Problem source

Three distinct points, $X_1$, $X_2$ and $X_3$, with position vectors ${\bf x}_1$, ${\bf x}_2$ and ${\bf x}_3$ respectively, lie on a   circle of radius 1 with its centre at the origin $O$. The point  $G$ has position vector $\frac13({\bf x}_1+{\bf x}_2+{\bf x}_3)$. The line through $X_1$ and $G$ meets the circle again at the point $Y_1$ and the points $Y_2$ and $Y_3$ are defined correspondingly.
Given that $\overrightarrow{GY_1} =-\lambda_1\overrightarrow{GX_1}$, where $\lambda_1$  is a positive scalar, show that
\[
\overrightarrow{OY_1}= \tfrac13 \big(
(1-2\lambda_1){\bf x}_1 +(1+\lambda_1)({\bf x}_2+{\bf x}_3)\big)
\]
and hence that
\[
\lambda_1 = \frac
{3-\alpha-\beta-\gamma} {3+\alpha -2\beta-2\gamma}
\,,\]
where $\alpha = {\bf x}_2 \,.\, {\bf x}_3$, $\beta = {\bf x}_3\,.\, {\bf x}_1$ and $\gamma = {\bf x}_1\,.\, {\bf x}_2$.
Deduce that $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} +
\dfrac {GX_3}{GY_3} =3
\,$.

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\a{2};
        \coordinate (O) at (0,0);
        \coordinate (A) at ({\a*cos(20)},{\a*sin(20)});
        \coordinate (B) at ({\a*cos(150)},{\a*sin(150)});
        \coordinate (C) at ({\a*cos(250)},{\a*sin(250)});
        
        % Calculate Centroid G
        \coordinate (G) at ($1/3*(A)+1/3*(B)+1/3*(C)$);

        % 1. Draw the circle and name it
        \draw[dashed, name path=circumcircle] (O) circle (\a);

        % 2. Create an invisible path that extends through A and G in BOTH directions.
        % Using -2 and 4 ensures it crosses the circle boundaries clearly.
        \path[name path=median_a] ($(A)!-1!(G)$) -- ($(A)!3!(G)$);
        \path[name path=median_b] ($(B)!-1!(G)$) -- ($(B)!3!(G)$);
        \path[name path=median_c] ($(C)!-1!(G)$) -- ($(C)!3!(G)$);

        % 3. Find intersections. 
        % We use 'sort by=median' to ensure point 1 is at one end and point 2 at the other.
        \fill [name intersections={of=circumcircle and median_a, sort by=median, by={Y1_other, Y1}}];
        \fill [name intersections={of=circumcircle and median_b, sort by=median, by={Y2_other, Y2}}];
        \fill [name intersections={of=circumcircle and median_c, sort by=median, by={Y3, Y3_other}}];

        % Draw the line through the triangle (A to Y1)
        \draw[blue] (A) -- (Y1);
        \draw[blue] (B) -- (Y2);
        \draw[blue] (C) -- (Y3);

        % Draw points and labels
        \filldraw (A) circle (1.5pt) node[right] {$X_1$};
        \filldraw (B) circle (1.5pt) node[left] {$X_2$};
        \filldraw (C) circle (1.5pt) node[below left] {$X_3$};
        \filldraw (G) circle (1.5pt) node[below] {$G$};
        \filldraw (O) circle (1.5pt) node[right] {$O$};
        \filldraw (Y1) circle (1.5pt) node[below left] {$Y_1$};
        \filldraw (Y2) circle (1.5pt) node[below right] {$Y_2$};
        \filldraw (Y3) circle (1.5pt) node[above] {$Y_3$};

    \end{tikzpicture}      
\end{center}

\begin{align*}
&& \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\
&&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 -   \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\
&&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\
&& 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\
&&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\
&&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\
\Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\
&&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\
&& 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\
\Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\
&&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\
\Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)}
\end{align*} 

as required.

Since $\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}$ we must have,

\begin{align*}
&& \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} +
\frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\
&&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\
&&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\
&&&= 3

\end{align*}