Brazilian Mathematical Olympiad

Let $\angle A$, $\angle B$ and $\angle C$ be the angles of $\triangle ABC$ and $\angle A'$, $\angle B'$ and $\angle C'$ be the angles of $\triangle A'B'C'$. Thus $\angle A = \angle A'$ and $\angle C = \angle C'$.

The circumcircle of $\triangle AA'C'$ meets the circumcircle of $\triangle CC'B'$ at $D \neq C'$. Thus $\angle A'DC' = \pi - \angle A$ and $\angle C'DB' = \pi - \angle C$. Moreover, $\angle A'DB' = 2\pi - (\pi - \angle A) - (\pi - \angle C) = \pi - \angle B$. Hence the circumcircle of $\triangle BB'A'$ passes through $D$.

It is easy to see that $\angle DAA' = \angle DC'A' = \alpha$ and $\angle DA'C' = \angle DAC' = \alpha$. Since $\angle A = \angle A'$ we conclude that $\angle DA'B' = \alpha$. Furthermore $\angle DBB' = \alpha$ as $BB'DA'$ is cyclic. Analogously $\angle DCC' = \alpha$.

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

As the angles $\angle A'DB'$, $\angle B'DC'$ and $\angle C'DA'$ are constant, the minimum area of $\triangle A'B'C'$ occurs when the lengths of $DA'$, $DB'$ and $DC'$ are minimum. Thus the lines $DA'$, $DB'$ and $DC'$ 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.