# Ratio of circumscribed/inscribed $(n{-}1)$-gons

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: 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!

• Why do you keep writing $n{-}1$ instead of $n-1$? Jan 29, 2012 at 0:22
• @Michael: Just a precisionist controlling exact spacing. :-) Jan 29, 2012 at 1:06
• The special case with all regular polygons would be interesting too — for that case finding the areas would boil down to finding the points of contact.
– user44143
Sep 9, 2022 at 13:20

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$(P)$.