n sided polygon.
I imagined $(n-<!--\; -\; -->r)$ vertices in a closed polygon. There are $(n-<!--\; -\; -->r)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $(\genfrac{}{}{0ex}{}{n-<!--\; -\; -->r}{r})$. But the correct answer wants me to multiply it with $\frac{n}{n-<!--\; -\; -->r}$. What is the need for the last step?

If the ratio of areas of two polygons is the square of the ratio of their perimeters, are the polygons similar?
It is known that for two similar polygons A and B, the ratio of the areas is the square of the ratio of the perimeters. That is, for example, if the ratio of perimeter A to perimeter B is 5:7, then the ratio of their areas is 25:49. However, is the converse to this statement true? That is,
If the ratio of areas of two polygons is the square of the ratio of their perimeters, are the polygons similar? What if we add the specification that the polygons have the same number of vertices?
I cannot think of a counterexample. Any help is appreciated!

Set of regular polygons
Defined set S of regular polygons on plane.
Known for every polygon: at least one of it's sides is parallel to x,and this side's length is rational number.
Also known- there is at least one vertex on this side that it's coordinates are rational numbers.
What is the cardinality of S?

Relationship between area of similar polygons and their corresponding lengths
It is known that ratio of the areas of two similar polygons is equal to the square of the ratio of the corresponding sides. Could one advise me how to prove this assertion? Do we divide the polygon into triangles?