75th NMO Selection Tests

From (1) and (2), we obtain that $△LBY\equiv △LCY$ 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 $△BLC$, 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 $△BLC$ 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 $\angle BAC=45^{\circ }$.