m \ n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8
2
| s = 1 + 1/√2 = 1.707+ s = 2 s = 1 + √2 = 2.414+ s = 2.724+ (MM) s = 1 + 3/√2 = 3.121+ s = 3.380+ (DH) s = 3.599+ (MM) after (DC) s = 3.819+ (MM) 3
| s=1+sin(π/12)=1.258+ s = (1 + √7)/2 = 1.822+ s = 2 s = 2.309+ (MM) s = 2.636+ (MM) after (EF) s = 2.846+ (MM) s = 2.999+ (DC) after (MM) s = 2 + 2/√3 = 3.154+ 4
| s = 1 s = √2 = 1.414+ s = 1 + 1/√2 = 1.707+ s = 1.942+ s=(3√2+√38)/5=2.081+ s = 2.375+ (DC) after (MM) s = 2.594+ (MM) s = 2.699+ (DC) after (MM) 5
| s = cos(π/20) = .987+ s = √2 = 1.414+ s = 1.673+ (MM) s=1+tan(π/5)=1.726+ (MM) s = 1.997+ (MM) after (DC) s = 2.176+ (DC) after (EF) s = 2.331+ (SH) and (TG) s = 2.439+ (DC) after (MM) 6
| s = cos(π/12) = .965+ s = 1.244+ (MM) s = 1.491 (MM) s = 1 + 1/√3 = 1.577+ (MM) s = 1.833+ (MM) s = 1.951+ (MM) s = 2.096+ (MM) s = 2.187+ (MM) 7
| s = cos(3π/28) = .943+ s = 1.109+ (MM) s = 1.340+ (DC) after (MM) s=1+tan(π/7)=1.481+ (MM) s = 1.685+ (MM) s = 1.815+ (MM) s = 1.940+ (MM) s = 2.021+ (MM) 8
| s = cos(π/8) = .923+ s = 1 s=(1+√2)/2=1.207+ s = 1.398+ (DC) after (MM) s = 1.554+ (DC) after (EF) s=√(14-8√2)=1.638+ (MM) after (EF) s = 1.786+ (DC) s = 1.915+ (MM) 9
| s = cos(5π/36) = .906+ s = cos(π/36) = .996+ s = 1.166+ s = 1.282+ (MM) s = 1.456+ (DC) after (JD) s = 1.551+ (MM) s = 1.704+ (DC) s = 1.798+ (JD) 10
| s = cos(3π/20) = .891+ s = cos(π/20) = .987+ s = 1.108+ (DC) after (JD) s = 1.184+ s = 1.371+ (DC) after (JD) s = 1.472+ (MM) s = 1.594+ (MM) s = 1.681+ (JD) |
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Bryce Herdt asks a couple questions:
Is there a limiting value for n=m? Seems to me it should be √π. Jeremy Galvagni sent the construction below, which might work. He also notes that if m/n → 2/π, the slices can be placed alternately up and down to fit inside a square of side approaching 1.
Clearly s is increasing in n and decreasing in m, but are these always strict? I'll guess yes. Does anyone have a counterexample?
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 1/26/12.