Let the second intersection of $\odot (ABC_1)$ and $BC$ be $X$, the second intersection of $\odot(CAB_1)$ and $BC$ be $Y$. Then it is easy to check $AX \parallel CA_1 \parallel PR$ and $AY\parallel BA_1 \parallel PQ$ by angle chasing. This shows that $\triangle AXY$ is an equilateral triangle. Therefore, $P$ and $Q$ are symmetric with respect to the perpendicular bisector of $AY$. Similarly, $P$ and $R$ are symmetric with respect to the perpendicular bisector of $AX$. Also, noticed that we only need to show the line connecting the centroids of $\triangle AXY$ and $\triangle PQR$ is parallel to $BC$ ($=XY$). The problem could be done by proving the following claim. Claim. Given an equilateral triangle $AXY$, for any point $P$, $Q$ and $R$ are defined as the symmetric points of $P$ with respect to the perpendicular bisector of $AY$ and $AX$, respectively. Then the Euler line of $\triangle PQR$ is the line passing the center of $\triangle AXY$ parallel to $XY$. Proof. First, it is obvious that the circumcenter of $\triangle PQR$ is the center of $\triangle AXY$. Noticed that $\angle QPR=60^\circ$, then this is done by a well-known fact: "The Euler line of triangle $ABC$ with $\angle A=60^\circ$ is parallel to the exterior angle bisector of $\angle A$." (One can show $AO=AH$.) This completes the proof.

Suppose the barycentric coordinates of points $P, Q, R$ are $(p_a, p_b, p_c), (q_a, q_b, q_c), (r_a, r_b, r_c)$, respectively. It suffices to prove that $\frac{1}{3}(p_a +q_a +r_a)=\frac{1}{3}$, which is equivalent as the following equation: $$| \triangle PBC| + | \triangle QBC| + | \triangle RBC| = | \triangle ABC|. $$By angle chasing, $\angle QCA=\angle QPA=\angle BA_1 A=\angle BA_1 P=\angle BCP$ and $\angle AQC=\angle BPC=120^\circ$, thus $\triangle QAC \sim \triangle PBC$. Likewise, $\triangle RBA \sim \triangle PBC$. Note that $$\frac{BP}{BC}=\frac{BR}{BA}\Rightarrow BR\cdot BC=BP\cdot BA$$and $$\angle RBC=\angle PBA$$implies that $|\triangle RBC|=|\triangle PBA|$. Likewise, $|\triangle QBC|=|\triangle PAC|$. By summing the areas, we get $$| \triangle PBC| + | \triangle QBC| + | \triangle RBC| = | \triangle PBC| + | \triangle PAC| + | \triangle PBA| = | \triangle ABC|, $$thus completes the proof.

USJL wrote: Let $ABC$ be a triangle. Let $ABC_1, BCA_1, CAB_1$ be three equilateral triangles that do not overlap with $ABC$. Let $P$ be the intersection of the circumcircles of triangle $ABC_1$ and $CAB_1$. Let $Q$ be the point on the circumcircle of triangle $CAB_1$ so that $PQ$ is parallel to $BA_1$. Let $R$ be the point on the circumcircle of triangle $ABC_1$ so that $PR$ is parallel to $CA_1$. Show that the line connecting the centroid of triangle $ABC$ and the centroid of triangle $PQR$ is parallel to $BC$. Proposed by usjl I think this holds for any point $P$ on the plane if $QR$ is defined by the reflection of $P$ to the lines passing through the circumcenters of $ABC_1$ and $ACB_1$ orthogonal to $CA_1$ and $BA_1$ Is that correct? I'll submit my solution later

this actually holds for any point $P$ on the plane if we redefine $QR$ as the reflection of $P$ wrt. the line passing through the circumcenter of $ABC_1$ perpendicular to $CA_1$ and the line passing through the circumcenter of $ACB_1$ perpendicular to $BA_1$ Prove: Let the midpoint of $BC$ be M, we then let M be the origin of the cartesian plane $(0,0)$. let $B(-1,0)$ $C(1,0)$ $A(a,h)$ then it is easy to tell that the centroid of $ABC$ has the coordinates $(\frac{a}{3},\frac{h}{3})$. by applying the 60 degree rotation matrix \[\begin{bmatrix}\frac{1}{2} & \frac{-\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2} \\\end{bmatrix}\]we can get $C_1(\frac{a}{2}-\frac{\sqrt{3}h}{2}-\frac{1}{2} , \frac{-\sqrt{3}a}{2}+\frac{h}{2}+\frac{\sqrt{3}}{2})$ , similarly we can get $B_1(\frac{a}{2}+\frac{\sqrt{3}h}{2}+\frac{1}{2} , \frac{-\sqrt{3}a}{2}+\frac{h}{2}+\frac{\sqrt{3}}{2})$. we then get the coordinates of the circumcenters, which are just centroids of equilateral triangles, by taking the arithmetic mean of the three vertices. Let $O_1$ be the circumcenter of $ABC_1$ and $O_2$ be the circumcenter of $ACB_1$ We have $O_1(\frac{a}{2}-\frac{\sqrt{3}h}{6}-\frac{1}{2} , \frac{-\sqrt{3}a}{6}+\frac{h}{2}+\frac{\sqrt{3}}{6})$ and $O_2(\frac{a}{2}+\frac{\sqrt{3}h}{6}+\frac{1}{2} , \frac{-\sqrt{3}a}{6}+\frac{h}{2}+\frac{\sqrt{3}}{6})$ , finally it's trivial that the slope of $A_1C$ is $\sqrt{3}$ , so the line passing through $O_1$ perpendicular to $A_1C$ should have the slope of $\frac{-\sqrt{3}}{3}$, we then get it's function $y = \frac{-\sqrt{3}x}{3}+\frac{\sqrt{3}a}{3}+\frac{h}{3}$. similarly the function of the line passing through $O_2$ perpendicular to $A_1B$ is $y = \frac{\sqrt{3}x}{3}-\frac{\sqrt{3}a}{3}+\frac{h}{3}$ we then consider a lemma of reflecting a point $(m,n)$ wrt. the line $y=ax+b$ Lemma: the $y$ coordinate of the reflection point of $(m,n)$ wrt. the line $y=ax+b$ is $\frac{2am+a^2n+2b-n}{a^2+1}$ we can prove that by setting up the line passing through $(m,n)$ perpendicular to $y=ax+b$, which is $y=\frac{-x+m}{a}+n$ we get the $y$ coordinate of the intersection of the two lines is $\frac{am+a^2n+b}{a^2+1}$, so we know the reflected $y$ coordinate is $\frac{2am+2a^2n+2b}{a^2+1}-n$ , which is $\frac{2am+a^2n+2b-n}{a^2+1}$ and we are done proving this lemma Now we take a random point $P(m,n)$ on the plane and reflect it wrt. $y = \frac{-\sqrt{3}x}{3}+\frac{\sqrt{3}a}{3}+\frac{h}{3}$ and $y = \frac{\sqrt{3}x}{3}-\frac{\sqrt{3}a}{3}+\frac{h}{3}$ , we say the two reflected points are $Q$ and $R$. the reflected $y$ coordinates would be $\frac{\frac{2\sqrt{3}m}{3}+(\frac{\sqrt{3}}{3})^2n+2(-\frac{\sqrt{3}a}{3}+\frac{h}{3})-n}{(\frac{\sqrt{3}}{3})^2+1}$ and $\frac{\frac{-2\sqrt{3}m}{3}+(\frac{\sqrt{3}}{3})^2n+2(\frac{\sqrt{3}a}{3}+\frac{h}{3})-n}{(\frac{\sqrt{3}}{3})^2+1}$ so the $y$ coordinate of the centroid of $PQR$ is $\frac{1}{3}(n+\frac{\frac{2\sqrt{3}m}{3}+(\frac{\sqrt{3}}{3})^2n+2(-\frac{\sqrt{3}a}{3}+\frac{h}{3})-n}{(\frac{\sqrt{3}}{3})^2+1}+\frac{\frac{-2\sqrt{3}m}{3}+(\frac{\sqrt{3}}{3})^2n+2(\frac{\sqrt{3}a}{3}+\frac{h}{3})-n}{(\frac{\sqrt{3}}{3})^2+1})$ , which very conveniently cancels out to $\frac{h}{3}$ , the same $y$ coordinate as the centroid of $ABC$. Therefore the line connecting the two centroids is parallel to $BC$. and we are done 申申申

It suffices to show that $\overrightarrow{AP}+\overrightarrow{BR}+\overrightarrow{CQ} \parallel BC$. WLoG, diagram as shown. Let $PR, PQ$ intersect $BC$ at $S, T$ respectively as shown, so that $\Delta PST$ is equilateral. It is easy to show via angle chasing that $\Delta BSR \stackrel +\sim \Delta BPA$ and $\Delta BSP \stackrel -\sim \Delta BPC$, and similarly for the other side. Now \begin{align*} &\mid \overrightarrow{BR} \mid _{\perp BC} + \mid \overrightarrow{CQ} \mid _{\perp BC} = BR \sin \angle SBR + CQ \sin \angle TCQ \\ &= \frac{BA \cdot BS}{BP} \sin \angle SBR + \frac{CA \cdot CT}{CP} \sin \angle TCQ = \frac{BA \cdot BP}{BC} \sin \angle PBA + \frac{CA \cdot CP}{CB} \sin \angle PCA \\ &= \frac{2([ABP]+[ACP])}{BC} = \frac{[ABPC]}{BC} = \mid \overrightarrow{AP} \mid _{\perp BC}, \end{align*}hence proved.