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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?

  Submitted by DJ    
Rating: 4.4545 (11 votes)
Solution: (Hide)
Yes.
(0,0), (1,0), (0,π)


Here is a proof by contradiction:
Suppose that after making the three stamps there is a point (x,y) which is not covered. If that is the case, then point (x,y) is a rational distance from each of the three center points. Then there are rational numbers a,b,c satisfying the following equations:
  x²   +   y²    = a
(x-1)² +   y²    = b
  x²   + (y-π)² = c
Subtract (2) from (1) and solve for x.
Thus, x is rational.
From equation (2), y is algebraic.
However, equation (3) implies that y is transcendental, wherein lies the contradiction.

So, every point on the Cartesian plane must be covered by one of the three stamps.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsSome thoughts on the problemK Sengupta2007-05-12 13:28:40
Continuum Hypothesis solved!Avin2005-05-18 14:11:51
General solutionLarry Settle2003-12-11 22:32:00
re(5):my proof - I stand corrected!SilverKnight2003-12-09 16:21:46
re(4):my proofBrian Wainscott2003-12-09 16:08:50
re(3):my proofSilverKnight2003-12-09 15:46:00
re(2):my proofBrian Wainscott2003-12-09 15:33:18
re:SilverKnight2003-12-09 15:16:50
No SubjectBrian Wainscott2003-12-09 15:02:55
re(9): Brian's solutionSilverKnight2003-12-09 12:55:07
re(8): Brian's solutionBrian Wainscott2003-12-09 12:36:11
re(7): Brian's solutionSilverKnight2003-12-08 19:43:52
re(6): Brian's solutionBrian Wainscott2003-12-08 19:10:21
re(5): Brian's solutionSilverKnight2003-12-08 17:28:25
re(6): Brian's solutionLarry Settle2003-12-08 15:44:18
re(5): Brian's solutionCory Taylor2003-12-08 14:39:10
re(4): Brian's solutionCory Taylor2003-12-08 14:30:46
re(4): Brian's solutionLarry Settle2003-12-08 14:29:15
re(3): Brian's solutionSilverKnight2003-12-08 12:20:44
re(2): Brian's solutionCory Taylor2003-12-08 12:09:11
re: Brian's solutionSilverKnight2003-12-08 10:44:48
Brian's solutionLarry Settle2003-12-08 10:26:42
Lateral thinking solutionPenny2003-12-04 07:54:17
re(2): SolutionCharlie2003-12-03 14:32:02
re: SolutionCory Taylor2003-12-03 12:19:56
SolutionBrian Smith2003-12-03 11:22:29
a solution??Cory Taylor2003-12-03 10:03:58
re(4): alterationCharlie2003-12-03 09:00:09
Questionre(5): alternation implies 1-1Charlie2003-12-03 08:46:51
re(3): alterationDJ2003-12-02 22:36:09
re(5): alternation implies 1-1SilverKnight2003-12-02 14:43:51
re(4): alternation implies 1-1Cory Taylor2003-12-02 14:36:29
re(3): alternation implies 1-1SilverKnight2003-12-02 14:15:28
re(2): alternation implies 1-1Cory Taylor2003-12-02 13:58:59
re: alternation implies 1-1SilverKnight2003-12-02 12:35:34
alternation implies 1-1Cory Taylor2003-12-02 12:20:25
re(3):Bryan2003-12-02 10:59:43
re(2):Charlie2003-12-02 08:57:43
re(2):Benjamin J. Ladd2003-12-02 00:04:53
re:Eric2003-12-01 23:41:24
Some Thoughtsre:DJ2003-12-01 23:34:51
SolutionNo SubjectBenjamin J. Ladd2003-12-01 22:49:08
Three pointsBrian Smith2003-12-01 14:28:30
Some ThoughtsthoughtsCharlie2003-12-01 13:58:24
This is a cool problem...SilverKnight2003-12-01 13:40:05
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