Find a convex polygon $P$ s.t. $P\subset Q \subset (1+\epsilon) P $.

Question

Problem : Assume that $Q$ is a compact convex plane figure containing $R$-ball of the center origin. Then prove that for any $\epsilon >0$, there is a convex polygon $P\subset Q$ s.t. $$ P\subset Q\subset (1+\epsilon)\cdot P $$

Proof : Define $A = \bigg\{ \delta (x,y) \bigg| x,\ y\in \mathbb{Z}\bigg\}$ From this the book says that $\epsilon = \frac{10\ {\rm diam}\ Q}{R}\delta,\ \delta < \frac{R}{10}$

Question : How can we obtain this ?

The following is my different approach :

(1) $x_i$ is point in $\partial Q$ s.t. $|x_i-x_{i+1} | \leq R$

Then ${\rm conv}\ \{x_i\}$ contains $\frac{R}{2}$-ball.

(2) Redefine $x_i\in \partial Q$ s.t. the arc $\widetilde{x_ix_{i+1}}$ in $\partial Q$ has a length $\leq M=\min\ \{\ R\epsilon , R\ \} \ (\ast)$.

If $$|o-x_i|,\ |o-x_{i+1}|\geq \frac{R}{2},$$ then consider a line containing $[x_ix_{i+1}]$ and a line containing $ (1+\epsilon )\cdot [x_i x_{i+1}]$.

Note that the distance between two lines is greater than $\frac{R}{2} \epsilon$.

From $\ast$ the arc $\widetilde{x_ix_{i+1}}$ is between two triangles $ox_ix_{i+1},\ o(1+\epsilon )x_i(1+\epsilon)x_{i+1}$.

Hence we have the desired $P ={\rm conv}\ \{x_i\}$.

Here in $\ast$ we do not have diameter term.

Reference : From Euclid to Alexandrov; a guided tour - Petrunin and Yashinski.

Answer 1

Fix $\varepsilon$. Define $A = \{ (m,n)\delta |m,\ n\in \mathbb{Z}\}$ We let $\delta = \frac{R}{\sqrt{2}}\frac{1}{M}$ for large positive integer $M$.

Then the convex polygon $C = {\rm conv} (A\cap Q)$ contains $\frac{R}{2}$-ball whose center is origin $o$.

For $x\in \partial Q$, let $y\in [xo] $ s.t. $$ |x-y|=d< \frac{R}{2} \varepsilon$$

Further there is $r>0$ s.t. a closed ball $B[r,y]$ is in $C$. Here $r$ is independent of $x,\ y$.

If $M$ is large, then each $B[r,y]$ contains at least nine squares generated from $A$.

Here we can assume that a square placed in center contains a point $y$. Hence $C$ contains $y$.

If $C\cap [ox]=[oz]$ where $z$ is a point in a line segment $[ox]$, then \begin{align*} |oz| (1+ \varepsilon )&\geq |oz| + \frac{R}{2}\varepsilon \\&\geq |oz| +d \\&\geq |ox| \end{align*} which finishes the proof.