Funny $\angle BUC = 90^{o}+\frac{\angle A}2$, so $\angle BO_{1}C=180^{o}-\angle A$, thus points $O_{1},O_{2}$ and $O_{3}$ ARE on the circumcircle of triangle $ABC$. Bye

I think this problem is very nice. I will call the incenter $I$: We know that $I$ is the orthocenter of the excentral triangle of $\Delta ABC$. Because $O_{1}O_{2},O_{2}O_{3},O_{1}O_{3}$ are the perpendicular bissectors of $CI,AI,BI$ respectively, we conclude that the homothecy $h$ with center $I$ and factor $2$ maps $\Delta O_{1}O_{2}O_{3}$ onto the excentral triangle of $\Delta ABC$. Let us call $O'$ the circumcentre of $\Delta O_{1}O_{2}O_{3}$, and it now follows that $h(O')=V$, the Bevan point of $\Delta ABC$. It is well known that $O$ is the midpoint of $[IV]$, so we conclude that $O' \equiv O$ Image not found

This problem is rather easy than nice, but you made it look hard Just one question, what is the Bevan point of triangle? edit: I found it on the net, never mind

Let $I_A, I_B, I_C$ be the excenters of the triangle $ABC$, we know that $U$ is the orthocenter of $I_AI_BI_C$ and $A, B, C$ are the feet of the altitudes of the triangle, thus the nine points circle of $I_AI_BI_C$ is the circumcircle of $ABC$ and it pass through the midpoints if segments $UI_A$, $UI_B$, and $UI_C$. As $UBCI_A$ is cycle quadrilateral of diameter $UI_A$, its center is the midpoint of $UI_A$, so it is equal to $O_1$, then the circumcircle of $ABC$ pass through $O_1$, simmilary it pass through $O_2$ and $O_3$. $\square$

Just note that: $O_1$, $O_2$, $O_3$ are midpoint of $\stackrel\frown{BC}$ not containing $A$, $\stackrel\frown{CA}$ not containing $B$, $\stackrel\frown{AB}$ not containing $C$ then: $O_1$, $O_2$, $O_3$ $\in$ ($ABC$)

Does anyone ever do the daily problem on the HSO+College feed, and is it the same for everyone? By Fact 5, $O_1,O_2,O_3$ lie on $(ABC)$, done.