Let $O$ be the center of the circle, $A$ and $B$ be the given points, and $X_0, X_1,..., X_{n-1}, X_n$ be the broken line where $X_0=A, X_n=B$. Let $\varepsilon$ be the ellipse with foci $A, B$ passing through $O$. Then for any $i\in\{1,...,n-1\}$ we have $AX_i+X_iB\le X_0X_1+X_1X_2+...+X_{n-1}X_n<AO+OB$ which means that $X_i$ lies inside $\varepsilon$. Hence the diameter of the circle that is tangent to $\varepsilon$ doesn't intersect the broken line which lies strictly inside the ellipse.

kiyoras_2001 wrote: Let $O$ be the center of the circle, $A$ and $B$ be the given points, and $X_0, X_1,..., X_{n-1}, X_n$ be the broken line where $X_0=A, X_n=B$. Let $\varepsilon$ be the ellipse with foci $A, B$ passing through $O$. Then for any $i\in\{1,...,n-1\}$ we have $AX_i+X_iB\le X_0X_1+X_1X_2+...+X_{n-1}X_n<AO+OB$ which means that $X_i$ lies inside $\varepsilon$. Hence the diameter of the circle that is tangent to $\varepsilon$ doesn't intersect the broken line which lies strictly inside the ellipse. What a nice problem and solution

Compare this with 2023 Germany R4 10.5.