Use variables to represent the position of the chair, and introduce plane graphics and coordinate system as shown in figure 1- 1. In the figure, A, B, C and D are the four legs of the chair, the origin of the coordinate system is selected as the center of the chair, and the coordinate axis is selected as the diagonal of the four legs of the chair. Therefore, according to Hypothesis 2, the moving position of the chair can be uniquely expressed by the angle Q at which the square rotates along the coordinate origin, and the vertical distance between the chair feet and the ground becomes a function of Q. Noting the central symmetry of the square, we can use the sum of the distances between the two opposite legs of the chair and the ground to express the corresponding distance relationship between the two legs and the ground. In this way, we can use a function to describe whether the legs of the chair touch the ground. In this problem, two functions are introduced to describe whether the four feet of the chair touch the ground. Remember that the function f(q) is the sum of the vertical distances between the legs A and C of the chair and the ground. The function g(q) is the sum of the vertical distances between the legs b and d of the chair and the ground. And then there's obviously f(q)? 0? 60、g(q)？ 0? 60, and they are all continuous functions of q (assumption 2). According to assumption 3, for any q, at least one of f(q) and g(q) is 0, so we assume that when q=0, f (0) >; 0, g(0)=0, so the problem 1 can be classified as the following mathematical proposition: it is known that f(q) and g(q) are nonnegative continuous functions of q, and for any q, there are f(q) g(q)=0, f (0) >; 0, g(0)=0, then q0 exists, so f(q0)= g(q0)=0. It is proved that when the chair rotates 90, diagonal AC and BD are interchanged, f (0) >; 0, g(0)=0 becomes f(p/2) =0, g (p/2) >; 0

If the constructor h(q)=f(q)-g(q), then there is h (0) >; 0 and h (p/2) < 0 and h(q) are also continuous functions. Obviously, it is continuous in the closed interval [0, p/2]. Judging from the zero theorem of continuous function, there must be q0? 0? 2(0, p/2), so that h(q0)=0, that is, q0 exists? 0? 2(0, p/2), so f(q0)= g(q0). For any q, there is f(q) g(q)=0, especially f(q0) g(q0)=0. So at least one of f(q0) and g(q0) is 0, so f(q0)= g(q0)=0. Complete the certificate.