, 2003

Solution:

The answer is (C). Let us denote by $M$ the midpoint of the segment $T S$ (it obviously remains fixed as $P$ varies on $\gamma$). The centroid $G$ of the triangle $P S T$ certainly lies on the median $P M$, and moreover $M G=\frac{1}{3} M P$ because the centroid divides the median into two parts, and the part containing the vertex is twice as long as the part containing the midpoint of the side. Therefore, if we perform a homothety (that is, a "dilation") of the plane centered at $M$ with scale $\frac{1}{3}$, the point $P$ is sent to $G$.

Homotheties send circles to circles, and the ratio between the radii is exactly the scale of the homothety. It follows that, as $P$ varies on the circle $\gamma$, the point $G$ varies on a circle $\gamma^{\prime}$ whose radius is a third of that of $\gamma$, and therefore the length of $\gamma^{\prime}$ is a third of that of $\gamma$.

## SECOND SOLUTION

Given an orthogonal Cartesian coordinate system $O x y$, we can choose $O$ so that it coincides with the center of $\gamma$, the unit of measure equal to the radius of $\gamma$, and the orientation of the axes so that the coordinates of $T$ are, for example, $(0,-1)$ and the line tangent to $\gamma$ has equation $y=-1$. A point $S$ belonging to this tangent will have coordinates $(a,-1)$, while a point $P$ of $\gamma$ will have coordinates $(\cos \vartheta, \sin \vartheta)$. The coordinates of the centroid $G$ will then be $\left(\frac{x_{P}+x_{S}+x_{T}}{3}, \frac{y_{P}+y_{S}+y_{T}}{3}\right)$, that is, $\left(\frac{\cos \vartheta+a}{3}, \frac{\sin \vartheta-2}{3}\right)$.

The parametric equations of $\gamma^{\prime}$ will therefore be $\left\{\begin{array}{l}x=\frac{1}{3} \cos \vartheta+\frac{a}{3} \\ y=\frac{1}{3} \sin \vartheta-\frac{2}{3}\end{array}\right.$, which represent a circle of radius $\frac{1}{3}$ and center $\left(\frac{a}{3},-\frac{2}{3}\right)$ (the Cartesian equation of this circle is $9 x^{2}+9 y^{2}-6 a x+12 y+3+a^{2}=0$).