Croatia Mathematical Competitions

Let us denote by $T$ the centroid of the triangle $ABC$, by $a$, $b$ and $c$ the lengths of the sides $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ respectively and by $v$ the length of the altitude to the side $\overline{AB}$. Let us denote $\alpha = \angle BAC$ and $\beta = \angle CBA$.
Since $T$ is the centroid of the triangle $ABC$, it follows that
$$|TA'| = |B'C| = \frac{1}{3}|CA| = \frac{1}{3}b, \quad |TB'| = |A'C| = \frac{1}{3}|BC| = \frac{1}{3}a, \quad |TC'| = \frac{1}{3}v.$$

$$\begin{align*} P(A'B'C') &= P(A'B'T) + P(B'C'T) + P(C'A'T) \\ &= \frac{1}{2} (|TA'| \cdot |TB'| + |TB'| \cdot |TC'| \sin(\pi - \alpha) + |TC'| \cdot |TA'| \sin(\pi - \beta)) \\ &= \frac{1}{18} (ab + av \sin \alpha + bv \sin \beta). \end{align*}$$
$$Since \$v = a \sin \beta\$, \$v = b \sin \alpha\$, \$a = c \sin \alpha\$, \$b = c \sin \beta\$ and \$c^2 = a^2 + b^2\$ hold, we get$$
$$\begin{align*} P(A'B'C') &= \frac{1}{18}(ab + a^2 \sin \alpha \sin \beta + b^2 \sin \alpha \sin \beta) \\ &= \frac{1}{18}(ab + c^2 \sin \alpha \sin \beta) = \frac{1}{18}(ab + ab) \\ &= \frac{1}{9}ab = \frac{2}{9}P(ABC). \end{align*}$$
$$$$
Hence $P(A'B'C') : P(ABC) = 2 : 9$.