Transpose inner polygon (2): Any Polygon contains an inner rectangle.

Then, we want to ask: Can an inner rectangle be found in any polygon? Interestingly, the answer is yes. But now, the two methods we used in the previous section are useless, and we need to use new methods. The following proves that it is actually a strange thing. I really don't know who came up with it.

For any two points A (x1, Y1) and B (X2, Y2) on the boundary of a polygon, make the points (X1 + x2)/2, (Y1 + y2)/2, √(X1-x2) ^ 2 + (y1-y2) ^ 2). In other words, place the Polygon on the horizontal plane Z = 0. For each unordered vertex A and B on the polygon, make a point at the top of the online vertex AB |. After adding the polygon itself, we get a closed surface in a three-dimensional space.

As you can see, in the example shown in the figure, this surface is intersecting with itself. This indicates that there are two groups of points A, B, C, and D on the boundary of the polygon. they satisfy the coincidence of the midpoint of the line AB and Cd, and the length of the two lines is the same. In this way, the Quadrilateral ABCD is an inner rectangle of the polygon. Next we will explain that this surface will surely intersect with itself.

It is easy to see that if both vertices A and B are on the same side of a polygon, all such points are a triangle in the corresponding point set of three-dimensional space; if a and B come from two different edges, the normal points of all possible line segments AB form a parallelogram. The point set corresponding to them in space is a surface with four edges. That is to say, this surface has an n edge shape (bottom surface) with N pentagons (corresponding to n edges of a polygon) and N (n-1) /Two Quadrilateral (corresponding to the N (n-1)/2 sides of the polygon ). Therefore, this surface has 1 + N (n-1)/2 faces, and its number of edges is

(1/2) * (N * 1 + 3 * n + 4 * n (n-1)/2) = n ^ 2 + n

In addition, each two vertices of a polygon correspond to a vertex on the surface, and N vertices exist on the base. Therefore, the surface has n + N (n-1)/2 vertices. Therefore, the Euler's number of representation of this surface v-e + F is equal

N + N (n-1)/2-N ^ 2-N + 1 + N (n-1)/2 = 1

This surface is non-oriented and cannot be embedded into a 3D space.