Let's construct a triangle, given the sides $a$ and $c$, and knowing that $\alpha=2 \beta$.
Consider the problem as solved. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_f1140c8aa65ba495cff3g-1.jpg?height=694&width=542&top_left_y=206&top_left_x=781)

If we extend the side $AC = b$ beyond $A$ by $AD = AB = c$, then the external angle at vertex $A$ of the resulting isosceles triangle $ABD$ is $2\beta$, and thus each of the angles at the base $BD$ is $\beta$. The triangle $BCD$ has two angles that are equal to two angles of the original triangle, and thus

$$
BCD \triangle \sim ACB \triangle
$$

Therefore,

$$
a:(b+c) = b: a
$$

or

$$
a^2 = b(b+c)
$$

The construction of the side $b$ is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_f1140c8aa65ba495cff3g-1.jpg?height=464&width=550&top_left_y=1416&top_left_x=771)

Here, we can refer to the well-known theorem that the length of the tangent segment is the geometric mean of the two segments of the secant drawn from the point of tangency, but knowledge of the Pythagorean theorem is also sufficient. Indeed, adding $\left(\frac{c}{2}\right)^2$ to both sides of (1),

$$
a^2 + \left(\frac{c}{2}\right)^2 = b^2 + bc + \left(\frac{c}{2}\right)^2 = \left(b + \frac{c}{2}\right)^2
$$

thus

$$
b = \sqrt{a^2 + \left(\frac{c}{2}\right)^2} - \frac{c}{2}
$$

The necessary and sufficient condition for solvability is that $a + b > c$, or $b > c - a$.

Substitute a value smaller than $b$, namely $(c - a)$, for $b$ on the right-hand side of (1):

$$
a^2 > (c - a)(2c - a) = 2c^2 - 3ac + a^2
$$

or

$$
3ac > 2c^2
$$

from which (considering that $a > 0$, $c > 0$),

$$
a > \frac{2}{3}c
$$