20040530, 01:22  #1  
May 2004
3 Posts 
How many punches?
I ran across this in a book and was wondering if anyone could help:
Quote:
Can anyone help me Oh yeah, I got this from "The Art and Craft of Problem Solving" by Paul Zeitz. 

20040611, 03:31  #2 
May 2004
3 Posts 
bump

20040611, 13:16  #3 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}·5·443 Posts 
Would it not matter most what the size of the punch is?
I would think that a plane would have infinite number of points, therefore you would need an infiniate number of punches. 
20040611, 14:15  #4 
Mar 2004
3·127 Posts 
My feeling says: 2 punches.
If you punch the second time at a distance of a transcendent number, every point might be removed. There is no point with a rational distance to centr a and center b if the distance between the centers is transcendent. Otherwise that distance would not be transcendent. By the way: the number needs to be transcendent. Sqrt(2) is wrong. In that case a number which has distance 1 from both punches will not be removed. Last fiddled with by biwema on 20040611 at 14:20 
20040611, 22:30  #5 
Mar 2004
3×127 Posts 
Sorry, I think I made a mistake. I thought again about the solution and came to the conclusion that it must be more than 2 punches.
If every punch eliminates all irrational numbers, infinitely many infinitely thin circles around the center remain. With 2 distinct punches, there are infinitely many distinct intersections (between two circles) remaining. but is should be possible to place a third center, whose circle does not hit any of the points (because it is possible to adjust the center with an infintely small difference). Rational points are very sparse in the irrational area. So I vote for 3 punches now. (with transcendent distance to each other) 
20040809, 03:53  #6 
May 2004
3 Posts 
It's Over!

20040810, 08:57  #7 
Mar 2004
101111101_{2} Posts 
Yes, 3 Punches! I got it...
