Maximum volume of a hexahedron

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Discussion Overview

The discussion revolves around determining the maximum possible volume of a hexahedron formed by a cuboid and an external point. Participants explore the geometric implications of replacing a vertex of the cuboid with the external point and how this affects the volume of the resulting solid.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes a formula involving the distance from the point to the cuboid and the dimensions of the cuboid to estimate the volume of the new hexahedron.
  • Another participant clarifies that replacing the closest vertex of the cuboid with the external point results in a solid with more than six sides, specifically a nine-sided solid.
  • There is a discussion about the implications of the Euler characteristic in relation to the number of faces, edges, and vertices after the transformation.
  • Some participants suggest that the volume calculation remains valid despite the change in the nature of the solid from a hexahedron to a different polyhedron.
  • Participants express uncertainty about the geometric properties and the correct classification of the resulting solid.

Areas of Agreement / Disagreement

Participants generally agree that the problem is more complex than initially presented, particularly regarding the classification of the resulting solid. However, there is no consensus on the exact nature of the solid or the implications for the volume calculation.

Contextual Notes

Participants note that the volume calculation may still hold despite the change in the solid's classification, but the discussion remains open regarding the geometric properties and the implications of the transformation.

huahaiy
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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 formed by P and R?

Is something like d*sqrt((x*y)^2 + (y*z)^2 + (x*z)^2) + x*y*z close?
 
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Welcome to PF!

huahaiy said:
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 formed by P and R?

Hello huahaiy! Welcome to PF! :smile:

I don't understand … what do you mean by the hexahedron (six-sided solid) formed by P and R? :confused:
 


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 volume. Is this clearer?

tiny-tim said:
Hello huahaiy! Welcome to PF! :smile:

I don't understand … what do you mean by the hexahedron (six-sided solid) formed by P and R? :confused:
 
ah!

huahaiy said:
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 volume. Is this clearer?

ah!

Hint: if the vertex A is closest to P, and A is connected to vertices B C and D, then the volume of the new hexahedron depends on the volume of tetrahedron PBCD (the rest is fixed).

And if you have two tetrahedra with the same base (BCD), how do you compare their volumes? :wink:
 


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 vertex A is closest to P, A is connected to vertices B, C and D. Both B and C are connected to vertex E. After P replaced A, the rectangle ABCE will be turned into two triangles PBE and PCE, which are not on the same plane! So we end up with a 9 sided solid.

If this is the case, the part that is fixed is only the tetrahedron to the opposite side of A.

tiny-tim said:
ah!

Hint: if the vertex A is closest to P, and A is connected to vertices B C and D, then the volume of the new hexahedron depends on the volume of tetrahedron PBCD (the rest is fixed).

And if you have two tetrahedra with the same base (BCD), how do you compare their volumes? :wink:
 


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.

huahaiy said:
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 vertex A is closest to P, A is connected to vertices B, C and D. Both B and C are connected to vertex E. After P replaced A, the rectangle ABCE will be turned into two triangles PBE and PCE, which are not on the same plane! So we end up with a 9 sided solid.

If this is the case, the part that is fixed is only the tetrahedron to the opposite side of A.
 
Hi huahaiy! :smile:
huahaiy said:
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:

I don't follow that :confused: … if the cuboid is made of twelve pieces of elastic joining eight vertices, then pulling one vertex out of position will still give eight faces …
E + F = V + 2. :smile:​
huahaiy said:
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?

Looks good! (except … isn't it 1/3 ?) :biggrin:
 
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?

tiny-tim said:
Hi huahaiy! :smile:


I don't follow that :confused: … if the cuboid is made of twelve pieces of elastic joining eight vertices, then pulling one vertex out of position will still give eight faces …
E + F = V + 2. :smile:​


Looks good! (except … isn't it 1/3 ?) :biggrin:
 
huahaiy said:
But when a vertex is pulled out of place, there will be 3 extra edges, so it become 9 faces. Right?

But the number of vertices is the same, so you've removed one vertex and replaced it by another.

And the removed vertex had three edges, so they've gone also.

So three edges have gone, and three new ones have arrived.

That's still 12 edges and 6 faces, isn't it? :smile:
 
  • #10
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:

tiny-tim said:
But the number of vertices is the same, so you've removed one vertex and replaced it by another.

And the removed vertex had three edges, so they've gone also.

So three edges have gone, and three new ones have arrived.

That's still 12 edges and 6 faces, isn't it? :smile:
 
  • #11
ping!

huahaiy said:
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:

ah … got it! … :biggrin:

(i can't think in 3D! :redface:)

yes, you're right … the four vertices of each of the three new "faces" aren't in a plane, so each "face" is really two faces, and there are three new edges along the base of our pyramid …

so the volume calculation is still correct, but it isn't a hexahedron! :smile:
 
  • #12


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

tiny-tim said:
ah … got it! … :biggrin:

(i can't think in 3D! :redface:)

yes, you're right … the four vertices of each of the three new "faces" aren't in a plane, so each "face" is really two faces, and there are three new edges along the base of our pyramid …

so the volume calculation is still correct, but it isn't a hexahedron! :smile:
 

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