63rd Czech and Slovak Mathematical Olympiad

Besides the Thales' circle $k$, denote as $l_A=AP$ and $l_B=BP$ the other two circles under consideration. Let us deal, for example, with the circle $l_B$ drawn in Fig. 4. In what follows the interior angles of $△ABC$ are denoted as usual.

Fig. 4

Let us explain that it is true indeed what the figure suggests. First of all, the point $Q$ lies in the half-plane $BCA$, because the local arc $BC$ of the circle $l_B$ has the following property: as its point $X$ moves from $C$ to $B$, the angle $AXB$ varies from an acute angle $\gamma$ to an obtuse angle $180^{\circ }-\beta /2$, and hence the Thales' circle $k$ meets the arc $BC$ in an interior point $Q$. Applying the properties of inscribed and subtended angles to the chord $BC$ of the circle $l_B$, we conclude that $|\angle BQC|=180^{\circ }-\beta /2$ and hence $|\angle AQB|+|\angle BQC|=270^{\circ }-\beta /2>180^{\circ }$. Consequently, $Q$ is a point of the half-plane $ACB$ which lies in the interior of the triangle $ABC$ and the convex angle $AQC$ equals $90^{\circ }+\beta /2$. As known, the last is a measure of the angle $AIC$, where $I$ denotes the incentre of $△ABC$ (indeed, $|\angle AIC|=180^{\circ }-\alpha /2-\gamma /2=90^{\circ }+\beta /2$). In this way, the congruence of the angles $AQC$ and $AIC$ is proven and hence the points $Q$ and $I$ lie on the same arc $AC$ of a new circle $k_B=ACI$. Therefore, the line $AQ$ is the radical axis of the circles $k$ and $k_B$. Analogously, the line $BP$ is the radical axis of the circles $k$ and $k_A=BC$.

It remains to note that the intersection point of the lines $AQ$ and $BP$ (treated as the above radical axes) has the same power with respect to the circles $k_A$ and $k_B$, whose radical axis is the line $CI$, i.e. just the bisector of the angle $ACB$. This completes our solution.

Fig. 4