Inside a convex $100$-gon are selected $k$ points, $2 \leq k \leq 50$. Show that one can choose $2k$ vertices of the $100$-gon so that the convex $2k$-gon determined by these vertices contains all the selected points.
1. **Initial Setup and Problem Restatement**:
   We are given a convex $100$-gon and $k$ points inside it, where $2 \leq k \leq 50$. We need to show that we can choose $2k$ vertices of the $100$-gon such that the convex $2k$-gon determined by these vertices contains all the selected points.

2. **Reduction to Smaller Polygons**:
   If $n = 2k$, the statement is trivial because we can simply take all $2k$ vertices. Therefore, we need to prove the statement for $n > 2k$. We will show that for any $n > 2k$, there exists an $(n-1)$-gon that contains the $k$ points. By repeatedly reducing the $n$-gon to $(n-1)$-gons, we will eventually reach $n = 2k$, at which point the $2k$-gon will contain all $k$ points.

3. **Example for Clarification**:
   Consider $n = 7$ and $2k = 6$. We need to show that we can reduce a $7$-gon to a $6$-gon while still containing the $k$ points.

4. **Finding an Ear**:
   An ear of a polygon is a vertex such that the line segment connecting its two adjacent vertices lies entirely inside the polygon. We need to show that we can always find an ear to remove.

5. **Case 1: $n$ is Even**:
   Let $n = 2m$ and $k \leq m-1$. Consider the $n$-gon. We can form $m$ disjoint ears. Since there are at most $m-1$ points, one of these ears must be empty. We can remove this ear, reducing the $n$-gon to an $(n-1)$-gon.

6. **Case 2: $n$ is Odd**:
   Let $n = 2m-1$ and $k \leq m-1$. Each point inside the $n$-gon is in at most $2$ of the $2m-1$ ears. Since there are at most $m-1$ points, at most $2(m-1) = 2m-2$ ears are used. Hence, there is at least one available ear that we can remove to reduce the $n$-gon to an $(n-1)$-gon.

7. **Conclusion**:
   By repeatedly finding and removing ears, we can reduce the $n$-gon to a $2k$-gon that contains all $k$ points.

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