the answer is $\measuredangle( BAC)=\frac{\Pi }{3}$ trekendeshi i vetem qe ploteson kete eshte trekendeshi barabrinjesh. shpresoj ta kem zgjidhur sakte

I think this is a little late for a 2013 TST problem.

Hi ; Let $ABC$ is acute angle Let $CH$ intersect $AB$ at $X$ and $Y$ is the midpoint of $AC$. From $\angle{AYO}=\angle{AXH}=90^{\circ}$ and $\angle{OAN}=\angle{HAX}=90^{\circ}-\angle{B}$ and $AH$=$AO$ we have : $\triangle AYO \simeq \triangle AXH $ so we have : $AX=AY=XY=YC \Rightarrow \triangle AXY$ equilateral , so $\angle{A}=60^{\circ}$ If $ABC$ obtuse triangle we can prove similary Best Regard

it's old problem. IMO SHORTLIST - 1989

$R=AO=AH=2R\cos{A}\Leftrightarrow \cos{\widehat{A}}=\dfrac{1}{2}\Leftrightarrow \widehat{A}=60^\circ$

$ AH=2R|cosA| $

Aiscrim wrote: $R=AO=AH=2R\cos{A}\Leftrightarrow \cos{\widehat{A}}=\dfrac{1}{2}\Leftrightarrow \widehat{A}=60^\circ$ anonymouslonely wrote: $ AH=2R|cosA| $ Hi ; We have $ AH=2R|cosA| $ So $cos{A}$ can be $\frac {-1}{2} , \frac{1}{2}$ and so $\angle A = 60^\circ , 120^\circ$ Best Regard

this problem is not in the level of mathematical Olympiads if $BB'$ is height then $AB=2AB'$ so $cos(BAC)=1/2$ so $angle(BAC)=60$

PLEASE, just check a picture when $ Angle(BAC)=120 $ at least one time...

$cos(A)=\frac{R}{2R}$ $=cos(A)=cos(\frac{\pi}{3})$ $=A=\frac{\pi}{3}$ $R$ circumradius

Drop perpendicular from $O$ on to $BC$ as $OF$. By Euler's theorem, $2OF = AH = AO = AC$. This gives, by $30 - 60 - 90$ or trignometry, $\angle OCF$ = 30. Therefore, $\angle FOC$ = $90 - \angle OCF$. = 60. There fore $\angle BAC$ = $\frac{\angle BOC}{2}$ = $\frac{120}{2}$ = $60$.

The answer is 60° and 120° Or only 60°??

https://artofproblemsolving.com/community/q1h1834354p12295133

It is a very easy TST Problem $$2sin\angle BAE=\frac{BC}{AH}=\frac{BE}{AE}=cot\angle ABE=tan\angle BAE=cos\angle BAE=1/2 \implies \angle BAE=60°$$

Olemissmath wrote: It is given a triangle $ABC$ whose circumcenter is $O$ and orthocenter $H$. If $AO=AH$ find the angle $\hat{BAC}$ of that triangle. It is known that $\angle BAH=\angle CAO$ As $AH=AO:$ Let be the feet of the perpendiculars of $C$(passes through $H$) and $O: X$ and $Y$ $\Rightarrow \triangle AXH \cong \triangle AYO$ $AY=YC=AX \Rightarrow AC=2AX$ As $C, X$ and $H$ are collineals $\Rightarrow \angle BAC=60^{o} _\blacksquare$

$Case\ 1: \bigtriangleup ABC\ is \ acute \\ \\ \frac{AH}{\sin \angle ABH} = \frac{AB}{\sin \angle BHA} \\ \\ \Leftrightarrow \frac{AH}{\sin (90-\angle A)} = \frac{2R\sin (\angle A + \angle B ))}{\sin (\angle A + \angle B )} \\ \\ \Leftrightarrow AH= \frac{2R\sin (\angle A + \angle B )\sin (90-\angle A)}{\sin (\angle A + \angle B ) } \ and \ we \ have \ AH=AO=R \ so: \\ \\ \frac{2R\sin (\angle A + \angle B )\sin (90-\angle A)}{\sin (\angle A + \angle B )} = R \\ \\ \Leftrightarrow sin (90-\angle A) = \frac{1}{2} \\ \\ \cos \angle A = \frac{1}{2} \Rightarrow \angle A =60+360n; 300 + 360 n \\ \\ \ and \ \angle A < 90 \\ \\ \Rightarrow \angle A = 60 \\ \\ Case\ 2: \bigtriangleup ABC\ is \ not acute\ - \ let \ \angle A>90 \\ Similarly: \\ \\ AH= \frac{2R\sin (\angle A + \angle B )\sin (\angle A - 90)}{\sin (\angle A + \angle B ) } \\ so \\ \sin (\angle A - 90)=\frac{1}{2} \ and \\ \angle A = 120$