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 $\frac{Area({P}_{m})}{Area(H)}}\ge {\displaystyle \frac{m}{2\pi}}\mathrm{sin}\left({\displaystyle \frac{2\pi}{m}}\right)$ and equality holds if and only if H is an ellipse.

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 $\frac{Area({P}_{m})}{Area(H)}}\ge {\displaystyle \frac{m}{2\pi}}\mathrm{sin}\left({\displaystyle \frac{2\pi}{m}}\right)$ and equality holds if and only if H is an ellipse.

Sanaa Hudson

Beginner2022-10-09Added 7 answers

Step 1

I think even a bound $\ge {\u03f5}_{m}$ has some value. I will provide an ${\u03f5}_{3}$.

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).

I think even a bound $\ge {\u03f5}_{m}$ has some value. I will provide an ${\u03f5}_{3}$.

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).

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