Problem of the Month (September 2017)

For 0 ≤ s ≤ 1, consider the shape you get by removing an isosceles right triangle with legs of length s from one corner of a unit square. For positive integer n, and n2 ≤ k ≤ 2n2, what is the smallest value of s so that k of that shape can be packed inside a square of side n?

Here are the best known solutions.

n=1
 k=1 s = 0 k=2 s = 1

n=2
 k=4 s = 0 k=5 s = 1/√2 = .707+ k=6 s = 2√2–2 = .828+ k=8 s = 1

n=3
 k=9 s = 0 k=10 s = (15√2–4)/28 = .614+Maurizio Morandi k=11 s = (24√2–18)/23 = .693+Maurizio Morandi k=12 s = 3/4 = .750 k=13 s = 2√2–2 = .828+Maurizio Morandi k=14 s = (19–11√2)/4 = .860+Maurizio Morandi k=15 s = (11√2–10)/6 = .926+Maurizio Morandi k=16 s = (8–3√2)/4 = .939+ k=18 s = 1

n=4
 k=16 s = 0 k=17 s = (33√2–28)/34 = .549+Maurizio Morandi k=18 s = (1+√2)/4 = .603+Maurizio Morandi k=19 s = (73√2–44)/89 = .665+Maurizio Morandi k=20 s = 1/√2 = .707+ k=21 s = 3/4 = .750 k=22 s = (4+√2)/7 = .773+ k=24 s = 2√2–2 = .828+ k=26 s = (24√2–9)/28 = .890+Maurizio Morandi k=28 s = (17+79√2)/137 = .939+Maurizio Morandi k=30 s = (90√2–105)/23 = .968+Maurizio Morandi k=32 s = 1

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 9/19/17.