This was the problem. Assume that $S$ is a circumference and $A$ a point outside it. Now draw tangents $AB$ and $AC$ to circle $S$ (Points $B$ and $C$ are on the circumference). Then from $B$ draw a perpendicular to $AC$ that intersects $S$ in $P$ and $AC$ in $D$. Suppose that $M$ is the midpoint $BC$ and that $MP$ is perpendicular to $BD$. Show that $P$ is the barycenter of $\triangle ABM$.
This would be a drawing for it:
I really couldn't do anything. I just "found" that the barycenter had to be in $MP$ because $AC$ is parallel to $MP$, therefore $\triangle BME \sim \triangle BCA$, and $E$ would be the midpoint of $BA$.
Thank you in advance.
Let $AB = a, MB = ar$. Then $BC = 2ar$ and by similarity, $CD = 2ar^2$. Thus,
$$BD = 2ar\sqrt{1-r^2}$$
Let $H = AM \cap BD$. Again by similarity, $\dfrac{HB}{MB} = \dfrac{CB}{DB} = \dfrac{1}{\sqrt{1-r^2}}$. Thus,
$$BH = \frac{ar}{\sqrt{1-r^2}}$$
also, $$BP = ar\sqrt{1-r^2}$$
Finally, by Power of a Point, $DP \cdot DB = (DC)^2$. Applying this to what we have, we get
$$r = \frac{1}{\sqrt{3}}$$
and
$$\frac{BP}{BH} = 1-r^2 = \frac{2}{3}$$
So $P$ is the barycenter.