Let $P$ be a cyclic polygon with circumcenter $O$ that does not lie on any diagonal, and let $S$ be the set of points on 2D plane containing $P$ and $O$. 

The $\textit{Matcha Sweep Game}$ is a game between two players $A$ and $B$, with $A$ going first, such that each choosing a nonempty subset $T$ of points in $S$ that has not been previously chosen, and such that if $T$ has at least $3$ vertices then $T$ forms a convex polygon. The game ends with all points have been chosen, with the player picking the last point wins. 

For which polygons $P$ can $A$ guarantee a win? 

[i]Proposed by Anzo Teh[/i]
1. **Case 1: \( P \) is an acute-angled triangle**

   - If \( P \) is an acute-angled triangle, then the circumcenter \( O \) lies inside \( P \).
   - In this scenario, \( A \) cannot choose \( T = S \) because \( S \) includes \( O \) and all vertices of \( P \), and \( O \) is inside \( P \).
   - Regardless of the subset \( T \) chosen by \( A \), there will be at most 3 points left (since \( P \) has 3 vertices and \( O \)).
   - Player \( B \) can then choose all the remaining points in one move and win the game.

2. **Case 2: \( P \) is not an acute-angled triangle**

   - If \( P \) is not an acute-angled triangle, we consider two sub-cases based on the position of \( O \).

   **Sub-case 2.1: \( O \) is outside \( P \)**
   - If \( O \) is outside \( P \), then the set \( S \) (which includes \( P \) and \( O \)) is convex.
   - Player \( A \) can choose \( T = S \) in the first move, which includes all points in \( S \).
   - Since \( A \) has chosen all points, \( A \) wins the game.

   **Sub-case 2.2: \( O \) is inside \( P \) and \( |P| \geq 4 \)**
   - If \( O \) is inside \( P \) and \( P \) has at least 4 vertices, we can consider an arbitrary triangulation of \( P \).
   - In this triangulation, there exists a triangle \( R \) whose interior contains \( O \).
   - Player \( A \) can choose \( T = S \setminus (R \cup \{O\}) \). This set \( T \) is convex because it is a subset of \( P \), which is convex.
   - Now, player \( B \) is left with the points in \( R \cup \{O\} \).
   - By the argument in the first paragraph, \( B \) cannot win because \( B \) cannot choose all remaining points in one move (since \( R \) has 3 vertices and \( O \)).

Conclusion:
- Player \( A \) can guarantee a win for all polygons \( P \) such that \( P \) is not an acute-angled triangle.

The final answer is all polygons \( \boxed{ P } \) such that \( P \) is not an acute-angled triangle.