You have an ink stamp that is so amazingly precise that, when inked and pressed down on the plane, it makes every circle whose radius is an irrational number (centered at the center of the stamp) black.
Is it possible to use the stamp three times and make every point in the plane black?
If it is possible, where would you center the three stamps?
(In reply to
re(3): Brian's solution by SilverKnight)
Im gonna try something new (to me). Whee. Im gonna try to link to a previous post. Very exciting (again, for me!!!)
Brian Smith's solution here
describes "x" as the distance between points A and B (which is also the distance between points B and C - my reference to equidistant). Since B is the mid-point of the segment created by points A and C, the three point A,B, and C are colinear. This is all in the post I've (hopefully) linked to, defined as part of Brians proof. And when you see it, you'll find that both your examples use an x value of one, as (π+2)-(π+1)=1.
Now I fully agree that there are all sorts of sets of three points (including both of your examples) that will not work, thats not the discussion.
I believe that Larry's claim is untrue, because although the probibility of choosing a particular set of three points that do not satisfy the problem requirements, there are an infinite number of such sets, thereby making the overall probibility uncertain. It might be interesting (and WAY WAY WAY over my head) to see what the probibility is for three truly random points... Anyone?....