What is the smallest number of disks radius ½ that can cover a disk radius 1?
Solution
Evidently we maximise ∠POQ by taking X on PQ to give 60o. Thus no one disk can cover more than 1/6 of the perimeter of the disk. If it covers 1/6, then we have ∠PXO = 90o and hence OX = √(1-1/4) = √3/2 > ½. The only way to use just 6 to cover the perimeter is to put all their centers on the circle radius √3/2, in which case they do not cover the center of the disk. So we need at least 7 disks.
We show that 7 suffice. Take 6 equally spaced disks with centers √3/2 from O. Take two adjacent disks, centers X and Y, with perimeters meeting at D inside the large disk. Then XYO is an equilateral triangle side √3/2. XD = ½, so DM = 1/4 (by Pythagoras). Hence OD = ½. So 7 disks are just sufficient.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03