5. In triangle $ABC$, the median from vertex $A$ is perpendicular to the bisector of angle $B$, and the median from vertex $B$ is perpendicular to the bisector of angle $A$. It is known that side $AB=1$. Find the perimeter of triangle $ABC$.
Answer: 5.

Solution. Let $A M$ be the median drawn from vertex $A$. Then, in triangle $A B M$, the bisector of angle $B$ is perpendicular to side $A M$, i.e., the bisector is also an altitude. Therefore, this triangle is isosceles, $A B = B M = 1$. Hence, $B C = 2 B M = 2$. Similarly, from the second condition, we get that side $A C$ is twice the length of $A B$, so the perimeter of triangle $A B C$ is $1 + 2 + 2 = 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b6522f95bc9413ab322cg-4.jpg?height=611&width=600&top_left_y=868&top_left_x=722)

## Grading Criteria.

- Correct solution - 7 points.
- It is determined that the triangle cut off by one of the medians is isosceles, but there is no further progress - 3 points.
- Only the correct answer is provided - 1 point.