What is the minimum number of coins of diameter $\sqrt 2$ needed to cover a $ 7 \times 7$ chessboard (made up of squares of length $1$) in such a way that each square contains at least one point covered by a coin ?
Piecemeal speaking the best you can do is intersect $6$ squares with one such coin, but this method leaves the margin of the board untouched.
Any ideas ?
Note that one coin can cover at most three consecutive squares, simply because its diameter is between 1 and 2. In other words, the squares touched by a single coin fall within a 3-by-3 bounding box.
Now consider the four corner squares, the middle squares on each side and the centre square of the chessboard:
A 3-by-3 box can cover at most one of these squares (marked by * above). Hence we need at least nine boxes, or coins, to cover every square of the board, which is also an upper bound:
where blocks of the same digit indicate the squares covered by one coin. Hence the answer to the original question is 9.