Let in triangle $ABC$ the inequality $1=AB<AC<BC$ holds. Then the median $AA'$ is equal to the height dropped from vertex $B$, and the median $BB'$ is equal to the height dropped from vertex $C$. Therefore, the distance from point $A'$ to the line $AC$ is half of $AA'$, which means $\angle A'AC=30^{\circ}$. Similarly, $\angle B'BA=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed1a9e9dd82795d4db2eg-35.jpg?height=695&width=824&top_left_y=1070&top_left_x=617)

Let $M$ be the point of intersection of the medians of the triangle. Since $\angle A'B'M=\angle B'BA=\angle B'AM$ (see the figure), triangles $B'A'M$ and $AA'B'$ are similar, which means

\[
(A'B')^2 = A'M \cdot A' = 3 A'M^2.
\]
Therefore, the angle $\angle A'MB'$ is either $120^{\circ}$ or $60^{\circ}$.

In the first case, triangle $ABC$ is equilateral, which contradicts the condition.

In the second case, $\angle B'A'A = \angle A'AB = 90^{\circ}$, $\angle BB'A = 30^{\circ}$, and $AB' = AB$. Thus, $AC = 2AB$ and $\angle A = 120^{\circ}$. By the cosine theorem, we find

\[
BC = \sqrt{7},
\]
and therefore, the median from vertex $C$ is $\frac{\sqrt{21}}{2}$, and the height from vertex $A$ is $\sqrt{\frac{3}{7}}$.

## Answer

$7: 2$.
Zaslavsky A.A.

In a non-isosceles triangle, two medians are equal to two altitudes. Find the ratio of the third median to the third altitude.