48th Kürschák 1947 Problem 3 What is the smallest number of disks radius ½ that can cover a disk radius 1?
Evidently we maximise ∠POQ by taking X on PQ to give 60 o . 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 = 90 o 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. 48th Kürschák 1947 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03