Comments on: Ponder This

By: Robbie

Robbie — Mon, 17 Apr 2006 00:32:26 +0000

If you have a coin on a table so that you cannot put another coin (with it’s centre) on the table without overlapping the first coin, then I think you’ll find four coins will be enough to cover the table.

By: syrjoe

syrjoe — Fri, 14 Apr 2006 17:05:43 +0000

I don’t think it’s possible for n = 1.

By: Ars Mathematica

Ars Mathematica — Mon, 01 Aug 2005 14:10:21 +0000

Ponder This

It’s the beginning of the month and the solution to last month’s Ponder This challenge is up, as well as the puzzle for August:
For K as large as possible, produce a K-digit integer M such that for each N=1,2,…,K, the integer given …

By: Robbie

Robbie — Tue, 26 Jul 2005 02:38:42 +0000

It's close enough to the end of the month ... here is my solution: Let x_1, x_2, ..., x_n be the centres of the coins and r be their radius. Since placing a new coin on the table will overlap with a coin already on the table, we have: for all x (on the table), there exists x_i such that |x - x_i| < r. Hence the n balls B(x_i, 2r) cover the table. Shrink this be a factor of 2 in each direction, and we get a cover of a L/2 x W/2 table by n balls of radius r. Four of these gives us 4n coins covering a table of size L x W.

By: Robbie

Robbie — Sun, 03 Jul 2005 18:34:23 +0000

You start with n coins on the table.

Actually, I liked this problem better than some of the previous ones. More metric spaces and less combinatorics

By: Walt

Walt — Sun, 03 Jul 2005 16:46:46 +0000

I don’t understand the problem. What’s the n in 4n?