Two triangles intersect to form seven finite disjoint regions, six of which are triangles with area 1. The last region is a hexagon with area \(A\). Compute the minimum possible value of \(A\).

[i]Proposed by Karthik Vedula[/i]
1. **Define Variables and Setup**: 
   Let the desired area of the hexagon be \( S \). We are given that there are six triangular regions each with area 1, and one hexagonal region with area \( S \). The total area of the two intersecting triangles is thus \( 6 + S \).

2. **Use of Convex Quadrilateral Property**:
   We use the fact that for any convex quadrilateral \( WXYZ \), if \( P \) is the intersection of the diagonals \( WY \) and \( XZ \), then:
   \[
   S_{PWX} \cdot S_{PYZ} = S_{PXY} \cdot S_{PZW}
   \]
   This property will help us relate the areas of the triangles formed by the intersection of the two main triangles.

3. **Assign Areas to Triangles**:
   Let the areas of the six triangular regions be denoted as follows:
   \[
   S_{AKF} = a, \quad S_{FLB} = f, \quad S_{BGD} = b, \quad S_{DHC} = d, \quad S_{CIE} = c, \quad S_{EJA} = e
   \]
   Notice that each of these areas is given to be 1.

4. **Apply the Convex Quadrilateral Property**:
   Using the property mentioned, we can write the areas of the larger triangles formed by the intersection points:
   \[
   S_{ALD} = \frac{(a+1)(b+1)}{f} \geq \frac{4\sqrt{ab}}{f}
   \]
   \[
   S_{FGC} = \frac{(f+1)(d+1)}{b} \geq \frac{4\sqrt{fd}}{b}
   \]
   \[
   S_{BHE} = \frac{(b+1)(c+1)}{d} \geq \frac{4\sqrt{bc}}{d}
   \]
   \[
   S_{DIA} = \frac{(d+1)(e+1)}{c} \geq \frac{4\sqrt{de}}{c}
   \]
   \[
   S_{CJF} = \frac{(c+1)(a+1)}{e} \geq \frac{4\sqrt{ca}}{e}
   \]
   \[
   S_{EKB} = \frac{(e+1)(f+1)}{a} \geq \frac{4\sqrt{ef}}{a}
   \]

5. **Sum of Areas**:
   Summing these inequalities, we get:
   \[
   3S + 6 = S_{ALD} + S_{FGC} + S_{BHE} + S_{DIA} + S_{CJF} + S_{EKB} \geq \sum \frac{4\sqrt{ab}}{f} \geq 24
   \]
   This implies:
   \[
   3S + 6 \geq 24 \implies S \geq 6
   \]

6. **Conclusion**:
   Therefore, the minimum possible value of \( S \) is 6.

The final answer is \( \boxed{6} \).