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In $R^3$, since the spheres are concentric, not only all faces are regular, but also all edges are of the same length, and all faces are inscribed in circles of the same radius, hence are congruent. A …

By Ivan's request, I describe here two algorithms for finding the maximum-size ball contained in a given $n$-dimensional convex polytope $P$. The first algorithm, requiring central symmetry of $P$, pr …

A modification of Victor Protsak's answer yields a short and almost elementary proof: Since the set of convex $n$-dimensional convex polytopes is dense in the space of all $n$-dimensional convex bodie …

The answer is $no$. In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the squ …

This won't be a complete answer to Q2, but something of a starting point, at least for all two-dimensional convex $P$. Assuming for simplicity that the area of $P$ is 1, $P$ contains a parallelogram o …

$\bullet$ Every convex region of area $1$ contains a triangle of area at least as large as the area of the equilateral triangle inscribed in the circle of area $1$. Moreover, if the region does not co …

Begin with a lemma: Let $\ T=\Delta oab$ be a triangle contained in a circle centered at the origin $o$ and let the arc $\alpha$ of the circle be such that the circle's sector $S$ based on $\alpha$ h …

A closely related problem, considering the $(n-1)$-dimensional surface area instead of the diameter of the body, has been solved by Keith Ball, Volume ratios and a reverse isoperimetric inequality. J. …

The drawing below illustrates a proof for the theorem as stated in your question; almost no words of explanation are necessary. To make things easy, I write a few lines. The whole proof is based on th …

A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows. The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respec …