Problem of the Month (May 2008)

Let n < m be positive integers. What is the largest shape with the property that n of them can be packed inside a square of area n, and m of them can be packed inside a square of area m ? Can you beat any of the results below?

Jeremy Galvagni noticed that if m=(a/b)2n, then area 1 could be trivially covered. He also showed that when n=m-k, that area √(1-k/m) can be covered using thin rectangles: He later improved this bound to area 2√(m/n) + 3√(n/m) – 4 when n is even, and √(n/m) + 2n√(n/m)/(n+1) + 2√(mn)/(n+1) – 4n/(n+1) when n is odd.

Joe DeVincentis noticed that if k is the smallest square number greater than n, then we can use n of k squares, for a fill fraction of n/k > n/(n+2√n), which approaches 1 as n approaches ∞.

Here are the best known solutions, together with the proportion of the area covered. Click on the pictures for the figures of m regions in squares of area m.

n=1
 m=2 2√2 – 2 = .828+ m=3 (2√3 + 3) / 8 = .808+(Gavin Theobald) m=4 1 m=5 .889+ m=6 7/√6 – 2 = .857+(Gavin Theobald) m=7 (25√7 – 49) / 20 = .857+(Gavin Theobald) m=8 √2 – 1/2 = .914+ m=9 1 m=10 .906+(Gavin Theobald) m=11 2√11 – 23/4 = .883+(Gavin Theobald) m=12 .897+(Gavin Theobald) m=13 .892+(Gavin Theobald) m=14 .913+(Gavin Theobald) m=15 .937+(Gavin Theobald) m=16 1

n=2
 m=3 2√3 – 5/2 = .964+ m=4 1 m=5 .948+(Gavin Theobald) m=6 (51 - 26√3)/6 = .994+(Gavin Theobald) m=7 .969+(Gavin Theobald) m=8 1 m=9 .973+(Gavin Theobald) m=10 √10 + √5 - √2 - 3 = .984+(Maurizio Morandi) m=11 .957+(Gavin Theobald) m=12 (36 - 5√6)/24 = 0.989+(Maurizio Morandi) m=13 .963+(Gavin Theobald) m=14 .986+(Maurizio Morandi) m=15 .978+(Gavin Theobald) m=16 1

n=3
 m=4 2√3 – 5/2 = .964+ m=5 .960+(Gavin Theobald) m=6 6√2 – 15/2 = .985+(Gavin Theobald) m=7 (224√2 – 35) / 289 = .975+(Gavin Theobald) m=8 (2√6 – 2) / 3 = .966+ m=9 .974+(Gavin Theobald) m=10 .975+(Maurizio Morandi) m=11 .961+(Maurizio Morandi) m=12 1 m=13 .964+(Maurizio Morandi) m=14 (224√2 – 35) / 289 = .975+(Gavin Theobald) m=15 .966+(Gavin Theobald) m=16 √3 – 3/4 = .982+(Gavin Theobald)

n=4
 m=5 .974+(Maurizio Morandi) m=6 (36 – 5√6) / 24 = .989+(Maurizio Morandi) m=7 63/64 = .984+(Joe DeVincentis) m=8 1 m=9 1 m=10 .986+(Maurizio Morandi) m=11 .981+(Maurizio Morandi) m=12 (36 - 7√3)/24 = .994+(Maurizio Morandi) m=13 .985+(Maurizio Morandi) m=14 .986+(Maurizio Morandi) m=15 .978+(Gavin Theobald) m=16 1

n=5
 m=6 .973+(Maurizio Morandi) m=7 35/36 = .972+ m=8 .975+(Maurizio Morandi) m=9 .977+(Maurizio Morandi) m=10 .974+(Maurizio Morandi) m=11 .9618+(Maurizio Morandi) m=12 .965+(Maurizio Morandi) m=13 .964+(Gavin Theobald) m=14 35/36 = .972+ m=15 .954+(Gavin Theobald) m=16 (4√5 – 4)/5 = .988+(Gavin Theobald)

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 12/16/09.