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The Amazing Stamp (Posted on 2003-12-01) Difficulty: 4 of 5
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?

See The Solution Submitted by DJ    
Rating: 4.4545 (11 votes)

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Solution No Subject | Comment 4 of 45 |
The real problem for me in approaching this problem has been in thinking of the circumference of circles intuitively as having depth and area (which they obviously don't), and not in considering them as symbolic representations of a fixed point and extended radius. If we take however, the differing circumferences of concentric circles, we will be able to examine such areas in terms of the line segment seperating them along the same radial line. Or, in other words, if we make a point anywhere on the paper and draw a radius coming out of that point such that it's full length extends to the outermosts edges, we can carefully look at the properties of segments of that radius that would yield either stampable or unstampable areas.

In fact, if we take such a segment there can only be (5) conceivable types, only two of which are possible. These are:

Case (1):
Case (1) would exist such that between the two points on this given segment, there stands an unbroken continuity of irrational numbers that would constitute an area that is inkable.

Case (2):
Case (2) is similar to case (1) in that between the two points on this given segment, there stands an unbroken continuity of rational numbers that would constitute an area that is uninkable.

Case (3):
Case (3) would exist such that between lengths along this given segment of unbroken irrational continuity there exists only singular points of rational numbers breaking up the otherwise continuous stream of irrational numbers (still inkable as shown below).

Case (4):
Case (4) would exist such that between lengths along this given segment of unbroken rational continuity there exists only singular points of irrational numbers breaking up the otherwise continuous stream of rational numbers (still uninkable as shown below).

Case (5):
And case (5) would contain composites of all of the above.

If case (1) exists, the area is inkable. Although it is mentally possible to form such a construct of length almost infinitesimally small whose entire construct is formed of entirely irrational numbers, this isn't at all feasible. However, if we consider that all rational points that interfere with this continuity are to be easily disposed of insofar as they do not form a continuous chain of unbroken rational numbers (and represent an area that is uninkable), we can treat case number (3) and case number (1) as identical. (As stated above, these points (whose circles are actually formed from the radius thus described) lack any depth and can be readily discarded.) If it can be determined that case (2) and case (4) are absolute impossibilities (and hence no uninkable portion can exist), even the existence of case (5) (the most probable outcome of randomly selecting a line segment) wouldn't matter, it would be shown to only exist as a type of case (1) or (3).

(Note: I can't go into the number theory, so I would gladly solicit some help at this point from better mathemeticians out there than myself.)

I say only hesitantly, therefore, that it is impossible to form a length of line that can only be composed of unbroken rational integers. If we take a point p/q there seems to me to exist an infinite amount of irrational numbers arbitrarily close (from either direction) that no rational number can get closer to this p/q. Again, I'm stating this somewhat axiomatically, and if someone can offer the formal proof of this here, I would be greatly at their disposal. This seems intuitively the case considering the number of transcendental numbers is far, far greater than the number of rational numbers. With this in mind, we can therefore disregard all possibilities of areas that are uninkable, (i.e. cases (2), (4), and (5) where the composites contain cases (2) and (4)), and arrive at the following answer:

For any given point on a sheet of paper, this unique stamp will ink the entire surface.

Well, these are my thoughts and I hope the proof will hold. In any case, I'm eager to hear your thoughts and comments on it, all. Great problem, DJ.
Edited on December 1, 2003, 11:48 pm
  Posted by Benjamin J. Ladd on 2003-12-01 22:49:08
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