3 ms·
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 lef
by 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 agoAh, 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 agoJust 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 :-)