perplexus.info :: Geometry : The Amazing Stamp

The Amazing Stamp (Posted on 2003-12-01)

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 thoughts on the problem	K Sengupta	2007-05-12 13:28:40
Continuum Hypothesis solved!	Avin	2005-05-18 14:11:51
General solution	Larry Settle	2003-12-11 22:32:00
re(5):my proof - I stand corrected!	SilverKnight	2003-12-09 16:21:46
re(4):my proof	Brian Wainscott	2003-12-09 16:08:50
re(3):my proof	SilverKnight	2003-12-09 15:46:00
re(2):my proof	Brian Wainscott	2003-12-09 15:33:18
re:	SilverKnight	2003-12-09 15:16:50
No Subject	Brian Wainscott	2003-12-09 15:02:55
re(9): Brian's solution	SilverKnight	2003-12-09 12:55:07
re(8): Brian's solution	Brian Wainscott	2003-12-09 12:36:11
re(7): Brian's solution	SilverKnight	2003-12-08 19:43:52
re(6): Brian's solution	Brian Wainscott	2003-12-08 19:10:21
re(5): Brian's solution	SilverKnight	2003-12-08 17:28:25
re(6): Brian's solution	Larry Settle	2003-12-08 15:44:18
re(5): Brian's solution	Cory Taylor	2003-12-08 14:39:10
re(4): Brian's solution	Cory Taylor	2003-12-08 14:30:46
re(4): Brian's solution	Larry Settle	2003-12-08 14:29:15
re(3): Brian's solution	SilverKnight	2003-12-08 12:20:44
re(2): Brian's solution	Cory Taylor	2003-12-08 12:09:11
re: Brian's solution	SilverKnight	2003-12-08 10:44:48
Brian's solution	Larry Settle	2003-12-08 10:26:42
Lateral thinking solution	Penny	2003-12-04 07:54:17
re(2): Solution	Charlie	2003-12-03 14:32:02
re: Solution	Cory Taylor	2003-12-03 12:19:56
Solution	Brian Smith	2003-12-03 11:22:29
a solution??	Cory Taylor	2003-12-03 10:03:58
re(4): alteration	Charlie	2003-12-03 09:00:09
re(5): alternation implies 1-1	Charlie	2003-12-03 08:46:51
re(3): alteration	DJ	2003-12-02 22:36:09
re(5): alternation implies 1-1	SilverKnight	2003-12-02 14:43:51
re(4): alternation implies 1-1	Cory Taylor	2003-12-02 14:36:29
re(3): alternation implies 1-1	SilverKnight	2003-12-02 14:15:28
re(2): alternation implies 1-1	Cory Taylor	2003-12-02 13:58:59
re: alternation implies 1-1	SilverKnight	2003-12-02 12:35:34
alternation implies 1-1	Cory Taylor	2003-12-02 12:20:25
re(3):	Bryan	2003-12-02 10:59:43
re(2):	Charlie	2003-12-02 08:57:43
re(2):	Benjamin J. Ladd	2003-12-02 00:04:53
re:	Eric	2003-12-01 23:41:24
re:	DJ	2003-12-01 23:34:51
No Subject	Benjamin J. Ladd	2003-12-01 22:49:08
Three points	Brian Smith	2003-12-01 14:28:30
thoughts	Charlie	2003-12-01 13:58:24
This is a cool problem...	SilverKnight	2003-12-01 13:40:05

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