The wobbly Table Theorem
The wobbly table theorem: you can turn a square table with four legs in such a way that all four legs are on the floor. |
Turn the table around a center in such a way that three legs A,B,C stay on the
surface. The distance of the fourth leg to the surface is a function f(x)
which depends on the angle x. The argument given below will show that if f(0)
is positive, then f(pi/2) is negative and that if f(0) is negative then f(pi/2) is positive. The
intermediate value theorem assures that there is an angle, where the fourth leg
is also on the surface. Here is the argument: 1) assume the center of the four points A,B,C,D is on the z axes. If we ask the three legs A,B,C to be on the ground and that the direction of the axes AC (when looking from above) is given by the angle x, then the position of the table is determined. 2) Define the function f(x) which is the signed vertical distance of the fourth leg from the ground. It is positive if it is above the surface and negative if below the surface. We need to show that there is an angle x, such that f(x) = 0. 3) Start with x=0 and assume the fourth leg is above the ground (f(0) is larger than 0) The other case is similar). We want to show that f(pi/2) is smaller than 0. The intermediate value theorem will then assure that there exists an angle, where f(x)=0. 4) Look at the position for x=0. First rotate the table around the axes AC so that both legs B,D have the same vertical distance from the ground. Now lower the table downwards until the two legs B,D are on the ground. The two legs A,C are now equal distance below the ground. Now rotate the table around the axes BD so that the leg C is on the ground. We have achieved that the three legs B,C,D are on the ground and the leg A is below the ground. But since this is the same situation as the position when the table is rotated by 90 degrees (keeping the same three legs on the ground), we see that f(pi/2), the distance of A from the ground is smaller than zero. |