Here's the question you clicked on:
nightwill
Say that there is a rough surface. Prove that it is always possible to put a square table so that all four legs rest on the surface. (The table can be tilted)
I'm sorry I don't get your question.
Any way possible. There isn't any restriction. That is all the things I got from the teacher.
I think this table problem may have to do with the postulate that through three noncollinear points there is exactly one plane. With the floor's being described as "rough," I assume that the fourth table leg may not lie in the same plane as the other three. In that scenario, someone sitting adjacent to that table leg can tilt the table so that it rests on the plane formed by the 3 noncollinear points. Or, place a wad of paper underneath the fourth leg.
I think the question asserts a perfect table, i.e. the 4 legs are the same length and are parallel.
Yep CeW. Thank you for clarifying that yes that's what I want.
What's being asked, in a mathematical sense, is to prove that for any function of 2 variables f(x,y), that f has the same value at the points: (x0, y0), (x0+d,y0), (x0,y0+d), (x0+d,y0+d), for at least one point (x0,y0) given a value of d. Here, x,y are the coordinates on a plane running through the mean height of the floor and z=f(x,y) is the height of the floor at that point (say, relative to its mean height). d is the square table side length.