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?

ANSWERS

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.