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

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Let $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$ be the medians of an isosceles triangle $A B C$, and $A A^{\prime \prime}, B B^{\prime \prime}$, and $C C^{\prime \prime}$ be its altitudes, with $B B^{\prime \prime}=A A^{\prime}$ and $C C^{\prime \prime}=B B^{\prime}$. Suppose that $A B<A C<B C$.

If $P$ is the projection of point $A^{\prime}$ onto the line $A C$, then $A^{\prime} P$ is the midline of the right triangle $C B B^{\prime \prime}$, so $A^{\prime} P=1 / 2 B B^{\prime \prime}=A A^{\prime}$, hence,

$\angle B^{\prime} A A^{\prime}=30^{\circ}$. Similarly, $\angle A B B^{\prime}=30^{\circ}$.

Let $M$ be the point of intersection of the medians of triangle $A B C$. Denote $M A^{\prime}=x$. Then $A A^{\prime}=3 x$. Triangles $A^{\prime} M B^{\prime}$ and $A^{\prime} B^{\prime} A$ are similar by two angles, since $\angle A^{\prime} B^{\prime} M=\angle A B B^{\prime}=30^{\circ}=\angle A^{\prime} A B^{\prime}$, and the angle at vertex $A^{\prime}$ is common, so $A^{\prime} B^{\prime}: A A^{\prime}=A^{\prime} M: A^{\prime} B^{\prime}$, which means $A^{\prime} B^{\prime 2}=A^{\prime} M \cdot A A^{\prime}=3 x^{2}$.

Denote $M B^{\prime}=y$. By the cosine theorem, $M A^{\prime 2}=M B^{\prime 2}+A^{\prime} B^{\prime 2}-2 M B^{\prime} \cdot A^{\prime} B^{\prime} \cos 30^{\circ}, x^{2}=y^{2}+3 x^{2}-3 x y, (y-x)(y-2 x)=0$.

If $y=x$, then triangle $A^{\prime} M B^{\prime}$ is isosceles, so triangle $A B C$ is also isosceles, which contradicts the condition.

For $y=2 x$, we get that triangle $A^{\prime} M B^{\prime}$ is right-angled, $\angle M A^{\prime} B^{\prime}=90^{\circ}$, hence, the similar triangle $A M B$ with a similarity coefficient of 2 is also right-angled, $M C^{\prime}=1 / 3 C C^{\prime}$ is its median. By the Pythagorean theorem, $M C^{\prime 2}=A M^{2}+A C^{2}=4 x^{2}+3 x^{2}=7 x^{2}, A^{\prime} B^{2}=A B^{2}+A A^{\prime 2}=12 x^{2}+9 x^{2}=21 x^{2}$, and since

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-05.jpg?height=71&width=883&top_left_y=1039&top_left_x=16)

Therefore, $C C^{\prime}: A A^{\prime \prime}=3 M C^{\prime}: A A^{\prime \prime}=7: 2$.

## Answer

3.5.