Play-over inner polygon (3): any convex polygon contains an inner rhombus.

When We further consider the inner diamond, there are some changes -- it turns out that the inner diamond in any polygon is not as easy as the first few problems. However, we can easily prove a relaxed proposition: arbitraryConvex PolygonThere is an inner diamond. The two different proofs of this proposition are given below, both of which are quite classic.

 

Proof 1: considering that a horizontal line segment in a convex polygon is swept from top to bottom, the trajectory formed by the midpoint of this line segment is a line segment connecting the top and bottom of a convex polygon. Similarly, considering a vertical line segment moving from left to right, its midpoint forms the line from the leftmost end to the rightmost end of the convex polygon. Obviously, the two links have an intersection. That is to say, we have found two lines perpendicular to each other and overlap the midpoint. Their four endpoints are obviously four vertices of a diamond.

 

Proof 2: considering any point P in a convex polygon, the horizontal and vertical straight lines are taken over P points, and the polygon is handed over to four points Pa, Pb, PC, and PD. F (p) indicates the center of gravity of PA, Pb, PC, and Pd (that is, the point of intersection between pa pc and Pb PD ). According to the convex shape, the center of gravity must be inside the convex polygon. In addition, it is easy to conclude that F is continuous. According to the Brouwer's fixed point theorem, there is a P point that makes P = f (p ), the PA, Pb, PC, and PD corresponding to this P point are obviously four vertices of a diamond.