Berolov s.l.

The incircle of triangle $ABC$ touches sides $AB$ and $AC$ at points $X$ and $Y$ respectively. Point $K$ is the midpoint of the arc $AB$ of the circumcircle of triangle $ABC$ (not containing point $C$). It turns out that line $XY$ bisects segment $AK$. What can the angle $BAC$ be?
Let segments $X Y$ and $A I$ intersect at point $S$. As known, $K I = K A$ (see problem $\underline{53119}$), which means the height $K T$ of triangle $A K I$ is also its median. Since $X Y \perp A I$, then $X Y \| K T$, and because $X Y$ bisects side $A K$, $X Y$ is the midline of triangle $A K T$. Therefore, $A S: S T = 1: 3$; and since $X S$ is the height of the right triangle $A X I$, we have $A S / S T = \left( \frac{A X}{X I} \right)^2$, $\frac{X I}{A X} = \sqrt{3}$. Thus, $\angle X A I = 60^{\circ}$ and $\angle B A C = 2 \angle X A I = 120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-27.jpg?height=563&width=1512&top_left_y=1805&top_left_x=273)

## Answer

$120^{\circ}$.