Explicit triangular-cylinder cover below half area

The tetrahedron \eqref{eq:tetrahedron} has $A_{\min}(K)=\sqrt2$.
For every $0<\eps\le1/2000$, it admits a covering by
$m=2\lceil2/\eps^2\rceil$ cylinders with compact triangular perpendicular
bases such that
\begin{equation}
 \frac1{\sqrt2}\sum_{i=1}^m|B_i|
 =\frac12-\frac{13}{6000}\eps^2+O(\eps^4).
\end{equation}
The remainder has absolute value at most $2\eps^4$, and the total base
area is strictly less than $A_{\min}(K)/2$.

Every nondegenerate tetrahedron $T\subset\R^3$ admits a finite cylinder
covering with compact triangular perpendicular bases for which
\[
 \sum_{i=1}^m\frac{|B_i|}{|\pi_{u_i^\perp}T|}<\frac12.
\]
Thus the 1-Codimensional Cylinder Covering Conjecture is false.

Let $h>0$, let $T\subset\R^2$ be bounded and measurable, and let
$g\in\R^2$. The image $F_h(\Cyl(T,g))$ is a cylinder whose perpendicular
base has area $h|T|J_h(g)$. A compact nondegenerate parallelogram tile
has a compact nondegenerate parallelogram base. An open tile gives an
open base. The formula is unchanged after a Euclidean isometry.

Let $D\subset\R^2$ be compact, let $V$ be $C^1$ on an open
neighborhood of $D$, and let $L,h>0$. Suppose that
\begin{equation}
 \tr DV(p)=0,\qquad -L^{-2}<\det DV(p)\le0
 \quad(p\in D).
\end{equation}
For every $\varepsilon>0$, there are finitely many compact
nondegenerate square tiles $T_i\subset\R^2$, with possibly different
orientations, and velocities $g_i\in\R^2$ such that
\begin{equation}
 E(D,V,L)\subset\bigcup_i\Cyl(T_i,g_i),\qquad
 \sum_i |T_i|J_h(g_i)
 \le\int_D J_h(V(p))\,dp+\varepsilon.
\end{equation}
Under $F_h$, the perpendicular bases are compact nondegenerate
parallelograms and have total area $h\sum_i|T_i|J_h(g_i)$.
No condition on the area of $\partial D$ or smallness of $\|DV\|$
is required.

Every regular tetrahedron $K\subset\R^3$ of positive edge length admits a finite
cylinder cover with compact parallelogram perpendicular bases such that
\[
 \sum_i|B_i|<\tfrac12 A_{\min}(K),
 \qquad
 \sum_i\frac{|B_i|}{|\pi_{u_i^\perp}K|}<\tfrac12.
\]
It also admits finite covers with bounded open measurable bases
satisfying both inequalities.

Every nondegenerate tetrahedron $T\subset\R^3$ admits a finite cylinder
cover $\mathcal C$ with compact parallelogram perpendicular bases and
$\mathcal R(\mathcal C;T)<1/2$. The same conclusion holds with bounded
open measurable bases.

For the data in \eqref{eq:radial-sweep}, suppose that
\[
 w'(t)=\lambda(t)(a+tb)\qquad(t\in J)
\]
for a continuous real function $\lambda$.
For any finite partition
\(\alpha=t_0<t_1<\cdots<t_N=\beta\), put \(I_i=[t_{i-1},t_i]\)
and choose arbitrary tags \(c_i\in I_i\). There are \(N\) cylinders
covering \(\mathcal S\), with axes parallel to \(e+w(c_i)\) and
nondegenerate compact triangular perpendicular bases. These covers
can be chosen so that their total base area tends to \(\mathcal A\)
as the mesh \(\delta=\max_i|I_i|\) tends to zero, uniformly over the
partitions and tags.

Alternatively, take each base to be the perpendicular projection of
the actual compact sweep with \(t\in I_i\), in the same frozen
direction \(e+w(c_i)\). These \(N\) cylinders cover \(\mathcal S\),
and their total base area also tends to \(\mathcal A\), uniformly
over the partitions and tags.

Every regular tetrahedron $K\subset\R^3$ of positive edge length has
a finite cylinder cover $B_i+\R u_i$, with unit axes $u_i$ and
nondegenerate compact triangular bases $B_i\subset u_i^\perp$, such that
\[
 \sum_i\area(B_i)<\frac{A_{\min}(K)}2,
 \qquad
 \sum_i\frac{\area(B_i)}{\area(\pi_{u_i^\perp}K)}<\frac12.
\]