## Task B-2.4.

The lengths of the sides of a triangle are three consecutive natural numbers, and the largest angle of the triangle is twice the smallest angle of the triangle. Determine the lengths of the sides of this triangle.
## First Solution.

Assume in triangle $ABC$, $\beta = 2\alpha$. Let $\overline{BD}$ be a part of the angle bisector of $\beta$ inside the triangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-11.jpg?height=537&width=851&top_left_y=428&top_left_x=611)

Then, with the labels as shown in the figure, according to the angle bisector theorem in a triangle, $\frac{|AD|}{|DC|} = \frac{c}{a}$, or $|DC| = \frac{a}{c}(b - |DC|)$. From this, we get $|DC| = \frac{ab}{a+c}$.

Triangle $BDC$ is similar to triangle $ABC$ by the $KK$ theorem, so $\frac{|DC|}{|BC|} = \frac{|BC|}{|AC|}$. Therefore,

$$
\frac{ab}{\frac{a+c}{a}} = \frac{a}{b}
$$

which simplifies to $a(a+c) = b^2$.

Since the lengths of the sides of the given triangle are consecutive natural numbers and $\alpha < \gamma < \beta$, we have $a < c < b$ and $c = a + 1$, $b = a + 2$. Substituting these into the previous equation, we get $a(a + a + 1) = (a + 2)^2$, or $a^2 - 3a - 4 = 0$. The positive solution to this equation is $a = 4$.

The lengths of the sides of the given triangle are $a = 4$, $c = 5$, $b = 6$.