Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$Prove that $YI$ is the angle bisector of $XYA$.
Problem
Source: Polish MO Finals P2 2023
Tags: geometry, incenter, angle bisector
v4913
29.03.2023 18:24
It suffices to show that <XYI = 60. Note that if Ma is the intersection of AI and (ABC), P is on (ABC) such that <AMaP = 60, and Z is the intersection of PMa and (BIC), then ZIMa is equilateral, X is the intersection of IZ with BC, and Y is the intersection of PX with (ABC), so X has equal power wrt (ABC) and (BIC), thus YIPZ is cyclic and <XYI = <PZI = 60.
timon92
29.03.2023 23:14
This problem was proposed by Burii.
starchan
21.09.2023 12:56
Good problem
Let $M$ denote the midpoint of minor arc $BC$ and let $\omega$ denote the circle $\odot(BIC)$. Note that this circle is centred at $M$. Suppose that line $IY$ meets $\omega$ again at a point $U$, so that triangle $IUM$ is equilateral. Let line $MU$ meet $\odot(ABC)$ again at a point $L$.
Observe that $\angle AML = 60^{\circ}$ and thus $\angle AXL = 120^{\circ}$, implying that $X, Y, L$ are collinear. From radical axis on $\odot(UIL), \odot(ABC)$ and $\omega$, we deduce that $XUIL$ is cyclic, whence \[\angle YXI = \angle LXI = \angle LUI = 60^{\circ} = \frac{1}{2} \angle AXY\]and thus $YI$ is the angle bisector of $\angle XYA$, as desired. $\square$