syk0526 wrote:

Let $O$ be the circumcenter of triangle $ABC$ , and let $l$ be the line passing through the midpoint of segment $BC$ which is also perpendicular to the bisector of angle $\angle BAC$ . Suppose that the midpoint of segment $AO$ lies on $l$ . Find $\angle BAC$ .

Solution.

Let $O'$ be the reflection of $O$ upon $\overline{BC},$ $X=AO\cap l,$ $M$ be the midpoint of $\overline{BC},$ $M_a$ be the midpoint of arc $\overarc{BC}.$ Beacuse $\overline{AX}=\overline{XO},$ so $l\parallel AO'.$ Therefore, $\angle O'AM_a=\angle M_aAC+\angle O'AC=\frac{\angle A}{2}+\left(90^{\circ}-\frac{\angle A}{2}\right)=90^{\circ}\implies O'\in\odot(ABC).$ Thus, it follows that,
$\begin{align*}\angle A&=\angle BO'C\\&=\angle BOC\\&=2\cdot\angle BM_aC\\&=360^{\circ}-2\cdot\angle A\end{align*}$ $\implies \boxed{\angle A=120^{\circ}}\qquad\blacksquare$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(17cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.410779605177924, xmax = 13.381513041381899, ymin = -8.311196221875706, ymax = 6.84538101673516; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen ffqqff = rgb(1,0,1); /* draw figures */draw((-3.25,3.08)--(-5.35,1.02), linewidth(1.2) + rvwvcq); draw((-5.35,1.02)--(4.33,0.96), linewidth(1.2) + rvwvcq); draw((-3.25,3.08)--(4.33,0.96), linewidth(1.2) + rvwvcq); draw(circle((-0.5272685231327364,-1.79598839874818), 5.584669154018941), linewidth(1.2) + linetype("4 4") + blue); draw((xmin, -3.891405216937257*xmin-9.567066955046085)--(xmax, -3.891405216937257*xmax-9.567066955046085), linewidth(1.2)); /* line */draw((xmin, 0.25697657896112225*xmin + 1.121058055270172)--(xmax, 0.25697657896112225*xmax + 1.121058055270172), linewidth(1.2) + wvvxds); /* line */draw((-3.25,3.08)--(-0.5272685231327364,-1.79598839874818), linewidth(1.2)); draw((-0.49273147686726315,3.7759883987481793)--(-0.5272685231327364,-1.79598839874818), linewidth(1.2) + linetype("4 4") + ffqqff); draw((-3.25,3.08)--(-0.49273147686726315,3.7759883987481793), linewidth(1.2) + linetype("4 4") + ffqqff); draw((-0.49273147686726315,3.7759883987481793)--(4.33,0.96), linewidth(1.2) + linetype("4 4") + ffqqff); /* dots and labels */dot((-3.25,3.08),dotstyle); label("$A$", (-3.170072550036374,3.2636028135808512), NE * labelscalefactor); dot((-5.35,1.02),dotstyle); label("$B$", (-5.281436543474696,1.2087932128239056), NE * labelscalefactor); dot((4.33,0.96),dotstyle); label("$C$", (4.408216069269031,1.152238820142522), NE * labelscalefactor); dot((-0.5272685231327364,-1.79598839874818),linewidth(4pt) + dotstyle); label("$O$", (-0.4554617013299602,-1.6377778854724132), NE * labelscalefactor); dot((-0.49273147686726315,3.7759883987481793),linewidth(4pt) + dotstyle); label("$O'$", (-0.41775877287570445,3.923404061530329), NE * labelscalefactor); dot((-0.51,0.99),linewidth(4pt) + dotstyle); label("$M$", (-0.4366102371028323,1.133387355915394), NE * labelscalefactor); label("$l$", (-8.316522284042284,-0.8271649237059118), NE * labelscalefactor,wvvxds); dot((-1.8855663582251463,0.6365116631292921),linewidth(4pt) + dotstyle); label("$X$", (-1.8127671256831672,0.794060999827091), NE * labelscalefactor); dot((-0.5618835760893031,-7.380550275740809),linewidth(4pt) + dotstyle); label("$M_a$", (-0.49316462978421594,-7.236662760929411), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$