TITLE: Ratio of circumscribed/inscribed $(n{-}1)$-gons
QUESTION [14 upvotes]: As a discrete analog of the MO question,
"Löwner-John Ellipsoid: incribed and circumscribed,"
I've been wondering what might be the maximum ratio
of this quantity?
Let $P$ be a convex polygon of $n$ vertices,
$P^+$ be a minimum area polygon of $n{-}1$ vertices circumscribing
$P$, and $P^-$ a maximum area polygon of $n{-}1$ vertices
inscribed in $P$.

Over all polygons $P$ of $n$ vertices, what is the maximum
  of the ratio area$(P^+)/$area$(P^-)$, as a function of $n \ge 4$?

Here are possible optimal (en/in)closures for a square and a regular pentagon:
     


[Some updates below.]
I believe that it is known that the circumscribing $(n{-}1)$-gon must have one
edge "flush" with $P$ (i.e., including an edge of $P$ as a subset), and
that this may be proved in a 1984 paper by
Chang and Yap
"A polynomial solution for potato-peeling problem"
(Discrete & Computational Geometry
Volume 1, Number 1, 155-182), which I cannot access at the moment.
This is certainly true for enclosing triangles.
I am not sure if there is any analogous characterization of inscribed $(n{-}1)$-gons.
A key result for minimum area circumscribing is by Victor Klee in 1986:
"Facet-centroids and volume minimization,"
Studia Scientiarum Mathematicarum Hungarica, Vol. 21, 143-147, 1986.
As the title indicates, he proved that each facet's centroid must touch $P$ (in any dimension).
The circumscribing polygons I drew above have their edge midpoints touching $P$.
For $n{=}4$, $a \times b$ rectangles have an area ratio of $(2 a b) / (a b / 2) = 4$,
illustrated with the square above;
perhaps this is the worst ratio over all $n$?
Any ideas would be appreciated, from a clean proof (or counterexample!) that one circumcribing
edge must be flush, to any constraints on the inscribed polygons, to an answer to the ratio
question, even for specific $n$, even for regular $n$-gons.  Thanks!

REPLY [2 votes]: Just a note about inscribed polygons. It is easy to show that area$(P^-)\ge$ area$(P)(1-1/(n-2))$. This follows from the fact that among the triangles formed by three vertices of $P$, (at least) one with minimum area must ``lie'' on a side of $P$. (Proof: Take minimum area triangle, use area$\cdot 2=$ height$\cdot$base and move vertices.) Now triangulate $P$ such that this minimum area triangle appears. Omitting it we get a polygon with $n-1$ vertices, done. This bound is also sharp for regular $n$-gons. (Proof is again by moving the vertices of maximum area $(n-1)$-gon.
I guess the next task is to determine the minimum of area$(P^+)/$area$(P)$.