Open Contests

Fig. 9 Fig. 10

Let $B^{\prime }$ and $C^{\prime }$ be respectively the second intersection points of the lines $AB$ and $AC$ with the circumcircle of the triangle $BCK$ (Fig. 9). Notice that $\angle BKC=180^{\circ }-\angle CBK-\angle BCK=180^{\circ }-\frac{\angle ABC}{2}-(90^{\circ }+\frac{\angle ACB}{2})=90^{\circ }-\frac{\angle ABC}{2}-\frac{\angle ACB}{2}$. Hence the central angle supported by the arc $BC$ has the size $180^{\circ }-\angle ABC-\angle ACB$. The central angle supported by the arc $B^{\prime }C$ has the size $2\angle ABC$ and the central angle supported by the arc $C^{\prime }B$ has the size $2\angle ACB$. Hence the central angle supported by the arc $B^{\prime }{C}^{\prime }$ has the size $360^{\circ }-(180^{\circ }-\angle ABC-\angle ACB)-2\angle ABC-2\angle ACB$, which equals $180^{\circ }-\angle ABC-\angle ACB$. Hence the arcs $BC$ and $B^{\prime }{C}^{\prime }$ have the same size. As $K$ is the midpoint of the arc $B^{\prime }C$, the point $B^{\prime }$ is the reflection of the point $C$ by the diameter drawn from point $K$, and $C^{\prime }$ is the reflection of point $B$. Hence the intersection point $A$ of $B{B}^{\prime }$ and $C{C}^{\prime }$ has to lie on the diameter drawn from the point $K$.

Remark. Among the used central angles there may also be angles of size greater than $180^{\circ }$ (reflex angles), such as the angle corresponding to the arc $B^{\prime }C$ in Fig. 10.

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Alternative solution.

Fig. 11

Let $L$ be the second endpoint of the diameter through $K$ of the circumcircle of the triangle $BCK$ (Fig. 11). Then $\angle KCL=90^{\circ }$, thus $CL$ is the bisector of the interior angle at vertex $C$ of the triangle $ABC$, and $\angle KBL=90^{\circ }$, so $BL$ is the bisector of the exterior angle at vertex $B$ of $ABC$. The bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle intersect at a common point. Thus the bisector of the exterior angle at vertex $A$ of the triangle $ABC$ passes through points $K$ and $L$. Thus point $A$ lies on $KL$.

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Alternative solution.

Fig. 12

Let us denote the angles of the triangle $ABC$ at vertices $A$, $B$ and $C$ respectively $\alpha$, $\beta$ and $\gamma$. Let the intersection point of the line $BC$ and the tangent to the circumcircle of triangle $BCK$ at point $K$ be $M$ (Fig. 12). By the property of inscribed angles, $\angle MKC=\angle MBK=\frac{\beta }{2}$. The bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle have a common point. Hence $AK$ is the bisector of the exterior angle at vertex $A$ of the triangle $ABC$. --- Thus $\angle CKA=180^{\circ }-\angle KAC-\angle KCA=180^{\circ }-\frac{180^{\circ }-\alpha }{2}-\frac{180^{\circ }-\gamma }{2}=\frac{\alpha }{2}+\frac{\gamma }{2}$. Hence $\angle MKA=\angle MKC+\angle CKA=\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\gamma }{2}=90^{\circ }$. As a tangent of a circle is perpendicular to the diameter drawn from the point of tangency we have that $A$ lies on the diameter drawn from point $K$.