XXI OBM

Let $\angle A$, $\angle B$ and $\angle C$ be the angles of $△ABC$ and $\angle A^{\prime }$, $\angle B^{\prime }$ and $\angle C^{\prime }$ be the angles of $△A^{\prime }{B}^{\prime }{C}^{\prime }$. Thus $\angle A=\angle A^{\prime }$ and $\angle C=\angle C^{\prime }$.

The circumcircle of $△A{A}^{\prime }{C}^{\prime }$ meets the circumcircle of $△C{C}^{\prime }{B}^{\prime }$ at $D\ne C^{\prime }$. Thus $\angle A^{\prime }D{C}^{\prime }=\pi -\angle A$ and $\angle C^{\prime }D{B}^{\prime }=\pi -\angle C$. Moreover, $\angle A^{\prime }D{B}^{\prime }=2\pi -(\pi -\angle A)-(\pi -\angle C)=\pi -\angle B$. Hence the circumcircle of $△B{B}^{\prime }{A}^{\prime }$ passes through $D$.

It is easy to see that $\angle DA{A}^{\prime }=\angle D{C}^{\prime }{A}^{\prime }=\alpha$ and $\angle D{A}^{\prime }{C}^{\prime }=\angle DA{C}^{\prime }=\alpha$. Since $\angle A=\angle A^{\prime }$ we conclude that $\angle D{A}^{\prime }{B}^{\prime }=\alpha$. Furthermore $\angle DB{B}^{\prime }=\alpha$ as $B{B}^{\prime }D{A}^{\prime }$ is cyclic. Analogously $\angle DC{C}^{\prime }=\alpha$.

Hence $D$ is a fixed point since it satisfies $$\angle DAB=\angle DBC=\angle DCA$$ and consequently does not depend on $A^{\prime }$, $B^{\prime }$ and $C^{\prime }$. Its construction is shown below.

As the angles $\angle A^{\prime }D{B}^{\prime }$, $\angle B^{\prime }D{C}^{\prime }$ and $\angle C^{\prime }D{A}^{\prime }$ are constant, the minimum area of $△A^{\prime }{B}^{\prime }{C}^{\prime }$ occurs when the lengths of $D{A}^{\prime }$, $D{B}^{\prime }$ and $D{C}^{\prime }$ are minimum. Thus the lines $D{A}^{\prime }$, $D{B}^{\prime }$ and $D{C}^{\prime }$ ought to be respectively orthogonal to the sides $AB$, $BC$ and $CA$.

Construction of point $D$

Let $E$ be the intersection point of the perpendicular bisector of $AB$ and the perpendicular to $BC$ through $B$. Thus the circle with center $E$ and radius $EA$ touches $BC$ at $B$. Therefore $\angle XAB=\angle XBC$ for each point $X$ in the shorter arc $AB$.

Similarly, define $F$ as the intersection point of the perpendicular bisector of $BC$ and the perpendicular to $CA$ through $C$. The circle with center $F$ and radius $FB$ touches $CA$ at $C$. Therefore $\angle XBC=\angle XCA$ for each point $X$ on the shorter arc $BC$.

The point $D$ is the intersection point of these two arcs. It is easy to see that the point $D$ is unique.