3. Given $\triangle A B C$ in which the lengths of the sides are consecutive natural numbers. The median from vertex $A$ is perpendicular to the angle bisector of the angle at vertex $B$. Calculate the perimeter of triangle $\triangle A B C$.
Solution. Let $D$ and $E$ be the intersection points of the median from $A$ and the angle bisector from $B$, with sides $BC$ and $AC$ respectively. From $\triangle ABD$ it follows that $\overline{AB} = \overline{BD}$, because $BE$ is the angle bisector of $\measuredangle B$, and it intersects $AD$ at a right angle. Thus, $\overline{BC} = 2 \cdot \overline{AB}$. Since $\overline{AB}, \overline{BC}, \overline{AC}$ are consecutive integers, the difference $\overline{BC} - \overline{AB}$ is 1 or 2. We will consider both cases separately:

![](https://cdn.mathpix.com/cropped/2024_06_05_50edc896fa4080786845g-1.jpg?height=500&width=488&top_left_y=1646&top_left_x=972)
a) If $\overline{BC} - \overline{AB} = 1$, then $2 \cdot \overline{AB} - \overline{AB} = 1$, i.e., $\overline{AB} = 1$ and $\overline{BC} = 2$. According to the problem's condition, the lengths of the sides are consecutive natural numbers, so $\overline{AC} = 0$ or $\overline{AC} = 3$. For $\overline{AC} = 0$, we have $\overline{AB} + \overline{AC} = 0 + 1 < 2 = \overline{BC}$, and for $\overline{AC} = 3$, we have $\overline{AB} + \overline{BC} = 1 + 2 = 3 = \overline{AC}$, which is not possible (the sum of the lengths of two sides is not greater than the length of the third side).

b) If $\overline{BC} - \overline{AB} = 2$, then $\overline{AB} = 2$. In this case, we have $\overline{BC} = 4$, and clearly $\overline{AC} = 3$.

Thus, the perimeter of the triangle is $L = 2 + 3 + 4 = 9$.