September 4th, 2017, 05:03 AM  #1 
Newbie Joined: Sep 2017 From: US Posts: 2 Thanks: 0  Tricky question of circles
Circle is a shape that cannot be tesselated. But how to optimize the number of circles fitted in a bounded region. Is there any relationship between the number of circles and the dimension of the regions. Design a rack with minimum area or perimeter to fit in 50 coke cans. Restrict the problem to racks in the shape of regular polygons. What kind of mathematical model can be used to make prediction?

September 4th, 2017, 06:28 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 830 Thanks: 244 
This is known as Malfatti's problem. Some references Excursions in geometry : Ogilvy : Oxford University Press Unsolved and Unsolvable Problems in Geometry : Meschowski : Oliver and Boyd The Seven Circles Theorem and Other New Theorems : Evelyn, MoneyCoutts and Tyrrell :Stacey International 
September 4th, 2017, 06:46 AM  #3 
Newbie Joined: Sep 2017 From: US Posts: 2 Thanks: 0 
THANKS sooo much!

September 4th, 2017, 08:12 AM  #4 
Senior Member Joined: Jun 2015 From: England Posts: 830 Thanks: 244 
You may like to know that Malfatti's problem is the dual or negative of yours. He wanted to cut cylinders out of regualr blocks of marble, with minimum wastage. You want to stack cylinders into minimum blocks. 
September 4th, 2017, 09:35 AM  #5  
Global Moderator Joined: Dec 2006 Posts: 19,162 Thanks: 1638  Quote:
Quote:
If the rack has the shape of an equilateral triangle large enough to contain a row of ten cans along one side, it can hold up to 55 cans and has perimeter 64.4 approximately. If the rack is square and uses square packing, it needs to have a side length of 16, and so will have a perimeter of 64. It will hold up to 64 cans. See this article and also this article if square packing needn't be used. If the rack has the shape of a regular hexagon and uses hexagonal packing, it needs to be large enough to contain a row of five cans along one side, it can hold up to 61 cans and I'll let you calculate its perimeter.  

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