Balti Tee 1995 · Ülesanne 17

Solution:

Let $h = \max \left(h_{A}, h_{B}, h_{C}\right)$ and $m = \min \left(m_{A}, m_{B}, m_{C}\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \leq m$.

Now let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A D| = h$ and $|B E| = m$. Let $F$ be the point on the line $B C$ such that $E F$ is parallel to $A D$. Then $m = |E B| \geq |E F| = \frac{h}{2}$, whence $h \leq 2 m$.

For an example with $h = 2 m$, consider a triangle where $D$ lies on the ray $C B$ with $|C B| = |B D|$. Hence the smallest such value is $\alpha = 2$.