Bulgarian Spring Tournament

a) Clearly $\angle BAC = \angle BAY + \angle CAY = 120^\circ$, respectively $\angle BOC = 360^\circ - 2\angle BAC = 120^\circ$. Thus, if $M$ is the midpoint of $BC$, then $OM \perp BC$ (because $BO = OC$), $\angle BOM = 60^\circ$ and $BM = \frac{BC}{2} = 1$. Let $BO = x$ and from triangle $BOM$ we have $OM = \frac{x}{2}$ from $\angle OBM = 30^\circ$ and $x^2 = (\frac{x}{2})^2 + 1^2$ from the Pythagorean theorem, respectively $x^2 = \frac{4}{3}$ and $x = \frac{2\sqrt{3}}{3}$.

b) We have $\angle ANB = \angle AOB = 180^\circ - 2\angle BAO = 180^\circ - 2\angle OBY = 180^\circ - 2(90^\circ - \angle YCB) = 2\angle YCB$. Hence $\angle ABY = \angle ABO + \angle OBY = 2\angle OAB = 180^\circ - \angle AOB = 180^\circ - 2\angle YCB$ and now from the other circle we calculate $\angle ANC = 180^\circ - \angle ACY = \angle ABY = 180^\circ - 2\angle YCB$. Therefore, $\angle ANB + \angle ANC = 180^\circ$, i.e. $N$ lies on $BC$. It remains to consider that $\angle NAC = \angle BCY = \angle BAY$ from the touching and $\angle YAN = \angle CAY - \angle CAN = \angle CAY - \angle BAY = 36^\circ$.