- #36
logics
- 137
- 0
reductio ad absurdum [[itex]\rightarrow\leftarrow[/itex]] and symmetry give you an instant proof by contradiction. I do not know if there is another proof, but surely it would be complex:vrmuth said:* a) shortest and the tallest are always on the opposite corners... b) h1+h3 = h2+h4 ( any proof ?)
Axis of rotation can be: space diagonal or face diagonal.
A plane rotating on a space diagonal [h1,h3] divides a cuboid (V=a²h3) into two symmetrical, complementary solids [h2 = h3-h4]: V1= a²h3/2 (case 1, h: 0,2,8,6 + 8,6,0,2)
Two such planes delimit a (Vw= 1/6...) wedge: V2 = V1 [itex]\pm[/itex] Vw (case 2, h: 0,2,8,5 + 8,6,0,3, hw =1)
If we are to exclude concave, convex or goofy surfaces, other cases are determined by: additional plane on axis [h2, h4], planes rotating on face diagonal axis etc.
The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h 1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1, brilliantly solved by Dave: V = a²h1+3/ 2
Edit: V = a²h(0-1)+3/2
Last edited: