There are at least two ways to mathematically describe a polyhedron, either by specifying its corners, or by specifying its bounding hyperplanes. This is illustrated using a cube of side length 1 with one corner at the origin and its three adjacent edges parallel to the three coordinate axes.

Plot the cube:

$$\begin{gathered} \mathrm{with}\left(\mathrm{plots}\right)\: \\ \mathrm{with}\left(\mathrm{plottools}\right)\: \end{gathered}$$

$\mathrm{display}\left(\mathrm{cuboid}\left(\left[0\,0\,0\right]\,\left[1\,1\,1\right]\right)\,\mathrm{axes}\=\mathrm{boxed}\,\mathrm{labels}\=\left[x\,y\,z\right]\right)$

Define this cube by bounding planes:

$\mathrm{c1}≔\left[x\ge 0\,x\le 1\,y\ge 0\,y\le 1\,z\ge 0\,z\le 1\right]$

Alternatively, define the vertices of the same cube:

$V≔\mathrm{Matrix}\left(\left[\left[0\,0\,0\right]\,\left[0\,0\,1\right]\,\left[0\,1\,0\right]\,\left[0\,1\,1\right]\,\left[1\,0\,0\right]\,\left[1\,0\,1\right]\,\left[1\,1\,0\right]\,\left[1\,1\,1\right]\right]\right)$

Suppose now that you are given the vertex representation and want to automatically derive the bounding planes representation. The cube is defined as the convex hull of all its vertices:

$\mathrm{op}\left(\mathrm{convert}\left({\mathrm{Vector}}_{\mathrm{row}}\left(\left[x\,y\,z\right]\right)\=~\mathrm{add}\left({\mathrm{\λ}}_i\cdot V_i\,i\=1..8\right)\,\mathrm{list}\right)\right)$

$\mathrm{add}\left({\mathrm{\λ}}_i\,i\=1..8\right)\=1$

$\mathrm{seq}\left({\mathrm{\λ}}_i\ge 0\,i\=1..8\right)$

$\mathrm{seq}\left({\mathrm{\λ}}_i\le 1\,i\=1..8\right)$

$\mathrm{sys}≔\left[\,\,\,\right]\:$

You can now use the LinearSolve command to eliminate all ${\mathrm{\λ}}_i$ from this system of equations and inequalities. The projection option is used to specify that elimination is required, and the canonical option to eliminate redundant equations and inequalities.

$\mathrm{with}\left(\mathrm{RegularChains}\right)\:\mathrm{with}\left(\mathrm{SemiAlgebraicSetTools}\right)\:$

$\mathrm{vars}≔\left[\mathrm{seq}\left({\mathrm{\λ}}_i\,i\=1..8\right)\,z\,y\,x\right]$

$\mathrm{LinearSolve}\left(\mathrm{sys}\,\mathrm{PolynomialRing}\left(\mathrm{vars}\right)\,\mathrm{canonical}\,\mathrm{projection}\=3\right)$

This agrees with the representation $\mathrm{c1}$ above.

For every convex 3-D polyhedron, there is a dual polyhedron with the roles of corners and faces swapped. Let us compute the dual polyhedron of the cube. Start with a representation of the cube with the origin as the center.

$\mathrm{c2}≔\left[x\le 1\,-x\le 1\,y\le 1\,-y\le 1\,z\le 1\,-z\le 1\right]$

Now rewrite this in matrix form.

$B≔\mathrm{Matrix}\left(\left[\left[1\,0\,0\right]\,\left[-1\,0\,0\right]\,\left[0\,1\,0\right]\,\left[0\,-1\,0\right]\,\left[0\,0\,1\right]\,\left[0\,0\,-1\right]\right]\right)$

$v≔\mathrm{Vector}\left(\left[x\,y\,z\right]\right)$

$c≔\mathrm{Vector}\left(\left[1\,1\,1\,1\,1\,1\right]\right)$

$B\.v\le ~c$

In general, if a bounded convex polyhedron is given by a set of inequalities of the form $B\.v\le 1$, then the vertices of its dual are exactly the columns of the matrix $B$. The convex hull of these six vertices defines an octahedron. Using the same method as before, compute the bounding planes for this octahedron.

$\mathrm{op}\left(\mathrm{convert}\left({\mathrm{Vector}}_{\mathrm{row}}\left(\left[x\,y\,z\right]\right)\=~\mathrm{add}\left({\mathrm{\λ}}_i\cdot B_i\,i\=1..6\right)\,\mathrm{list}\right)\right)$

$\mathrm{add}\left({\mathrm{\λ}}_i\,i\=1..6\right)\=1$

$\mathrm{seq}\left({\mathrm{\λ}}_i\ge 0\,i\=1..6\right)$

$\mathrm{seq}\left({\mathrm{\λ}}_i\le 1\,i\=1..6\right)$

$\mathrm{sys}≔\left[\,\,\,\right]\:$

$\mathrm{vars}≔\left[\mathrm{seq}\left({\mathrm{\λ}}_i\,i\=1..6\right)\,z\,y\,x\right]$

$S≔\mathrm{LinearSolve}\left(\mathrm{sys}\,\mathrm{PolynomialRing}\left(\mathrm{vars}\right)\,\mathrm{canonical}\,\mathrm{projection}\=3\right)$

The first two inequalities represent that projection of the octahedron onto the $x$ axis:

$S_{1..2}$

Together with the following four inequalities, which involve only $x$ and $y$ but not $z$, this defines the projection of the octahedron onto the $x\,y$-plane:

$S_{1..2}\; S_{3..6}$

The following plot shows this projection (blue) together with the four boundary lines.

$\mathrm{inequal}\left(S_{3..6}\,x\=-2..2\,y\=-2..2\right)$

Finally, the last 8 inequalities, i.e., the ones containing $z$, define the bounding planes of the octahedron.

$S_{-8..-1}$

The following 3-D plot displays the tetrahedron as well as the last of these 8 bounding planes. Rotate the plot with the mouse in order to see that the plane actually touches one of the faces.

$\mathrm{plane}≔\mathrm{Student}:-\mathrm{LinearAlgebra}:-\mathrm{PlanePlot}\left(\left(\mathrm{lhs}-\mathrm{rhs}\right)\left(S_{-1}\right)\=0\,\left[x\,y\,z\right]\,\mathrm{caption}\=''''\,\mathrm{shownormal}\=\mathrm{false}\,\mathrm{showpoint}\=\mathrm{false}\,\mathrm{planeoptions}\=\left[\mathrm{transparency}\=0.3\right]\right)\:$

$\mathrm{display}\left(\mathrm{octahedron}\left(\left[0\,0\,0\right]\right)\,\mathrm{plane}\,\mathrm{axes}\=\mathrm{boxed}\,\mathrm{labels}\=\left[x\,y\,z\right]\right)$

RegularChains[SemiAlgebraicSetTools][LinearSolve]