Problem 46: EGMO 2016, Day 2, Problem 1

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Joined: Mar 10, 2015

by henderson, Aug 2, 2016, 4:26 PM

$\color{red}\bf{Problem \ 46}$ Two circles $\omega_1$ and $\omega_2,$ of equal radius intersect at different points $X_1$ and $X_2.$ Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2.$ Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega.$ $($ My solution

$\color{red}\bf{My \ solution}$ Let $O, O_1$ and $O_2$ be the centers of $\omega, \omega_1$ and $\omega_2,$ respectively. Observe that $O_1, T_1, O$ and $O_2, O, T_2$ are collinear. Let $S=X_2T_2 \cap \omega_2,$ $\quad$ $T_1X_1X_2=\alpha,$ $\quad$ $T_1OS=2\beta$ and $OT_2S=\theta.$ Then,

$\color{red}{1)}$ $\triangle OT_2S \sim \triangle O_2T_2X_2,$ so $OS\parallel O_2X_2.$
$\color{red}{2)}$ $\angle X_2T_1S=180^\circ-(\angle O_1T_1X_2+\angle OT_1S)=180^\circ-(180^\circ-\alpha-\beta)=\alpha+\beta.$

Since $T_1X_1X_2=\alpha,$
$\color{red}{2)}$ implies that
$X_1, T_1,S$ are collinear $\iff$ $\angle X_1X_2T_1=\beta$ $\iff$ $\angle X_1O_1T_1=2\beta$ $\iff$ $X_1O_1\parallel OS$ $($ since $\angle T_1OS=2\beta$ $)$ $\iff$ $X_1O_1\parallel X_2O_2$ $($ remember that $OS\parallel O_2X_2$ $),$ which is true, because $X_1O_1X_2O_2$ is a rhombus