A square is called [i]proper[/i] if its sides are parallel to the coordinate axes. Point $P$ is randomly selected inside a proper square $S$ with side length 2012. Denote by $T$ the largest proper square that lies within $S$ and has $P$ on its perimeter, and denote by $a$ the expected value of the side length of $T$. Compute $\lfloor a \rfloor$, the greatest integer less than or equal to $a$.

[i]Proposed by Lewis Chen[/i]
1. Let's consider a proper square \( S \) with side length 2012. We can place \( S \) in the coordinate plane with vertices at \((0,0)\), \((0,2012)\), \((2012,2012)\), and \((2012,0)\).

2. Point \( P \) is randomly selected inside \( S \). We need to find the expected value of the side length of the largest proper square \( T \) that lies within \( S \) and has \( P \) on its perimeter.

3. Let \( P \) have coordinates \((x, y)\) where \( 0 \leq x, y \leq 2012 \). The side length of \( T \) will be determined by the minimum distance from \( P \) to the sides of \( S \). Specifically, the side length of \( T \) will be \( 2 \min(x, 2012 - x, y, 2012 - y) \).

4. To find the expected value of the side length of \( T \), we need to compute the expected value of \( 2 \min(x, 2012 - x, y, 2012 - y) \).

5. Consider the random variable \( Z = \min(x, 2012 - x, y, 2012 - y) \). The side length of \( T \) is \( 2Z \), so we need to find \( \mathbb{E}[2Z] = 2 \mathbb{E}[Z] \).

6. To find \( \mathbb{E}[Z] \), we need to integrate over the possible values of \( x \) and \( y \). The probability density function for \( Z \) can be derived by considering the geometry of the problem.

7. The cumulative distribution function (CDF) of \( Z \) is given by:
   \[
   F_Z(z) = \mathbb{P}(Z \leq z) = \left( \frac{z}{2012} \right)^2
   \]
   for \( 0 \leq z \leq 1006 \).

8. The probability density function (PDF) of \( Z \) is the derivative of the CDF:
   \[
   f_Z(z) = \frac{d}{dz} F_Z(z) = \frac{2z}{2012^2}
   \]

9. The expected value of \( Z \) is:
   \[
   \mathbb{E}[Z] = \int_0^{1006} z f_Z(z) \, dz = \int_0^{1006} z \cdot \frac{2z}{2012^2} \, dz = \frac{2}{2012^2} \int_0^{1006} z^2 \, dz
   \]

10. Compute the integral:
    \[
    \int_0^{1006} z^2 \, dz = \left[ \frac{z^3}{3} \right]_0^{1006} = \frac{1006^3}{3}
    \]

11. Therefore:
    \[
    \mathbb{E}[Z] = \frac{2}{2012^2} \cdot \frac{1006^3}{3} = \frac{2 \cdot 1006^3}{3 \cdot 2012^2}
    \]

12. Simplify the expression:
    \[
    \mathbb{E}[Z] = \frac{2 \cdot 1006^3}{3 \cdot (2 \cdot 1006)^2} = \frac{2 \cdot 1006^3}{3 \cdot 4 \cdot 1006^2} = \frac{2 \cdot 1006}{12} = \frac{1006}{6} = 167.6667
    \]

13. The expected value of the side length of \( T \) is:
    \[
    \mathbb{E}[2Z] = 2 \cdot 167.6667 = 335.3334
    \]

14. The greatest integer less than or equal to \( 335.3334 \) is:
    \[
    \lfloor 335.3334 \rfloor = 335
    \]

The final answer is \(\boxed{335}\).