How many punches?

dm21mayo · 2004-05-30, 01:22

I ran across this in a book and was wondering if anyone could help:

Quote:

A special paper punch has been created that removes all points in the plane that are an irrational distance from the center. How many punches are required to remove all points from the plane?

For a line, this is easy: punch once at 0 and once at any irrational number, say Sqrt(2). However, I can't extend this to the entire plane.

Can anyone help me

Oh yeah, I got this from "The Art and Craft of Problem Solving" by Paul Zeitz.

dm21mayo · 2004-06-11, 03:31

bump

Uncwilly · 2004-06-11, 13:16

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.

biwema · 2004-06-11, 14:15

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.

biwema · 2004-06-11, 22:30

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)

dm21mayo · 2004-08-09, 03:53

I have the answer!

http://www.kalva.demon.co.uk/putnam/psoln/psol904.html

Whatever that means.

biwema · 2004-08-10, 08:57

Yes, 3 Punches! I got it...

2004-06-11, 14:15	#4
biwema 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 2004-06-11 at 14:20

2004-08-09, 03:53	#6
dm21mayo May 2004 3 Posts	It's Over! I have the answer! http://www.kalva.demon.co.uk/putnam/psoln/psol904.html Whatever that means.

2004-06-11, 13:16	#3
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 10011000111010₂ 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.

2004-06-11, 22:30	#5
biwema Mar 2004 17D₁₆ 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)

2004-08-10, 08:57	#7
biwema Mar 2004 3×127 Posts	Yes, 3 Punches! I got it...