A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \ge 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.
To determine the largest integer \( m \) such that no planar \( n \)-gon splits into less than \( m \) triangles, we need to analyze the relationship between the number of sides \( n \) and the number of triangles \( t \) in a splitting.

1. **Understanding the Problem:**
   - A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon.
   - We need to find the minimum number of triangles \( t \) required to split an \( n \)-gon.

2. **Example for \( n = 6 \):**
   - Consider a concave hexagon (6-gon). It can be split into 2 triangles as shown in the example.
   - This suggests that for \( n = 6 \), the minimum number of triangles is 2.

3. **General Case:**
   - We need to prove that for any \( n \)-gon, the minimum number of triangles \( t \) required is \( \lceil n/3 \rceil \).

4. **Proof:**
   - Let \( t \) be the number of triangles in a splitting of an \( n \)-gon.
   - Each triangle has 3 vertices, and each vertex of the \( n \)-gon must be a vertex of at least one triangle.
   - Therefore, the total number of vertices in all triangles is at least \( 3t \).
   - However, since each vertex of the \( n \)-gon is counted at least once, we have:
     \[
     n \leq 3t
     \]
   - Solving for \( t \), we get:
     \[
     t \geq \frac{n}{3}
     \]
   - Since \( t \) must be an integer, we take the ceiling function:
     \[
     t \geq \lceil \frac{n}{3} \rceil
     \]

5. **Conclusion:**
   - The largest integer \( m \) such that no planar \( n \)-gon splits into less than \( m \) triangles is \( \lceil n/3 \rceil \).

\(\blacksquare\)

The final answer is \( \boxed{ \lceil \frac{n}{3} \rceil } \)