Let's construct a triangle, given two sides, $a$ and $b$, and knowing that (using the usual notation) $\alpha=3 \beta$.
Solution. Let's start from the completed diagram and use its notation. Let $K$ be the internal point of side $BC$ such that $\angle KAB = \beta$. Then the triangle $ABK$ is isosceles, since the angles at $A$ and $B$ on side $AB$ are both $\beta$. Therefore, $AK = KB$. The angle $\angle CAK$ is clearly $\alpha - \beta$, and since $\alpha = 3\beta$, we have $\angle CAK = \alpha - \beta = 3\beta - \beta = 2\beta$. The angle $\angle AKC$ is an exterior angle of triangle $AKB$, so it is equal to the sum of the two non-adjacent interior angles: $\angle AKC = \beta + \beta = 2\beta$. Therefore, the triangle $AKC$ is also isosceles, since the angles at $A$ and $C$ on side $AK$ are both $2\beta$. Hence, $CK = AC = b$, and thus $AK = KB = BC - CK = a - b$.

![](https://cdn.mathpix.com/cropped/2024_05_02_0475bd0a5166b8954ebeg-1.jpg?height=301&width=545&top_left_y=424&top_left_x=779)

We now know all three sides of triangle $AKC$: $AC = CK = b$, and $AK = a - b$; thus, triangle $AKC$ can be constructed from its three sides, and from this, the vertex $B$ of triangle $ABC$ can be constructed by measuring $(a - b)$ from $K$ along the extension of segment $CK$ beyond $K$.

In the resulting triangle, $AC = b$, $BC = b + (a - b) = a$, and by reversing the above reasoning, it follows that $\alpha = 3\beta$.

Since $\alpha = 3\beta$, it is necessary that $a > b$, because in a triangle, the larger angle is opposite the longer side. Then $a - b$ is positive, and the segment of length $a - b$ can be constructed. By the triangle inequality,

\[
\begin{aligned}
& AC + CK > AK, \quad \text{that is} \quad b + b > a - b, \quad \text{rearranging:} \quad 3b > a, \\
& \left.\begin{array}{l}
AC + AK > CK \\
CK + AK > AC
\end{array}\right\} \quad b + (a - b) > b, \quad \text{that is} \quad a > b
\end{aligned}
\]

If these conditions are satisfied, then the triangle $AKC$ is essentially uniquely determined. The second step, constructing the vertex $B$, can then be carried out uniquely.

If $b < a < 3b$, then the problem has exactly one solution. If this condition is not met, then no such triangle $ABC$ can be constructed.