Relations between a convex polygon and an ellipse Let $F$ be bounded convex polygon on plane.

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*How can we justify that there exists circle $K$ s.t. if an ellipse contains $F$ and is not contained in $K$, then field of this ellipse is bigger than $1$?


*How to show that there exists at least one ellipse whose field is the least and which contains polygon $F$?
I haven't got any sensible ideas.
 A: For 1: Draw a small circle $C$ inside of $F$. Let $r$ be the radius of $C$. Any ellipse that contains $F$ must contain $C$. Further, the semi-minor axis of any ellipse that contains $C$ must be at least $r$. Now draw $K$ concentric with $C$, so that the radius $R$ of $K$ is at least $\frac{1}{\pi r}$. Now for an ellipse to not be completely contained in $K$, it must extend a distance at least $R$ from the center. And any ellipse that contains $F$ must contain the center as well. So an ellipse that both contains $F$ and is not contianed in $K$ must have a semi-major axis of at least $R$, since the semi-major axis is the farthest that two points in an ellipse can be from each other. So the ellipse must have a semi-major axis of at least $\frac{1}{\pi r}$ and a semi-minor axis of at least $r$, and so its area must be at least $\pi r \frac{1}{\pi r} = 1$.
For 2: I will just give a hint, so you get a chance to think about the problem (and also I am lazy). Think about stretching the polygon in one direction and squeezing it in the other by the same amount. This transformation conserves area. It also turns one convex polygon into another convex polygon. For any ellipse, there is an area-preserving stretching/squeezing transformation that turns it into a circle. Also this: https://en.wikipedia.org/wiki/Extreme_value_theorem Hopefully that is enough to give you some ideas.
