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## 6 thoughts on “Kapikachhu Online Without Prescription”

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

2. 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 |xx_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.

3. Pingback: Ars Mathematica

4. 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.