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Not reading very carefully, but at least this oVlad wrote: However, observe that for any $i$, $OA_i=\frac{r}{\cos \theta_i}$ is wrong, since $OA_i= \frac{r}{\sin (\theta_i /2)}$ actually.

You are right, I made a little confusion. Anyways, here is an official solution: Let $K$ be a convex $n$-gon (interior and boundary) containing a disc $D$ of radius $r$, centrednat some point $O$. Let $a_1,...,a_n$ be the vertices of $K$, labelled in a circular order around the boundary and let $2\alpha_i$ be the angular span of $D$ from $a_i$, $i=1,...,n.$ Note that $\text{diam} K>r+\text{dist}(O,a_i)=r\bigg(1+\frac{1}{\sin \alpha_i}\bigg)$ for each index $i$, so \[\text{diam} K>r\Bigg(1+\frac{1}{n}\sum_{i=1}^{n}\frac{1}{\sin \alpha_i}\Bigg)\geq r\Bigg(1+\frac{1}{\sin\frac{\alpha_1+\alpha_2+...+\alpha_n}{n}}\Bigg)\]by Jensen's inequality. Since $K$ is convex, the angular span $2\alpha_i$ does not exceed that of the internal angle of $K$ at $\alpha_i$ so $\frac{\alpha_1+\alpha_2+...+\alpha_n}{n}$ does not exceed $\frac{\pi}{2}-\frac{\pi}{n}$, and \[\sin\frac{\alpha_1+\alpha_2+...+\alpha_n}{n}\leq\sin\bigg(\frac{\pi}{2}-\frac{\pi}{n}\bigg)=\cos\frac{\pi}{n}.\] Consequently, \[\text{diam} K>r\Bigg(1+\frac{1}{\sin\frac{\alpha_1+\alpha_2+...+\alpha_n}{n}}\Bigg)\geq r\Bigg(1+\cos\frac{\pi}{n}\Bigg).\]