IMO 1960 - Problem 4

Problem: Construct a triangle $ABC$ whose lengths of heights $h_a$ and $h_b$ (from $A$ and $B$, respectively) and length of median $m_a$ (from $A$) are given.
Solution: Analysis. Let $A'$ and $B'$ be the feet of the perpendiculars from $A$ and $B$, respectively, to the opposite sides, $A_1$ the midpoint of $BC$, and let $D'$ be the foot of the perpendicular from $A_1$ to $AC$. We then have $AA_1=m_a$, $AA'=h_a,\angle{AA'A_1}=90^0,A_1D'=\frac{h_b}{2}$ and $\angle{AD'A_1}=90^0$.
Construction. We construct the quadrilateral $AD'A_1A'$( starting from the circle with diameter $AA_1$). Then $C$ is the intersection of $A'A_1$ and $AD'$, and $B$ is on the line $A_1C$ such that $CA_1=A_1B$ and $\mathscr{B}(B,A_1,C)$.
Dicussion. We must have $m_a\ge h_a$ and $m_a\ge \frac{h_b}{2}$. The number of solutions is $0$ if $m_a=h_a=\frac{h_b}{2},1$ if two of $m_a,h_a,\frac{h_b}{2}$ are equal, and $2$ otherwise.