10. From a point $S$ outside a circle $\gamma$ of radius 1, draw a tangent line to $\gamma$ and let $T$ be the point of tangency. As a point $P$ varies on $\gamma$, the centroid of the triangle $P S T$ describes a curve $\gamma^{\prime}$. What is the ratio between the length of $\gamma^{\prime}$ and that of $\gamma$?
(A) $\frac{1}{\pi}$
(B) $\frac{1}{2}$
(C) $\frac{1}{3}$
(D) $\frac{4}{3} \pi$
(E) depends on the position of $S$.

## Numerical Answer Problems - 8 points
10. The answer is (C). Let $M$ be 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 $M G = \frac{1}{3} M P$ 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 (i.e., a "dilation") of the plane centered at $M$ and with scale $\frac{1}{3}$, the point $P$ is mapped to $G$.

![](https://cdn.mathpix.com/cropped/2024_04_17_9dd5da716b9e5be06201g-17.jpg?height=339&width=333&top_left_y=2113&top_left_x=1595)
Homotheties map circles to circles, and the ratio of 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 one-third of that of $\gamma$, and thus the length of $\gamma^{\prime}$ is one-third of that of $\gamma$.

## SECOND SOLUTION

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

![](https://cdn.mathpix.com/cropped/2024_04_17_9dd5da716b9e5be06201g-18.jpg?height=372&width=415&top_left_y=199&top_left_x=1603)

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 $9 x^{2} + 9 y^{2} - 6 a x + 12 y + 3 + a^{2} = 0$).