According to http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 post #2, the radical center of the excircles of triangle ABC is the Spieker point of triangle ABC, that is, the incenter of triangle A'B'C', where A', B', C' are the midpoints of the sides BC, CA, AB of triangle ABC. Thus, it remains to show that the incenter of triangle A'B'C' lies on the line GI. Now, it is well-known that the homothety with center G and ratio $-\frac12$ maps the points A, B, C to the points A', B', C'. Hence, this homothety maps the incenter I of triangle ABC to the incenter of triangle A'B'C'. Since the center of the homothety is G, this yields that the incenter of triangle A'B'C' lies on the line GI. Proof complete. Actually, this problem is way too well-known for an exam imho... darij

this was posted also on the italian forum here, this is the edriv's solution: Let the A-excircle touch AB on L an the B-excircle touch AB on K. Then as well-know the radical axes of $ \Gamma_a$ and $ \Gamma_b$ pass through the point M that is the midpoint of LK that is the midpoint of AB. Also we have that the radical axes is perpendicular to $ I_aI_b$ that is parallel to the internal angle bisector of $ \angle CAB$ and the radical axes of $ \Gamma_a$ and $ \Gamma_b$ is the internal M-angle bisector of the medial triangle and equally for the others, from which the radical center P of the excircles is the incenter of the medial triangle (the Spieker point of ABC) so for an homothety with center G it lies on IG with $ IG = 2 GP$. ok It's the same solution of darij

Omid Hatami wrote: Prove that in triangle $ ABC$, radical center of its excircles lies on line $ GI$, which $ G$ is Centroid of triangle $ ABC$, and $ I$ is the incenter. I have a solution Maybe it is same of you Let $ D$ is midpoint of $ AC$ $ a$ is radical axis of $ O_1)$ and $ (O_3)$ Easy to prove $ I$ is orthocenter of $ O_1O_2O_3$ and $ AE=CF$ So $ DE=DF \Rightarrow P_{D|(O_3)}=P_{D|(O_1)}$ so $ D \in a$ and $ a // BI$ $ K=GI\cap a$ then $ 2GK=IG$ Similar we have desire