Let $ R ~,~ r $ , $ X ~,~ Y $ be the radii and centres of $ \omega_1 ~,~ \omega_2 $ respectively , Consider the inversion w.r.t. $ \omega_2 $ , it sends the nine-point circle $ (Z) $ of $ \Delta DEF $ to the circumcircle of $ \Delta ABC ~,~ \omega_1 $ , so $ Z $ is on $ XY $ and $ XY : ZY = R : r/2 $ , since $ G $ is on the same line and $ GZ:GY = 1:2 $ , we can see that $ G $ is a fixed point .

simplependulum wrote: Consider the inversion w.r.t. $ \omega_2 $ , it sends the nine-point circle $ (Z) $ of $ \Delta DEF $ to the circumcircle of $ \Delta ABC ~,~ \omega_1 $ How did you get this, it means the ninepointcircle of $DEF$ and $\omega_1$ their common extouchlines are collineair in $Y$ (so again a bad picture made by me) How did you prove that inversion?

Let $I_a$ , $r_a$ be the A-excenter and its radius . Consider inversion with center $I_a$ and ${r_a}^2$ as radius . It sends midpoints of $DE,EF,FD$ to $C,A,B$ and It means nine point circle $(N_a)$ of $\triangle DEF$ is image of ${\omega}_1$ under this inversion and so it's fix. But it means its center , $N_a$, is fixed and as we know $\frac{N_a G}{GI_a}$ is fixed we get $G$ is fixed . Note that in the same way we can prove orthocenter of $\triangle DEF$ is fixed .

SCP wrote: simplependulum wrote: Consider the inversion w.r.t. $ \omega_2 $ , it sends the nine-point circle $ (Z) $ of $ \Delta DEF $ to the circumcircle of $ \Delta ABC ~,~ \omega_1 $ How did you get this, it means the ninepointcircle of $DEF$ and $\omega_1$ their common extouchlines are collineair in $Y$ (so again a bad picture made by me) How did you prove that inversion? A useful property of inversion is that for a point chosen to undergo inversion , sending a circle to another circle , the external common tangents of these two circles intersect at this point . To prove it , we can draw a tangent line from the point to the first circle , we know that the image of this circle is also tangent to the image of the tangent line which is itself , so the common tangents pass through this point . It we can't draw the line because the circle contains this point , we still have the collinearity ( $ X,Y,Z $ ) .

Consider the inversion about $ \omega_1. $ It clearly takes $ \omega_1 $ to the nine-point circle of $ \triangle{DEF} $ which implies that the nine-point center $ N $ of $ \triangle{DEF} $ fixed. Since the circumcenter $ O $ of $ \omega_2 $ is fixed as well and since the dilation with center $ O $ and ratio $ \frac{2}{3} $ takes $ N $ to the centroid of $ \triangle{DEF} $ we have that this centroid is fixed as desired. What a troll problem.

I am sorry to bump, but I was just wondering, what is the motivation for performing inversion with the excircle? Thanks!

Wolstenholme wrote: Consider the inversion about $ \omega_1. $ It clearly takes $ \omega_1 $ to the nine-point circle of $ \triangle{DEF} $ which implies that the nine-point center $ N $ of $ \triangle{DEF} $ fixed. Since the circumcenter $ O $ of $ \omega_2 $ is fixed as well and since the dilation with center $ O $ and ratio $ \frac{2}{3} $ takes $ N $ to the centroid of $ \triangle{DEF} $ we have that this centroid is fixed as desired. What a troll problem. It should be $\omega_2$ in the first line.

We can easily proof that w1 , w2 and the nine-point circle of DEF are coordinate circles ( just by homogeneous ratio and power of points ) . Now because the radius of nine-point is fix and it's on the line OIa , N should be fix , too . and because the ratios of G , N , Ia( the center of DEF ) are fix , G is fix , too .