Given two sides of a triangle $a$ and $b$, and we know the angles opposite these sides are in the ratio $\alpha: \beta=1: 2$. Construct the triangle. What is the condition for solvability?
I. Solution: Imagine the problem is solved and draw the circumcircle of the triangle. (Figure 1.)

![](https://cdn.mathpix.com/cropped/2024_05_02_12dc302657953613f23dg-1.jpg?height=496&width=466&top_left_y=218&top_left_x=819)

Let the angle bisector of $\beta$ intersect this circle at $D$ and the side $AC = b$ at $N$. According to the inscribed angle theorem, $AD = DC = BC = a$ because the arcs corresponding to these three chords all subtend an inscribed angle of $\alpha$. The trapezoid $ABCD$ is thus isosceles, and besides the upper sides, we also know the diagonal $AC = BD$. From these data, the trapezoid and with it the sought triangle can be easily constructed, provided that the triangle $ABC_{\triangle} \left(\cong BCD_{\triangle}\right)$ is constructible, i.e., provided that $a + a > b$, or $a > b / 2$. On the other hand, since $\alpha > 0^{\circ}$ and $120^{\circ} > \beta > 0^{\circ}$.