Two non-rolling circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ and radii $2R$ and $R$, respectively, are given on the plane. Find the locus of the centers of gravity of triangles in which one vertex lies on $C_1$ and the other two lie on $C_2$. (B. Frenkin)
Problem
Source: 2006 Oral Moscow Geometry Olympiad grades 10-11 p3
Tags: geometry, Locus, Centroid
vanstraelen
01.01.2021 19:37
Given the circle $C_{1}$, midpoint $O_{1}(0,0)$ and radius $2r$, see drawing. Given the circle $C_{2}$, midpoint $O_{2}(a,0)$ and radius $r$. Point $A \in C_{1}$, points $B \in C_{2},\ C \in C_{2}$. $\triangle ABC$ with center of gravity $F$. Locus of the point $F$, if $A$, $B$ and $C$ move on their circles: those are three independent variables. We try to eliminate one variable; if the points $B$ and $C$ move, the midpoint $D$ lies on and inside the circle $C_{2}$. Instead of $B$ and $C$, we take a point on the circle $D_{0}(a+r\cos \alpha,r\sin \alpha)$. First $A_{1}(2r,0)$ (in blue). We use $A_{1}F=2 \cdot FD_{0}$, giving $\left\{\begin{array}{ll} 3x_{F}=x_{A_{1}}+2x_{D_{0}} \\ 3y_{F}=y_{A_{1}}+2y_{D_{0}} \end{array}\right.$ Eliminating, $(\frac{3x}{2}-a)^{2}+(\frac{3y}{2}-r)^{2}=r^{2}$ (in blue), all points inside this circle belong to the locus. Secondly $A_{2}(0,2r)$ (in green), same method, $(\frac{3x}{2}-a-r)^{2}+(\frac{3y}{2})^{2}=r^{2}$ (in green). Idem, $A_{3}$ and circle (in red) and $A_{4}$ and circle (in light blue). The wanted locus is the circle (in black), midpoint $O(\frac{2a}{3},0)$ and radius $\frac{4}{3}r$.
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