# 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=1s = 0 k=2s = 1

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

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

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