[Inscribed and circumscribed circles ] $[\quad$ Induction in geometry $]$

$n$ points are located at the vertices of a convex $n$-gon. Inside this $n$-gon, $k$ points are marked. It turns out that any three of the $n+k$ points do not lie on the same line and are vertices of an isosceles triangle. What can the number $k$ be?
Answer: $k=1$ or 0. Lemma. Let point $O$ lie inside triangle $\triangle A B C$, and points $A, B, C, O$ satisfy the condition of the problem. Then point $O$ is the center of the circumcircle of triangle $\triangle A B C$. Proof of the lemma. Consider triangle $\triangle A B C$, by the condition of the problem it is isosceles. Suppose $A B=B C$, then note that $B O<A B$ and $B O<B C$. By the condition of the problem, triangles $\triangle A B O$ and $\triangle B C O$ are isosceles, which means $O B=O A$ and $O B=O C$. We have obtained that $O A=O B=O C$, so $O$ is the center of the circumcircle of triangle $\triangle A B C$.

We will prove that $k \leq 1$ by induction on $n$. If $n=3$, then all $k$ points lie inside the triangle and by the lemma all coincide with the center of the circumcircle of this triangle. We get that $k \leq 1$. If $n \geq 4$, then in this $n$-gon there is a right or obtuse angle $\angle A_{\mathrm{i}-1} A_{\mathrm{i}} A_{\mathrm{i}+1}$. By the proved lemma, inside triangle $A_{\mathrm{i}-1} A_{\mathrm{i}} A_{\mathrm{i}+1}$ there cannot be any of the $k$ points. Therefore, by cutting off triangle $A_{\mathrm{i}-1} A_{\mathrm{i}} A_{\mathrm{i}+1}$ from our polygon, we get an $(n-1)$-gon, inside which there are $k$ points. By the induction hypothesis, $k \leq 1$. Thus, $k \leq 1$.

The example for $k=0$ is obvious. For $k=1$, it is sufficient to take an isosceles triangle with the center of its circumcircle.