Let $A_1B_1C_1$ be a triangle with $A_1B_1 = 16, B_1C_1 = 14,$ and $C_1A_1 = 10$. Given a positive integer $i$ and a triangle $A_iB_iC_i$ with circumcenter $O_i$, define triangle $A_{i+1}B_{i+1}C_{i+1}$ in the following way:

(a) $A_{i+1}$ is on side $B_iC_i$ such that $C_iA_{i+1}=2B_iA_{i+1}$.
(b) $B_{i+1}\neq C_i$ is the intersection of line $A_iC_i$ with the circumcircle of $O_iA_{i+1}C_i$.
(c) $C_{i+1}\neq B_i$ is the intersection of line $A_iB_i$ with the circumcircle of $O_iA_{i+1}B_i$.

Find \[ \left(\sum_{i = 1}^\infty [A_iB_iC_i] \right)^2. \]

Note: $[K]$ denotes the area of $K$.

[i]Proposed by Yang Liu[/i]
1. **Understanding the Problem:**
   We are given a triangle \( A_1B_1C_1 \) with sides \( A_1B_1 = 16 \), \( B_1C_1 = 14 \), and \( C_1A_1 = 10 \). We need to define a sequence of triangles \( A_iB_iC_i \) and find the sum of their areas squared.

2. **Using Heron's Formula to Find the Area of \( A_1B_1C_1 \):**
   Heron's formula for the area of a triangle with sides \( a \), \( b \), and \( c \) is:
   \[
   \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \( s \) is the semi-perimeter:
   \[
   s = \frac{a + b + c}{2}
   \]
   For \( A_1B_1C_1 \):
   \[
   s = \frac{16 + 14 + 10}{2} = 20
   \]
   Thus, the area is:
   \[
   [A_1B_1C_1] = \sqrt{20 \cdot (20-16) \cdot (20-14) \cdot (20-10)} = \sqrt{20 \cdot 4 \cdot 6 \cdot 10} = \sqrt{4800} = 40\sqrt{3}
   \]

3. **Establishing Similarity of Triangles:**
   By the problem's construction, each triangle \( A_{i+1}B_{i+1}C_{i+1} \) is similar to \( A_iB_iC_i \). This is due to the angles being preserved through the construction process, as indicated by the angle chasing argument in the solution.

4. **Finding the Ratio of Similarity:**
   Since the triangles are similar, the ratio of their areas is the square of the ratio of their corresponding sides. Let \( r \) be the ratio of the sides of \( A_{i+1}B_{i+1}C_{i+1} \) to \( A_iB_iC_i \). The area ratio is then \( r^2 \).

5. **Calculating the Ratio \( r \):**
   From the given solution, the ratio \( r \) is found to be \( \frac{1}{\sqrt{3}} \). Therefore, the area ratio is:
   \[
   r^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}
   \]

6. **Summing the Areas of the Infinite Sequence:**
   The areas form a geometric series with the first term \( a = 40\sqrt{3} \) and common ratio \( r^2 = \frac{1}{3} \). The sum of an infinite geometric series is given by:
   \[
   S = \frac{a}{1 - r^2}
   \]
   Substituting the values:
   \[
   S = \frac{40\sqrt{3}}{1 - \frac{1}{3}} = \frac{40\sqrt{3}}{\frac{2}{3}} = 60\sqrt{3}
   \]

7. **Squaring the Sum:**
   Finally, we need to find the square of this sum:
   \[
   \left(60\sqrt{3}\right)^2 = 3600 \times 3 = 10800
   \]

The final answer is \(\boxed{10800}\)