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Let $C_1,C_2$ be two concentric circles, the radius of $C_2$ being twice the radius of $C_1$. From a point $P$ on $C_2$ tangents to $C_1$, $PA$ and $PB$, are drawn. Prove that the centroid of $\triangle ABP$ lies on $C_1$. The circles are concentric thus the equations are $x^2+y^2+2fx+2gy+c_1/c_2=0$. Now I have found $c_2=0$, $c_1=-3r^2$. But I dont think its of any use. Is there any way to prove it.

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Note that if $O$ is the center of the circles, smaller one having radius $r$, then $PO=2r$, $AO=BO=r$.Also, $PB \perp BO$ and $PA \perp AO$. It follows that the $PA=PB$ , and because $PAO$ is a right angles triangle, it follows that $PA^2 = 4r^2-r^2=3r^2$.

Since $PA=PB$, the triangle $PAB$ is an isosceles triangle. Suppose we draw the median from $P$ to $AB$ intersecting at $E$. Note that this line is bisecting $AB$. Furthermore, because $PAB$ is isosceles, this median in particular divides $PAB$ into two congruent triangles $PAE$ and $PBE$ $(PE=PE,PA=PB,AE=BE)$. It follows that this line is a perpendicular bisector of $AB$.

But then, every perpendicular bisector of a chord of a circle passes through the center. It follows that $O$ lies on $PE$.

Let $AE=a,EO=b,PE=c$. The following equations then are true: $a^2+b^2=r^2$ (From triangle $AOE$) $a^2+c^2=3r^2$(From triangle $PAE$) $b+c=2r$ ($OE+PE=PO$) Solving these equations, we get $c=\frac{3r}{2}$ and $b=\frac{r}{2}$(a doesn't matter).

It's known that the centroid divides medians in the ratio $2:1$. Let the centroid of $PAB$ be $Q$, then this says that $PQ:QE=2:1$. Hence, since $PE=\frac{3r}{2}$, $QE=\frac{r}{2}$. Finally, $QO=QE+EO$ (they are on the same straight line) , hence $QO=r$ and $Q$ lies on the smaller circle.

Please ask if any doubts.

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The right triangle $\triangle OPA$ has $OP=2OA$, thus $\angle OPA=30^\circ$. Similarly, $\angle OPB=30^\circ$. So $\angle APB=60^\circ$.

Since $PA=PB$, $\triangle PAB$ is an equilateral triangle.

From that fact, there are many ways to draw the conclusion. For example:

Let $G$ be the centroid of $\triangle PAB$, it follows that $\angle AGB=120^\circ$. Thus it lies on the small arc $AB$ of the circle $C_1$.

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