Recent content by huahaiy

I can't too. I carefully drew a picture with a ruler to figure it out :smile:

On top of three edges gone and coming, three extra edges have arrived. They were the diagonal lines of the three rectangles facing P, now have been pulled out of the planes to become edges of their own :smile:

Oh, Euler characteristic, I forget to use it. :shy: V + F = E + 2. For a cuboid, it's 8+6=12+2. But when a vertex is pulled out of place, there will be 3 extra edges, so it become 9 faces. Right?

I think I got it. The maximum volume seems to be 2/3*d*S + x*y*z, where S is the area of the triangle BCD. Is that right? Thanks a lot for the hint.

Look at the height? Thanks for the hint. But the rest doesn't seem to be fixed. My question was in fact wrong. The problem is more complicated than that: if A is replaced by P, to keep the resulting solid convex, a hexahedron won't do. We will get a solid with 9 sides instead of 6: Say...

Thanks for reply. Sorry I didn't make it clear. Basically, I mean the vertex of R that is closest to P is now replaced by P, this gives a new hexahedron which is bigger. Depending on where P is, the volume of this new hexahedron will be different. I would like to know what is the maximum...

Hello everyone, may I ask a question? I've been struggle with it for a few days. There's a cuboid (rectangular prism) R, with dimensions of x, y, z respectively. A point P is outside of R. The shortest distance between P and R is d. What is the maximum possible volume of the new hexahedron...