IMO Shortlist

Let $\delta_A$, $\delta_B$, $\delta_C$ be the circumcircles of $\triangle AA_1A_2$, $\triangle BB_1B_2$, $\triangle CC_1C_2$. The general strategy of the solution is to find two different points having equal power with respect to $\delta_A$, $\delta_B$, $\delta_C$.

Claim. $A_1$ is the circumcentre of $A_2BC$ and cyclic variations.

Proof. Since $A_1$ lies on the perpendicular bisector of $BC$ and inside $\triangle BA_2C$, it suffices to prove $\angle BA_1C = 2 \angle BA_2C$. This follows from
$$\begin{aligned} \angle BA_2C &= \angle A_2BA + \angle BAC + \angle ACA_2 \\ &= \frac{1}{2}\left((180^\circ - \angle AC_1B) + (180^\circ - \angle CB_1A)\right) + 60^\circ \\ &= 240^\circ - \frac{1}{2}(480^\circ - \angle BA_1C) \\ &= \frac{1}{2} \angle BA_1C \end{aligned}$$
$\square$

The circumcentres above give
$$\angle B_1B_2C_1 = \angle B_1B_2A = \angle B_2AB_1 = \angle C_1AC_2 = \angle AC_2C_1 = \angle B_1C_2C_1$$
and so $B_1C_1B_2C_2$ is cyclic. Likewise $C_1A_1C_2A_2$ and $A_1B_1A_2B_2$ are cyclic. Note that hexagon $A_1B_2C_1A_2B_1C_2$ is not cyclic since
$$\angle C_2A_1B_2 + \angle B_2C_1A_2 + \angle A_2B_1C_2 = 480^\circ \neq 360^\circ.$$
Thus we can apply radical axis theorem to the three circles to show that $A_1A_2$, $B_1B_2$, $C_1C_2$ concur at a point $X$ and this point has equal power with respect to $\delta_A$, $\delta_B$, $\delta_C$.

Let the circumcircle of $\triangle A_2BC$ meet $\delta_A$ at $A_3 \neq A_2$. Define $B_3$ and $C_3$ similarly.

Claim. $BCB_3C_3$ cyclic.

Proof. Using directed angles
$$\begin{align*} \npreceq BC_3C &= \measuredangle BC_3C_2 + \not C_2C_3C \\ &= \measuredangle BAC_2 + \not C_2C_1C \\ &= 90^\circ + \nmid(C_1C, AC_2) + \not C_2C_1C \tag{1} \\ &= 90^\circ + \nmid C_1C_2B_1. \end{align*}$$
Similarly $\nsucceq CB_3B = 90^\circ + \measuredangle B_1B_2C_1$. Hence, using $B_1C_1B_2C_2$ cyclic
$$\text{BB 3C = 90 + C 1B 2B 1 = 90 + C 1C 2B 1 = BC 3C}$$
as required. $\square$

Similarly $CAC_3A_3$ and $ABA_3B_3$ are cyclic. $AC_3BA_3CB_3$ is not cyclic because then $AB_2CB_3$ cyclic would mean $B_2$ lies on $\odot ABC$ which is impossible since $B_2$ lies inside $\triangle ABC$. Thus we can apply radical axis theorem to the three circles to get $AA_3$, $BB_3$, $CC_3$ concur at a point $Y$ which has equal power with respect to $\delta_A$, $\delta_B$, $\delta_C$.

We now make some technical observations before finishing.
- Let $O$ be the centre of $\triangle ABC$. We have that
$$\angle BA_1C = 480^\circ - \angle CB_1A - \angle AC_1B > 480^\circ - 180^\circ - 180^\circ = 120^\circ.$$
so $A_1$ lies inside $\triangle BOC$. We have similar results for $B_1$, $C_1$ and thus $\triangle BA_1C$, $\triangle CB_1A$, $\triangle AC_1B$ have disjoint interiors. It follows that $A_1B_2C_1A_2B_1C_2$ is a convex hexagon thus $X$ lies on segment $A_1A_2$ and therefore is inside $\delta_A$.
- Since $A_1$ is the centre of $A_2BC$ we have that $A_1A_2 = A_1A_3$ so, from cyclic quadrilateral $AA_2A_1A_3$ we get that lines $AA_2$ and $AA_3 \equiv AY$ are reflections in line $AA_1$. As $X$ lies on segment $A_1A_2$, the only way $X \equiv Y$ is if $A_1$ and $A_2$ both lie on the perpendicular bisector of $BC$. But this forces $B_1$ and $C_1$ to also be reflections in this line meaning $A_1B_1 = A_1C_1$ contradicting the scalene condition.

Summarising, we have distinct points $X, Y$ with equal power with respect to $\delta_A$, $\delta_B$, $\delta_C$ thus these circles have a common radical axis. As $X$ lies inside $\delta_A$ (and similarly $\delta_B$, $\delta_C$), this radical axis intersects the circles at two points and so $\delta_A$, $\delta_B$, $\delta_C$ have two points in common.