We used two clever methods to prove this proposition: arbitraryConvexThe polygon contains an inner rhombus. Using the climbing theorem we talked about last time, we can prove a stronger proposition:Arbitrary PolygonThere is an inner diamond.
The general idea of proof is as follows: select a point u outside the polygon. Write the point closest to U on the polygon as Y and the point farthest from U on the polygon as Z. The points Y and Z divide the boundary of the whole Polygon into two parts.
Recall the content of the climbing theorem: for the "broken line function" f (x) and g (x) whose two function values are continuously changed from 0 to 1 ), we can always adjust the positions of X1 and X2 consecutively, so that F (X1) and g (X2) are always equal. They start from 0 at the beginning and reach 1 at the same time. Applying the climbing theorem, we can draw this conclusion: we can let point X1 start from Y and move the upper part of the graph to Z, point X2 starts from Y and moves down the lower part of the graph to Z, and ensures that the distance from X1 to U is always equal to the distance from X2 to U (to take care of each other, if necessary, X1 and X2 may go back ). In this way, U, X1, and X2 can always be three vertices of a diamond. We record the fourth vertex of the diamond AS v. It is easy to prove that the trajectory of V is continuous.
When X1 and X2 are sufficiently close to Y, V points are obviously inside the polygon, but when X1 and X2 run near Z, V obviously runs outside the polygon. In this process, the V point inevitably passes through the boundary of the polygon. At this time, U, x1, x2, and V constitute such a diamond, and the three vertices behind it are on the polyon.
Now, if we fix the y point and gradually move the V point closer to the Y point, the corresponding diamond will change continuously. It is easy to think that the limit of this process will converge to a fixed Diamond (which can be proved), which is the internal diamond we want.