2 ms·
Neat visualisations and very erudite article. Thanks! My observation is that the final formula is equivalent to two tetrahedra joined on a shared face. There m
by pcwelder 4y ago
Neat visualisations and very erudite article. Thanks!
My observation is that the final formula is equivalent to two tetrahedra joined on a shared face. There must be a good way to manipulate that directly to get the constant sums. I can find a way but that requires relative rotation of the second tetrahedron. There might be a better way.
- kilovoltaire 4y agoOnly just saw this comment, that's a great point! (Since the formula is two tetrahedra minus one triangle.) Now very curious why that shape corresponds to sum of cubes / why it's equivalent to a pyramid...