Search Results

Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties: $M(a,a)=a\qquad$ (identity) $M(a,b)=M(b,a)\qquad$ (commutativity). and possibly $M(M(a,b),M(a,c …

CLAIM. The only fair cutting with at least one quadrilateral is the square grid (2). It was already shown in the original answer (below) that there cannot be $n$-agons for $n\ge 5$. LEMMA. No fair cut …

Almost exactly the same picture can be drawn for the 2-dimensional and 3-dimensional cases (simpler than the OP's) and then it's easy to generalize to higher dimensions by induction. Here $I$ indicate …

Robert Israel's answer is best possible. As pointed out there, the problem is equivalent to finding a vector in $\mathbb{R}^d$ with minimal separation $1$ among sums of subsets of entries, and with th …

Here is one construction. On the horizontal xy plane place a forest of vertical cylinders of radius r<1/2 (or =1/2 if we allow contacts) centered at each point in $(\mathbb{Z} \backslash{\lbrace0\rbra …

Adding to Zeb's proof that the tetrahedron can be deformed, one should notice that any tassellation of the sphere containing at least one hexagon (fullerene type) won't be rigid either. In fact alrea …

Since Anton's beautiful solution makes use of the symmetry of the sphere, I wonder how similar results could be proven, or counterexamples given, for any other convex shape, including 2-dimensional on …

It seems that both the 2-agon and the octahedron (which after all is a collection of 3 somewhat constrained 2-agons) can be shrunk off the sphere, but with 0 derivative at the start, which means that …

Reid, excellent proof. It works for the cube too and even kills the icosahedron. In this last case I don't know what the angles of a triangle are exactly, so let's just say 60+ each. Then joining ea …

This addresses only the easy part. The unit $(n-1)$-sphere in $\mathbb{R}^n$ is covered by finitely many disks (sectors) each portending an acute cone from the origin. Each projection to a half-space …