A diversion: On square circles

25 November, 2007 at 9:18 am (mathematics) (, , , )

This is a semiformal proof. A formal one would be hard to understand without pictures and would require me checking out the English translations of a number of words, which I am not inclined to do.

A square is defined as an ordered set of four points, no three of which are on the same line, and further that the angles thus created are all right (there is a bit more, but it is not too relevant). Circle is defined as a set of points that are from given distance (radius) from a given point. Because a square does not exist in hyperbolic geometry, it is sufficient to think about Euclidean geometry. Assume that there is a square circle. Let A and B be two points of the circle that are part of the same edge (i.e. that edge does not go through the circle’s center). Let C be a point that is between A and B. Distance from The center of the circle to C is lesser than the distance to A or B, so it is also lesser than the radius. Accordin to one axiom I am a bit too lazy to check out, it is possible to find D so that D is behind C when observed from the circle’s center and D’s distance from the center is the circle’s radius. Hence D is part of the circle. D is not anyof the original points that defined the circle, because if it was either of the unnamed ones, the definition of rectangle would get messy, and if it were A or B, C would also be A or B, which contradicts the choice of C. Hence, there are no square circles. Qued est demonstratum.

This diversion due to another discussion.

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