3 ms·

See response to pfedak.

• bzbarsky 10y ago
  I'm still not quite following. Is the "perimeter" here the sum of the lengths of projections of the edges, then how is that different from "the sum of its projections"? Anyway if you define "perimeter" that way then for the outer box the "perimeter" is in fact 4 * (a+b+c), where a, b, c are the dimensions of the box. I also agree that for the inner box the "perimeter" is at least 4 * (a'+b'+c'), by the triangle inequality. What is not at all clear to me is why the "perimeter" of the inner box is no larger than the "perimeter" of the outer box in this setup. You say it's easy to check, but it doesn't seem very obvious to me.
  • cousin_it 10y ago
    Let's say the projections of the inner box's edges onto the X axis have three distinct lengths a,b,c (all positive). Also let's say the distance between the leftmost and rightmost projected vertices is d. The nontrivial fact is that d=a+b+c, and not say a-b+c. That leads to the desired inequality, I think.
    • bzbarsky 10y ago
      Ah, I see. Yes, I agree that given that fact you get the desired inequality. But this fact is, as you say, nontrivial. It's certainly false for various non-box-like shapes afaict, so there's something special to boxes that needs proving here.
      • cousin_it 10y ago
        Just note that all combinations ±a±b±c can be realized by paths on the box. I'm sorry, I see now that my original comment glossed over tons of stuff that was clear only to me. Instead of saying "easy to check" I should've posted my napkin sketch with the easy check. Anyway, my solution is still simpler and more natural than the one in that pdf :-)