Taiwan IMO Selection Camp

If, in the similarity relation between $\triangle ABC$ and $\triangle P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$, $BC$ corresponds to the longest side of $\triangle P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$, then $BC \ge AB \ge AC$. The condition $BC \ge AB$ is equivalent to $(x + 1)^2 + y^2 \le 4$, while $AB \ge AC$ holds for any point in the first quadrant. Hence we first define:
$$S := \{(x, y) : (x + 1)^2 + y^2 \le 4, x \ge 0, y \ge 0\}.$$
Note that $S$ is the intersection of a disk and the first quadrant, so it is a bounded convex set, and we may choose any point in it as $T$ to satisfy condition (i). For any point $A$ in $S$, $BC \ge AB \ge AC$ always holds. Hence the point $A$ satisfying condition (ii) can be uniquely determined, and its existence follows from the construction above.

Next, if, in the similarity relation between $\triangle ABC$ and $\triangle P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$, $BC$ corresponds to the second longest side of $\triangle P_{\sigma(1)}P_{\sigma(2)}P_{\sigma(3)}$, then $A'B \ge BC \ge A'C$. These two inequalities are respectively equivalent to $(x + 1)^2 + y^2 \ge 4$ and $(x - 1)^2 + y^2 \le 4$.
$$S' := \{(x, y) : (x+1)^2 + y^2 \ge 4, (x-1)^2 + y^2 \le 4, x \ge 0, y \ge 0\}.$$
That $S'$ is bounded and satisfies condition (ii) can be proved using an argument similar to that for $S$. For condition (i), note that $S'$ consists of points that lie inside the disk $(x - 1)^2 + y^2 \le 4$ and outside the disk $(x + 1)^2 + y^2 \ge 4$, so we may take $T' = (1, 2)$ to satisfy condition (i); this point is the highest point (the point with the largest $y$-coordinate) on the circle $(x - 1)^2 + y^2 = 4$.

Finally, we check that $BA \cdot BA'$ is a constant value. Suppose triangle $P_1P_2P_3$ satisfies $P_1P_2 \ge P_2P_3 \ge P_3P_1$. By similarity, we know:
$$BA = BC \cdot \frac{P_2P_3}{P_1P_2} \quad \text{and} \quad BA' = BC \cdot \frac{P_1P_2}{P_2P_3}.$$
Therefore $BA \cdot BA' = BC^2 = 4$, and this value is naturally independent of the triangle $P_1P_2P_3$. This completes the proof.