Expert Assistance for Polygons: Comprehensive Resources and Practice Problems

Minkowski difference of two convex polygons
I just want to make sure that the following algorithm is correct for computing the Minkowski difference of two shapes A,B:
$\text{Minkowski}(A,B)=\text{ CH }\{x:x=a-<!--\; -\; -->b\text{ for }a\in <!--\; \in \; -->A,b\in <!--\; \in \; -->B\}$
Where CH(S) is the convex hull of the set S and a,b are the vertices of the two polygons.

Regular polygons meeting at a point
N regular polygons with E edges each meet at a point with no intervening space. Show that $N=\frac{2E}{E-<!--\; -\; -->2}E=\frac{2N}{N-<!--\; -\; -->2}$.
I did this part considering that the internal angles must add up to $2\mathrm{\pi <!--\; \pi \; -->}$, i.e. $(1-<!--\; -\; -->{\displaystyle \frac{2}{E}})\mathrm{\pi <!--\; \pi \; -->}\times <!--\; \times \; -->N=2\mathrm{\pi <!--\; \pi \; -->}$. I am unable to explain the following:
Using the result given above, show that the only possibilities are $E=3,4,6$

If I have a circle that has ten dots on it and I have to make as many polygons using those dots, the answer should be 10. Connecting each dot starting with triangle, to square, all the way to the max of a 10-sided polygon should give me only 10 polygons that I can make. Am I missing something here?

Does the intersection of two concave polygons without holes result in a set of several polygons without holes?
I did not find a proof anywhere, so here is my assumption :
If I have 2 concave and simple polygons without holes, then the intersection yields a set of different polygons without any holes.
If it can result to a polygon with a hole inside, could you provide an example?
Edit: the polygons are simple. edges of one polygone don't intersect, except consecutive edges, which intersect in their common vertex. The polygons in input do not have holes. I consider the intersection of the interiors of the polygons. The resulting set is all the points that are included both in interior of polygon P1 and the interior of the polygon P2.

Least Squares Circumcenter of Polygons
It is well known that the circumcenter of a polygon exists if and only if the polygon is cyclic.
I would like to extend the definition of an circumcenter for noncyclic polygons. Namely, let us define the least squares circumcenter as the point A $(x_0,y_0)$ such that the point A minimizes the sum of the squares of the residuals.
Let us consider the case for a noncyclic quadrilateral with vertices P $(x_1,y_1)$, Q$(x_2,y_2)$, R$(x_3,y_3)$, and S $(x_4,y_4)$. Let us also define the origin O(0,0).
How would we solve for the point A in this case? I was thinking of using matrices and solving $A^{\mathsf{T}}A\stackrel{^<!--\; ^\; -->}{x}=A^{\mathsf{T}}b$, although any methods are welcome.

Centroid of contiguous polygons
Say that I know which are the centroids of two polygons. These polygons share a number of edges (they belong to a planar subdivision). I want to compute the union of the two polygons and also to know the centroid of the new polygon which covers the other two. I know I could recompute the centroid taking the points of the new polygon. But, is there a way to obtain the centroid by knowing the other centroids, without the need for recomputing everything from scratch?

Number of regular polygons possible
Thirty six equally spaced points are plotted on a circle, and some of these points are joined successively to form a polygon. How many distinct such regular polygons are possible.
What I thought that the answer should be ${}^{36}{C}_3{+}^{36}{C}_4{+}^{36}{C}_5{+}^{36}{C}_6{+}^{36}{C}_7+\cdots <!--\; \cdots \; -->{+}^{36}{C}_{36}$ but this is not the correct answer that has been provided. What am I doing wrong?

Question regarding polygons
Can you prove, that if a equilateral lattice n-gon is constructible, then there will be such a polygon for which the sides have minimal length?

Midpoint polygons?
A midpoint polygon is formed by taking the midpoints of each side of a polygon, and making a new polygon out of those points. The end result is the midpoint polygon inscribed in the polygon you started off with.
Given an arbitrary polygon, is it always the midpoint polygon of some other polygon? And if so, is this other polygon unique? Is there a simple construction to find it? And also, are there any special cases of the original question which are easier to answer?

Covering the plane with convex polygons?
I have got the following task here:
Prove, that you can't cover the "Plane" with convex polygons, which have more than 6 vertices!
The answer is pretty obvious for $n=3$ vertices, because $6\cdot <!--\; \cdot \; -->{60}^{\circ <!--\; \circ \; -->}={360}^{\circ <!--\; \circ \; -->}$.
For $n=4$ it works too, because $4\cdot <!--\; \cdot \; -->{90}^{\circ <!--\; \circ \; -->}={360}^{\circ <!--\; \circ \; -->}$.
I think that $n=6$ is good too, but how do I prove, that other than that, it isn't possible to do that?

Inner angles in polygons
A point of a polygon is called convex if it's inner angle is less than 180 degrees. Prove that in every simple polygon there is at least one convex point.

Lebesgue measure using special polygons
I want to prove that the following set satisfies $m(G)=1$.
$G=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:x>0,0<y<e^{-<!--\; -\; -->x}\}$
I have to prove this without using the Lebesgue integral.
This set is open, so its measure is defined as it follows:
$m(G)=sup\{m(P):P\subset <!--\; \subset \; -->G\}$, where P is a special polygon.
We can construct a sequence of special polygons $P_n\subset <!--\; \subset \; -->G$:
$P_n=\underset{j=1}{\overset{n}{\bigcup <!--\; \bigcup \; -->}}[0,log\left(\frac{n+1}{j}\right)]\times <!--\; \times \; -->[\frac{j-<!--\; -\; -->1}{n+1},\frac{j}{n+1}].$
The measure of one of these special polygons is $m(P_n)=\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}.$
As the sequence of measures $\{m(P_n)\}$ is increasing, the supremum of this sequence is its limit:
$sup(\{m(P_n)\})=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}log\left(\frac{n+1}{j}\right)\frac{1}{n+1}=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\log <!--\; \; -->{\left(\frac{(n+1{)}^n}{n!}\right)}^{\frac{1}{n+1}}.$
I know that this limit is 1, so the logarithm argument must tend to e, but I don't know how to prove this analytically. I've tried working on the final expression to get the definition of e, but I haven't been able to do so. Could yo help me?

Hyperbolic Angle Measure of Polygons
Is there a way to determine the angle measure of a regular polygon in hyperbolic space? I know that this depends on the length of the sides. A an example, I know that for an equilateral triangle with side length a and angle A, then $\sec <!--\; \; -->A=1+\frac{2e^a}{e^{2a}+1}$.
Is there a similar formula for higher regular polygons?

Scan line algorithm for intersecting polygons
Given two sets of polygons $P_1=\{s_1,...,s_m\}$ and $P_2=\{s_m+1,...,s_n\}$ with total number of n segments, the previous and next segment on it's polygon can be determined in O(1). Describe a scan-line algorithm that computes all points of $P_1\cap <!--\; \cap \; -->P_2$ in O(n).

Using trigonometric ratios to express area of regular polygons
I am very confused. My book just asked me to use trigonometric ratios to express the area of a regular polygon with 9 sides and lengths of 8.
So far I have learned how to use the sin cos and tan in right triangles and have no idea how this applies to all polygons. Can someone please explain this to me

Constructible polygons
I know certain polygons can be constructed while others cannot. Here is Gauss' Theorem on Constructions:
$\cos <!--\; \; -->(2\pi /n)$ is constructible iff $n=2^r{p}_1{p}_2···p_k$, where each $p_i$ is a Fermat prime.
Can this is be used to determine the constructibility of a regular $p^2$ polygon? If so how and what would be the $\cos <!--\; \; -->(2\mathrm{\pi <!--\; \pi \; -->}/n)$ here?

“Area metric” and “Hausdorff metric” are not equivalent on all closed polygons, but equivalent on convex closed polygons
Suppose X is the set of all closed polygons, $d_{\mathrm{\Delta <!--\; \Delta \; -->}}$ is the “area metric” defined by the area sum of the symmetric difference of two closed polygons, and dH is the Hausdorff metric on X. How should I prove that $d_{\mathrm{\Delta <!--\; \Delta \; -->}}$ and $d_H$ do not generate the same topology on X? Also why do they generate the same topology on the subset of convex polygons? I tried to visualize how a typical open ball in both metrics looks but this seems rather impossible.

The area of a triangular sail for a boat is 176 square feel. If the base of the sail is 16 feet long, find its height.

Create a nest of 5 sets whose largest set in the set of all polygons. For each set but the largest set in the nest, write a deffinition that precisly describs the objects in that set in terms of the objects in the next larger level.
my solution to this qesttion is:
all polygons
Quadrilalateral
tarpezoid
Isosceles Traingle
Rectangle
But i'm not sure about that. Also, i need their deffinitions?

A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude 25ft and base edge of 10ft. how much will the surface rise when 6,779 liters of water is added?