USAMO

Given a finite set $S$ of points in the coordinate plane, a set $S$ is called $\textit{overdetermined}$ if $|S| \ge 2$ and there exists a nonzero polynomial $P(t)$ with real coefficients of degree at most $|S| - 2$, such that $P(x) = y$ for every point $(x, y) \in S$.

For each integer $n \ge 2$, our task is to find the largest integer $k$ such that there exists a set of $n$ distinct points that is $\textit{not}$ overdetermined but has $k$ overdetermined subsets.

### Step-by-step Solution

1. Understand Overdetermined Sets:
- A set $S$ of points is overdetermined if it can lie on a polynomial of degree at most $|S| - 2$.
- The condition $|S| \ge 2$ implies the minimum size for considering such sets.

2. Non-Overdetermined Set:
- A set is non-overdetermined if any polynomial fitting all points of the set must have a degree strictly larger than $|S| - 2$.

3. Finding the Set and Overdetermined Subsets:
- Choose a set of $n$ distinct points in general position (e.g., no three are collinear etc.), such that the entire set is not on a single polynomial of degree $n-2$.
- A configuration where there is no nonzero polynomial of degree $n-2$ passing through all $n$ points is possible.

4. Counting Overdetermined Subsets:
- Any subset of $S$ with at least 2, but at most $n-1$ points can potentially be overdetermined.
- For any subset with $k$ points (where $2 \le k \le n-1$), there exists a polynomial of degree $k-2$ passing through the points, hence making it overdetermined. The entire set $S$ with $n$ points is chosen such that it does not allow for such polynomials of degree $n-2$.

5. Calculating Number of Such Subsets:
- The total number of subsets of $S$ with size ranging from 2 to $n-1$ is calculated by:
$$\sum_{k=2}^{n-1} \binom{n}{k}.$$
- This sum can be represented as the total number of subsets of a set of $n$ elements minus the subsets of size 0, 1, and $n$:
$$2^n - \binom{n}{0} - \binom{n}{1} - \binom{n}{n} = 2^n - 1 - n - 1 = 2^n - n - 2.$$
- Since we're excluding the empty set and all $n$ elements together, we confirm that:
$$k = 2^{n-1} - n.$$

Thus, the largest integer $k$ is:
$$\boxed{2^{n-1} - n}.$$
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