Convex Polyhedron: How many corners maximum? How many corners can a $n$-dimensinal convex polyhedron have at tops? Is it the same as the number of corners a $n$-dimensional simplex has?
EDIT:
By polyhedron $P$, I mean, that for some matrix $A \in \mathbb{R}^{m,n}$, $P = \{ x\in \mathbb{R}^n \mid A x \leq b\}$ where $Ax \leq b$ is meant as $a_i^T x \leq b$.
An alternative charakterization would be $P = \cap_{H \text{ is hyperplane}} H_{+}$, where $H_{+}$ is the half-room $\{x \in \mathbb{R}^n \mid \langle x, u \rangle \geq b\}$.
ADDED convex
 A: A vertex of the polyhedron $P = \{x \in {\mathbb R}^n: Ax \le b\}$ is defined by the solution of $n$ of the equations $(A x)_i = b_i$ corresponding to linearly independent rows of $A$.
In general, not all of those will be vertices, because they may violate other constraints.  But the number of vertices is at most $m \choose n$.  
EDIT: P. McMullen improved the upper bound to $${{m - \lfloor (n+1)/2 \rfloor} \choose {m-n}} + {{m - \lfloor (n+2)/2 \rfloor} \choose {m-n}}$$
which turns out to be best possible.  See
 P. McMullen, "The maximum number of faces of a convex polytope",
Mathematika 17 (1970), pp. 179–184, and
D. Gale, "Neighborly and cyclic polytopes", in  Proc. Sympos. Pure Math., Vol. VII, Amer. Math. Soc., Providence, R.I., 1963, pp. 225–232,
 http://books.google.ca/books?hl=en&lr=&id=MuEFJR7Ek4EC&oi=fnd&pg=PA225 
A: An $n$-dimensional polyhedron can have an arbitrarily large number of vertices. For regular convex polytopes the maxima are
Dimension   Maximum vertices

    2         unlimited
    3         20 (dodecahedron)
    4        600 (120-cell)
    5+       2^n (n-cube)

