Shrink this arrangement of coin centers by 1/2. Now every point (of the shrunken rectangle) is within R of a coin center, i.e. every point is covered. The original table can be tiled with 4 copies of this.

]]>4 coins of radius R are placed with their centers on the corners of a square table having a diagonal of 4R. (This is my 'best' worst case'. ie. it ALMOST meets requirements because a coin will fit in the central hole. Now to prove that this square of sides of 2R * root 2 can be painted with 16 coins. As it sits now the square has an area of 8r^2 square units. Utilizing only area of the largesr squares contains a coin it is clear the area may be tiled with the coins available. ]]>