IMO 1961 - Problem 5

Problem: Construct a triangle $ABC$ if the following elements are given: $AC=b,AB=c$ and $\angle{AMB}=\omega(\omega<90^0)$, where $M$ is the midpoint of $BC$. Prove that the construction has a solution if and only if $btan(\frac{\omega}{2})\le c<b$.
In what case does equality hold>
Solution: Analysis. Let $C_1$ be the midpoint of $AB$. In $\triangle{AMB}$ we have $MC_1=\frac{b}{2},$ $AB=c$ and $\angle{AMB}=\omega$. Thus, given $AB=c$, the point $M$ is at the intersection of the triangle $k(C';\frac{b}{2})$ and the set of points $e$ that view $AB$ at an angle of $\omega$. The construction of $ABC$ is now obvious.
Dicussion. It suffices to establish the conditions for which $k$ and $e$ intersect. Let $E$ be the midpoint of one of the arcs that make up $e$. A necessary and sufficient condition for $k$ to intersect $e$ is
$\frac{c}{2}=C'A\le \frac{b}{2}\le C'E=\frac{c}{2}cot\frac{w}{2}\iff btan\frac{\omega}{2}\le c<b$.