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 $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} +
\dfrac {GX_3}{GY_3} =3
\,$.
