Problem 1752

To prove that if the union of all the sets in the sequence is finite, then $\mathcal{A}_{i} = \mathcal{A}_{1}$ for any $i \geq 1$, we will proceed by contradiction and use properties of geometric configurations.

1. Assume the union of all sets is finite:
Suppose $\bigcup_{j=0}^{\infty} \mathcal{A}_j$ is finite. This implies that there exists some $j$ such that $\mathcal{A}_j = \mathcal{A}_{j+1} = \ldots$. Let this set be $\mathcal{A}_j$ and denote it by $\mathcal{A}$.

2. **Consider the configuration of points in $\mathcal{A}$:**
Since $\mathcal{A}$ is finite, we can analyze the geometric properties of the points in $\mathcal{A}$. We will consider two cases:
- There exists a set of four points in $\mathcal{A}$ such that one of them is strictly inside the non-degenerate triangle formed by the other three.
- There exists a set of four points in $\mathcal{A}$ forming a non-degenerate convex quadrilateral.

3. Case 1: A point inside a triangle:
Suppose there are four points $A, B, C, P \in \mathcal{A}$ such that $P$ is strictly inside the triangle $\triangle ABC$. Consider the points $A' = AP \cap BC$, $B' = BP \cap CA$, and $C' = CP \cap AB$. These points are intersection points and hence belong to $\mathcal{A}_{j+1}$, which means they are also in $\mathcal{A}$.

- Since $P$ is inside $\triangle ABC$, it follows that $P$ is also inside $\triangle A'B'C'$, and the area of $\triangle A'B'C'$ is less than the area of $\triangle ABC$.
- This contradicts the minimality assumption of the area of $\triangle ABC$ containing $P$.

4. Case 2: A non-degenerate convex quadrilateral:
Suppose there are four points $A, B, C, D \in \mathcal{A}$ such that $ABCD$ is a non-degenerate convex quadrilateral. Let $P = AC \cap BD$. Since $P$ is an intersection point, $P \in \mathcal{A}$.

- Assume $ABCD$ is not a parallelogram. If $AB$ is not parallel to $CD$, then $Q = AB \cap CD$ is in $\mathcal{A}$. From the convexity of $ABCD$, $P$ must lie inside one of the triangles formed by $Q$ and two of the vertices of $ABCD$, leading to a contradiction as in Case 1.
- Therefore, $ABCD$ must be a parallelogram.

5. Parallelogram configuration:
If $ABCD$ is a parallelogram with $P = AC \cap BD$, consider any other point $Q \in \mathcal{A}$. If $Q$ is not on $AC$ or $BD$, then $QP$ intersects some side of $ABCD$ in its interior, leading to a contradiction as in Case 1.

- If $Q \in AC$ or $Q \in BD$, then $Q$ must lie on one of the diagonals, and the configuration remains a parallelogram.

6. Conclusion:
Since $\mathcal{A}$ cannot contain any configuration other than a set of collinear points or a parallelogram with its intersection point, it follows that $\mathcal{A}$ must be the same as $\mathcal{A}_1$. Therefore, $\mathcal{A}_i = \mathcal{A}_1$ for any $i \geq 1$.

$$\boxed{\mathcal{A}_i = \mathcal{A}_1 \text{ for any } i \geq 1}$$