Problem 1653

1. Define the sequence of triangles:
Let $A_0B_0C_0$ be the initial triangle. For each positive integer $n \geq 1$, define the points $A_n$, $B_n$, and $C_n$ as follows:
- $A_n$ is on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2:1$.
- $B_n$ is on the segment $C_{n-1}A_{n-1}$ such that $C_{n-1}B_n : A_{n-1}B_n = 2:1$.
- $C_n$ is on the segment $A_{n-1}B_{n-1}$ such that $A_{n-1}C_n : B_{n-1}C_n = 2:1$.

2. Nested compact sets:
Each triangle $A_nB_nC_n$ is a compact set, and the sequence of triangles is nested, i.e., $A_{n+1}B_{n+1}C_{n+1} \subseteq A_nB_nC_n$ for all $n$. By the finite intersection property of nested compact sets, there exists at least one point $P$ that lies in the intersection of all these triangles.

3. Uniqueness of the intersection point:
We need to show that the intersection of all triangles $A_nB_nC_n$ is a single point. Assume for contradiction that the intersection contains more than one point. Then there exist subsequences of $A_n$, $B_n$, and $C_n$ (denoted again by $A_n$, $B_n$, and $C_n$ for simplicity) that converge to distinct points $A$, $B$, and $C$, respectively.

4. **Case 1: $A$, $B$, and $C$ are distinct:**
Without loss of generality, assume $\angle BAC \leq 60^\circ$. For sufficiently large $n$, the points $A_n$, $B_n$, and $C_n$ are close to $A$, $B$, and $C$, respectively. Consider the next triangle $A_{n+1}B_{n+1}C_{n+1}$. The side $B_{n+1}C_{n+1}$ is far from $A$, implying that $A$ is outside $\triangle A_{n+1}B_{n+1}C_{n+1}$. This contradicts the assumption that $A$ is a limit point of the sequence $A_n$.

5. **Case 2: $B = C \neq A$:**
In this case, $\angle B_nA_nC_n < 60^\circ$ (it tends to 0 as $n \to \infty$). Applying the same argument as in Case 1, we find that $A$ cannot be a limit point of the sequence $A_n$, leading to a contradiction.

6. Conclusion of uniqueness:
Since both cases lead to contradictions, the points $A_n$, $B_n$, and $C_n$ must converge to a common point $P$. Therefore, there is a unique point $P$ that lies in the intersection of all triangles $A_nB_nC_n$.

7. Interior of the triangles:
Finally, we need to show that $P$ lies in the interior of all triangles $A_nB_nC_n$. Assume for contradiction that $P$ lies on some side, say $A_nB_n$, for some $n$. Then $P$ would be outside $\triangle A_{n+2}B_{n+2}C_{n+2}$, which contradicts the fact that $P$ is in the intersection of all triangles. Hence, $P$ must lie in the interior of all triangles $A_nB_nC_n$.

8. Generalization:
The proof remains valid if we only require that $A_{n+1}$, $B_{n+1}$, and $C_{n+1}$ are on the sides $B_nC_n$, $A_nC_n$, and $A_nB_n$, respectively, and not too close to the vertices $A_n$, $B_n$, $C_n$. Specifically, the distance from $A_{n+1}$ to both points $B_n$ and $C_n$ should be greater than $\varepsilon \cdot |B_nC_n|$ for some fixed $\varepsilon > 0$, and similarly for $B_{n+1}$ and $C_{n+1}$.

$\blacksquare$