Note: The contestants were given 10 minutes to think about the problem(s), and then they were supposed to present their initial ideas and proof strategies.

Let $\omega$ be the circle through the midpoints of $AA',BB',CC'$. Consider moving $X$ uniformly on $(ABC)$. Note that $X'$ should lie on the image $\Omega$ of $\omega$ under homothety at $X$ with ratio $2$, as well as $(A'B'C')$, so the unique existence of $X'$ implies $\Omega$ and $(A'B'C')$ should be tangent to each other. However, $\Omega$ has fixed radius and its center moves along a circle with the same angular velocity as $X$ does, so it is easy to see that $X'$ should move on $(A'B'C')$ with the same angular velocity as well. Noting that the map $X \mapsto X'$ sends $A \mapsto A', B \mapsto B', C \mapsto C'$, we conclude that $\Delta ABC$ and $\Delta A'B'C'$ are similar. (Actually, they should be homothetic as well for the condition to hold, but we are not asked to prove that.)