## Task 3 - 010733

In a plane, there are three intersecting lines $g_{1}, g_{2}$, and $g_{3}$ at a point $S$, and a point $A$ on $g_{1}$.

Construct a triangle that has $A$ as a vertex and the intersection point $S$ as the circumcenter, with $B$ on $g_{2}$ and $C$ on $g_{3}$ or vice versa! How many different triangles can be constructed this way?
Construction:

A circle with center $S$ and radius $S A$ is drawn, intersecting the other two lines $g_{2}$ and $g_{3}$ at points $B_{1}, B_{2}$ and $C_{1}, C_{2}$, respectively (Figure a).

Thus, $S A = S B_{i} = S C_{j}, (i, j = 1, 2)$, and $S$ is indeed the circumcenter of all possible triangles $A B_{i} C_{j}$.

Since the lines intersect the circle at two points each, there are two possibilities for $B$ and $C$, resulting in a total of four $(=2 \cdot 2)$ triangles: $A B_{1} C_{1}, A B_{1} C_{2}$ (Figure b), $A B_{2} C_{1}$ (Figure c), and $A B_{2} C_{2}$ (Figure d).