miniliv4

2022-10-08

Polygon of maximum area contained in compact, convex subset of ${\mathbb{R}}^{2}$?
Let H be a compact, convex subset of ${\mathbb{R}}^{2}.$. For a given $m\ge 3,$, let ${P}_{m}$ be a polygon of maximum area which contained in H and has atmost m sides. Then $\frac{Area\left({P}_{m}\right)}{Area\left(H\right)}\ge \frac{m}{2\pi }\mathrm{sin}\left(\frac{2\pi }{m}\right)$ and equality holds if and only if H is an ellipse.

Sanaa Hudson

Step 1
I think even a bound $\ge {ϵ}_{m}$ has some value. I will provide an ${ϵ}_{3}$.
Consider H compact convex body. Take $\mathrm{\Delta }{A}_{1}{A}_{2}{A}_{3}$ of largest area. The points ${A}_{i}$ are on the boundary. Because we cannot enlarge the area by moving ${A}_{1}$, there exists a supporting line to H through ${A}_{1}$ that is parallel to ${A}_{2}{A}_{3}$. Similarly for the other vertices. Thus we can inscribe H in a larger triangle $\mathrm{\Delta }{A}_{1}^{\prime }{A}_{2}^{\prime }{A}_{3}^{\prime }$ for which the ${A}_{i}$'s are the midpoints. Therefore $\mathrm{Area}\mathrm{\Delta }{A}_{1}{A}_{2}{A}_{3}\ge \frac{1}{4}\mathrm{Area}H$.
Step 2
Notice that if H were the ellipse tangent to the sides of the larger triangle at the midpoints ${A}_{i}$'s then we would have the exact bound stated ( the figure would be affinely equivalent to a circle inscribed in a equilateral triangle).

Do you have a similar question?