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?

ANSWERS

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:

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:

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 .854+ (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 .890+ (Gavin Theobald)	m=13 (10√13 – 29) / 8 = .881+ (Gavin Theobald)
m=14 .911+ (Gavin Theobald)	m=15 15/16 = .937+	m=16 1

n=2

m=3 2√3 – 5/2 = .964+	m=4 1	m=5 (11√10 – 33) / 2 = .892+	m=6 4√3 – 6 = .928+
m=7 .952+ (Gavin Theobald)	m=8 1	m=9 9√2 / 14 = .909+ (Joe DeVincentis)	m=10 4√5 – 8 = .944+ (Gavin Theobald)
m=11 33/2 - 11√2 = .943+	m=12 (2√6 – 2) / 3 = .966+	m=13 3√26 / 16 = .956+ (Joe DeVincentis)	m=14 (319 – 96√7) / 72 = .902+ (Gavin Theobald)
m=15 15/16 = .937+	m=16 1

n=3

m=4 2√3 – 5/2 = .964+	m=5 15/16 = .937+	m=6 11√2 / 4 – 3 = .889+ (Gavin Theobald)	m=7 15/16 = .937+ (Joe DeVincentis)
m=8 (2√6 – 2) / 3 = .966+	m=9 (6√3 – 3) / 8 = .924+ (Jeremy Galvagni)	m=10 8√30 / 15 – 2 = .921+ (Gavin Theobald)	m=11 .919+ (Gavin Theobald)
m=12 1	m=13 (√39 + 3) / 10 = .924+ (Joe DeVincentis)	m=14 √(6/7) = .925+	m=15 75/81 = .925+ (Joe DeVincentis)
m=16 √3 – 3/4 = .982+