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#2 is super interesting! I wonder if it generalizes to higher dimensions.
by panic 10y ago
#2 is super interesting! I wonder if it generalizes to higher dimensions.
- alanbernstein 10y agoI don't quite follow the solution, can anyone explain a bit more? What's the motivation for considering large epsilon?
- yorwba 10y agoThe rigorous version would be: 0 < (Vol(Bε) - Vol(Aε))/ε² = ((a+b+c) - (a'+b'+c'))π + 2((ab + ac + bc) - (a'b' + a'c' + b'c'))/ε + (abc - a'b'c')/ε² The limit for ε to infinity must be >= 0 and is ((a+b+c) - (a'+b'+c'))π, therefore a+b+c >= a'+b'+c'. "large epsilon" is a standard way to express "consider the asymptotic behavior for epsilon to infinity and you will see what I mean".
- alanbernstein 10y agoThanks, I get that it means asymptotic behavior - that's the thing I don't get the motivation for.
- panic 10y agoI think the intuition is that we want to focus on the edges of the box (where the lengths sum together like the answer we're looking for), not the volume (where the lengths multiply together and cause problems). So we expand the edges until they completely dominate the equation.