Two lines intersect inside a unit square, splitting it into four regions. Find the maximum product of the areas of the four regions.

[i]Proposed by Nathan Ramesh
1. Let the areas of the four regions be \( a, b, c, \) and \( d \). Since the lines intersect inside a unit square, the sum of the areas of these regions must be equal to the area of the square:
   \[
   a + b + c + d = 1
   \]

2. We aim to maximize the product \( abcd \). By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality), we have:
   \[
   \frac{a + b + c + d}{4} \geq \sqrt[4]{abcd}
   \]

3. Substituting \( a + b + c + d = 1 \) into the inequality, we get:
   \[
   \frac{1}{4} \geq \sqrt[4]{abcd}
   \]

4. Raising both sides to the fourth power to eliminate the fourth root, we obtain:
   \[
   \left( \frac{1}{4} \right)^4 \geq abcd
   \]

5. Simplifying the left-hand side, we find:
   \[
   \frac{1}{256} \geq abcd
   \]

6. Therefore, the maximum value of \( abcd \) is:
   \[
   \max(abcd) = \frac{1}{256}
   \]

7. This maximum value is achievable when \( a = b = c = d = \frac{1}{4} \). To verify, we substitute these values back into the product:
   \[
   abcd = \left( \frac{1}{4} \right) \left( \frac{1}{4} \right) \left( \frac{1}{4} \right) \left( \frac{1}{4} \right) = \left( \frac{1}{4} \right)^4 = \frac{1}{256}
   \]

Thus, the maximum product of the areas of the four regions is indeed \(\frac{1}{256}\).

The final answer is \(\boxed{\frac{1}{256}}\).