9.2. The bisectors of triangle $A B C$ intersect at point $I$, and the external bisectors of its angles $B$ and $C$ intersect at point $J$. Circle $\omega_{b}$ with center at point $O_{b}$ passes through point $B$ and is tangent to line $C I$ at point $I$. Circle $\omega_{c}$ with center at point $O_{c}$ passes through point $C$ and is tangent to line $B I$ at point $I$. Segments $O_{b} O_{c}$ and $I J$ intersect at point $K$. Find the ratio $I K / K J$.

(L. Emelyanov, I. Bogdanov)
Answer: $1 / 3$.

First Solution. In circle $\omega_{b}$, draw the diameter $I X$, and in circle $\omega_{c}$, draw the diameter $I Y$. Note that $\angle I B J=90^{\circ} = \angle I C J$, since the internal and external bisectors of an angle are perpendicular. Therefore, point $X$ lies on $B J$, and point $Y$ lies on $C J$ (see Fig. 1). Moreover, $I X \perp I C$, because $\omega_{b}$ is tangent to $I C$ at point $I$, so $I X \| C J$. Similarly, $I Y \| B J$. Therefore, quadrilateral $I X J Y$ is a parallelogram. Let the diagonals intersect at point $M$. Then $I M = M J$, and segment $O_{b} O_{j}$ is the midline of triangle $I X Y$, so point $K$ is the midpoint of segment $I M$. Thus, $I K = I M / 2 = I J / 4$, from which it follows that $I K / K J = 1 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aeebc79ceb55f7038dcbg-2.jpg?height=766&width=535&top_left_y=552&top_left_x=189)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_aeebc79ceb55f7038dcbg-2.jpg?height=887&width=587&top_left_y=485&top_left_x=720)

Fig. 2

Second Solution. Let $N$ be the midpoint of the arc $B A C$ of the circumcircle $\Omega$ of triangle $A B C$, and let $S$ be the midpoint of the other arc $B C$. Let the ray $N I$ intersect $\Omega$ again at point $P$ (see Fig. 2). Since $S N$ is the diameter of circle $(A B C)$, we have $\angle N P S = 90^{\circ}$.

By a known lemma about the trident, we have $S I = S C = S J$, in particular, $S$ is the midpoint of segment $I J$. Since $\angle B A C = \angle B N C$ and $B N = N C$, we get $\angle N B C = \angle N C B = \frac{1}{2} \angle A B C + \frac{1}{2} \angle A C B$.

Extend ray $C I$ to intersect $A B$ at point $L$. Since $\angle L I B$ is the external angle of triangle $B I C$ and quadrilateral $B N C P$ is cyclic, we obtain $\angle L I B = \angle I B C + \angle I C B = \frac{1}{2} \angle A B C + \frac{1}{2} \angle A C B = \angle N C B = \angle I P B$, so the circle $(I B P)$ is tangent to line $C I$ at point $I$. Also, this circle passes through $B$, so it is the circle $\omega_{b}$. Similarly, circle $\omega_{c}$ is circumscribed around triangle $I P C$.

Thus, $I P$ is the common chord of circles $\omega_{b}$ and $\omega_{c}$, and therefore $O_{b} O_{c}$ is the perpendicular bisector of segment $I P$. Since $\angle I P S = 90^{\circ}$, we get that $O_{1} O_{2}$ passes through the midpoint of segment $I S$, that is, $K I = K S$, and then $I K / K J = 1 / 3$.

Remark. Point $S$ in the second solution coincides with point $M$ from the first solution.