IMO 1959 - Problem 4

Problem: Construct a right-angled triangle whose hypotenuse $c$ is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle.
Solution: Analysis. Let $a$ and $b$ be the other two sides of the triangle. From the conditions of the problem we have $c^2=a^2+b^2$ and $\frac{c}{2}=\sqrt{ab}\iff \frac{3}{2}c^2=a^2+b^2+2ab=(a+b)^2\iff \sqrt{\frac{3}{2}}c=a+b$. Given a desired $\triangle{ABC}$ let $D$ be a point on ($AC$ such that $CD=CB$). In that case, $AD=a+b=\sqrt{\frac{3}{2}}c,$ and also, since $BC=CD$, it follows that $\angle{ADB}=45^0$.
Construction. From a segment of length $c$ we elementarily construct a segment $AD$ of length $\sqrt{\frac{3}{2}}c$. We then construct a ray $DX$ such that $\angle{ADX}=45^0$ and a circle $k(A,c)$ that intersects the ray at point $B$.
Finally, we construct the perpendicular from $B$ to $AD$; point $C$ is the foot of that perpendicular.
Proof. It holds that $AB=c$ and since $CB=CD$, it also holds that $AC+CB=AC+CD=AD=\sqrt{\frac{3}{2}}c$. From this it follows that $\sqrt{AC.CB}=\frac{c}{2}$. Since $BC$ is perpendicular to $AD$, it follows that $\angle{BCA}=90^0$. Thus $ABC$ is the desired triangle.
Dicussion. Since $AB\sqrt{2}=\sqrt{2}c>\sqrt{\frac{3}{2}}c=AD>AB$, the circle $k$ intersects the ray $DX$ in exactly two points, which correspond to two symmetric solutions.