So let's assume (all three legs can always land at the same time)
(1) The ground is a continuous surface.
(2) The four legs of the square table have the same length.
(3) Compared with the bending degree of the ground, the legs of the square table are long enough.
(4) As long as the legs of the square table touch the ground slightly, they will touch the ground.
Now, let's prove that if the above hypothesis holds, then the answer is yes. Take the center of the square table as the coordinate origin and the rectangular coordinate system as shown in the figure. The four legs of the square table are at A, B, C and D respectively. The initial positions of A and C are on the X axis, and B and D are on the Y axis. When the square table rotates around the center 0, the included angle between diagonal AC and X axis is recorded as θ.
It is easy to see that when all four legs are not on the ground, the distance between the legs and the ground is uncertain. In order to eliminate this uncertainty, let f(θ) be the sum of ground distances of A and C, and g(θ) be the sum of ground distances of B and D, and their values are uniquely determined by θ. Suppose (1), f(θ) and g(θ) are continuous functions of θ. And from the assumption (3), three legs can always land at the same time, so f(θ)g(θ)=0 must hold (θ). Let f (0) = 0 and g (0) >; 0 (if g(0) is also 0, four legs have landed at the initial moment, so there is no need to rotate), so the problem boils down to:
It is known that f(θ) and g(θ) are continuous functions of θ, f (0) = 0 and g (0) >; 0 and f(θ)g(θ)=0 For any θ, it is proved that there is a certain θ0, so that f(θ0)=g(θ0)=0.
(Prove 1) When θ=π/2, AC and BD exchange positions, so f(π/2)>0, g(π/2)=0. Let h(θ)=f(θ)-g(θ). Obviously, h(θ) is also a continuous function of θ.
(Proof 2) F (π/2) can be obtained from the same proof 1 > 0 and g(π/2)=0. Let θ θ o = sup {θ| f (zeta) = 0 zeta) = 0, 0 ≤ zeta; 0, there is always δ > 0 and δ; 0。 Because f(θ0+δ)g (θo+δ)=0, there must be g (θ0+δ)=0, and δ can be arbitrarily small and G is continuous, so there must be g (θ0)=0. Prove 2 not only uses the continuity of f and g, but also uses the property of supremum.
There is a picture, I don't know how to post it. )