NMO Selection Tests

From (1) and (2), we obtain that $\triangle LBY \equiv \triangle LCY$ by the SSS criterion, and thus $\angle BLY = \angle CLY$, which implies that $LY$ is the angle bisector of $\angle BLC$ (3).
On the other hand, $\overline{AX} = \overline{BX}$ implies that $\angle ACX = \angle XCB$, so $XC$ is the angle bisector of $\angle LCB$ (4).

Let $I$ be the intersection point of the segments $LY$ and $CX$. From (3) and (4), we conclude that $I$ is the incenter of triangle $\triangle BLC$, hence $BI$ is the angle bisector of $\angle LBC$ (5).
From the hypothesis, $\angle XLY = 90^\circ$, and since $LY$ is the angle bisector of $\angle BLC$ (from (3)), it follows that $LX$ is the external angle bisector of $\angle BLC$ (6).
From (4) and (6), we deduce that $X$ is the excenter of triangle $\triangle BLC$ corresponding to vertex $C$, hence $XB$ is the external angle bisector of $\angle LBC$. Combining this with (5), we obtain that $XB \perp BI$, i.e., $\angle XBI = 90^\circ$ (7).

Let $D$ be the intersection of $BI$ with the circumcircle $\Gamma$, and let $\alpha = \angle IBC = \angle IBL$. Then $\angle ABL = \angle LBC = 2\alpha$. Since $\angle LBC = \angle LCB = 2\alpha$, we also have $\angle LCX = \angle XCB = \alpha$. Therefore:
$$\widehat{AD} = 2\angle ABD = 2(\angle ABL + \angle LBD) = 2(2\alpha + \alpha) = 6\alpha.$$
Also, $\widehat{AX} = 2\angle ACX = 2\alpha$, so combining the two gives $\widehat{DX} = 8\alpha$.
On the other hand, $\widehat{DX} = 2\angle XBD = 180^\circ$, by (7). Combining these two relations gives $8\alpha = 180^\circ$, hence $\angle ACB = 2\alpha = 45^\circ$, $\angle ABC = 4\alpha = 90^\circ$, and thus $\angle BAC = 45^\circ$.