Light inside a polyhedron I have two questions the same as Mostafa's Question:
Visibility of vertices in polyhedra
Suppose $P$ is a closed polyhedron in space (i.e. a union of polygons which is homeomorphic to $S^2$) and $X$ is an interior point of $P$.
We know that maybe $X$ can't see any entire face or edge of $P$, because in O'Rourke or Khezeli's examples $X$ can't see any vertex of $P$. But how about just part of faces or edges in $X$'s view. At least how many faces or edges can $X$ see, even part of them?
So let $X$ is a light point inside a simple polyhedron $P$:


*

*Is it true that $X$ can see part of at least four different faces of $P$?

*Is it true that $X$ can see part of at least six different edges of $P$?


We proved Q1, I think the answer of Q2 is Yes too, but I have no simple idea yet.
 A: At least Q2 is definitely wrong!
Just consider the great dodecahedron. It is non-convex, faces are (convex) regular pentagons, edges are the same ones as those of its covex hull (an icosahedron). But its kernel is a tiny dodecahedron, which touches only onto the faces, but neither onto the edges nor vertices.
The same holds for the great icosahedron, which even has (regular) triangular faces only (as you required). Here the edges are the same as for the stellated dodecahedron, but the kernel is a tiny (convex) icosahedron, which touches only onto the faces, but neither onto the edges nor vertices. (It is only that due to even higher density that the inscribed icosahedron becomes less easy to visualize in this case.)
Edit according to comment discussion:
you well could consider the zigzag push-pleated prism, cf. the also added pic: that one clearly is continously transformable into a fully convex body, i.e. homeomorphic to $S^2$, but still no bit of its edges at B are visible from a point A! - And if you'd contract the lower triangle into a single point (i.e. would use a push-pleated pyramid instead) you get that the point A could see 3 of the edges of P only!

--- rk
A: The answer is yes:
Place a small sphere $S$ around $X$ and project the visible faces and (if any) edges and vertices onto this sphere.  Each face must subtend a maximum angle strictly less than $\pi$ since it is bounded and does not contain $X$.  From Q1 the number of faces on $S$ satisfies $F\geq 4$.  Each edge (boundary between faces) on $S$ corresponds to an edge of $P$, not necessarily between the relevant faces (possibly between the nearer face and an invisible face).  A line segment projected to a sphere becomes an arc of a great circle.  Each pair of distinct great circles intersects exactly twice, at opposite points, thus subtending an angle $\pi$.  Because each face subtends a maximum angle less than $\pi$ it must be bounded by more than two edges.  Choose a face with at least three vertices on $S$ (which need not correspond to vertices on $P$).  This does not divide four faces on $S$, so there must be at least one further vertex, so the number of vertices on S satisfies $V\geq4$.  Thus from the Euler formula for $S$, the number of edges is
$E=V+F-2\geq 4+4-2=6$.
If $P$ is a tetrahedron we have exactly $6$ edges, so the bound is tight.
