Triangle inscribed in a circle,2 points fixed and 1 moving. The track of centroid makes a circle but how to prove it without cartesian coordinate

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Triangle ABC and circle O. A and B are fixed, but C is moving on the circle.

So I have triangle ABC and circle O. A and B are fixed on the circle, but C is moving around the circle. Let G is the centroid of ABC, G' is the centroid of OAB, and $r$ is the radius of O. Then the track of G makes a circle, and its center is G' and radius is $\frac{r}{3}$.

It is easy to prove with Cartesian coordinate. Let O($0,0$), A($a_x,a_y$), B($b_x,b_y$), C($c_x,c_y$), G($g_x,g_y$). Then $$a_x^2+a_y^2=r^2$$ $$b_x^2+b_y^2=r^2$$ $$c_x^2+c_y^2=r^2$$
Since G is the centroid of ABC, $$g_x=\frac{a_x+b_x+c_x}{3}\quad \therefore c_x=3g_x-a_x-b_x$$ $$g_y=\frac{a_y+b_y+c_y}{3}\quad \therefore c_y=3g_y-a_y-b_y$$Then $$c_x^2+c_y^2=(3g_x-a_x-b_x)^2+(3g_y-a_y-b_y)^2=r^2$$ $$(g_x-\frac{a_x+b_x}{3})^2+(g_y-\frac{a_y+b_y}{3})^2=(\frac{r}{3})^2$$ so G$(g_x,g_y)$ makes a circle, center of which is $(\frac{a_x+b_x}{3},\frac{a_y+b_y}{3})$ and radius $\frac{r}{3}$. Also, $(\frac{a_x+b_x}{3},\frac{a_y+b_y}{3})$ is the centroid of triangle OAB.

But there must be a way that proves this without cartesian coordinate but with pure geometry. Problem is, I know little of geometry and can't find the way. Could you enlighten me and show me the way?

Best Answer

This is not too hard to see using ordinary geometry.

Let $A$, $B$ be fixed points on circle with center $O$, and $C$ any other point on the circle. In triangle $ABC$, bisecting $CB$, $AB$ at $E$, $F$, and joining $AE$, $CF$, then $G$ is the centroid of $\triangle ABC$.

In fixed triangle $AOB$, bisect $AO$ at $D$, and join $BD$, $OF$, giving $K$ the centroid of $\triangle AOB$. Finally, join $GK$.

centroid, 12/14/19

Assuming as well known, that the centroid divides the median lines of a triangle in a $\frac{2}{1}$ ratio, then$$\frac{CG}{GF}=\frac{OK}{KF}=\frac{2}{1}$$Hence$$\frac{CF}{GF}=\frac{OF}{KF}=\frac{3}{1}$$Therefore$$GK\parallel CO$$whence$$\triangle CFO\sim\triangle GFK$$and$$\frac{CO}{GK}=\frac{3}{1}$$And since $CO$ has a fixed length for all positions of $C$, so does $GK$.

Therefore, since $K$ is fixed in position, $G$ lies always on the circumference of a circle centered on $K$.

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