Junior Balkan Mathematical Olympiad Shortlist

Solution:

As $AD$ is the radical axis of the circumcircles of $\triangle ADC$ and $\triangle BAD$, we have that $O_{1}O_{2} \perp AD$, therefore $\widehat{DAB} = 90^{\circ}$. Let $F$ be the midpoint of $[CD]$ and $[CE]$ be a diameter of the circumcircle of $\triangle ADC$. Then $ED \perp CD$ and $EA \perp CA$, so $ED \parallel AH$ and $EA \parallel DH$ since $AH \perp CD$ and $DH \perp AC$ ($H$ is the orthocenter of $\triangle ADC$), and hence $EAHD$ is a parallelogram. Therefore $O_{1}O_{2} = AH = ED = 2O_{1}F$, so in $\triangle FO_{1}O_{2}$ with $\widehat{O_{1}FO_{2}} = 90^{\circ}$ we have $O_{1}O_{2} = 2FO_{1} \Rightarrow \widehat{ABC} = \widehat{O_{1}O_{2}F} = 30^{\circ}$.

Then we get $\widehat{ACB} = 45^{\circ}$ and $\widehat{BAC} = 105^{\circ}$.

Solution:

As $AD$ is the radical axis of the circumcircles of $\triangle ADC$ and $\triangle BAD$, we have that $O_{1}O_{2} \perp AD$, therefore $\widehat{DAB} = 90^{\circ}$ and $O_{2}$ is the midpoint of $[BD]$.
Take $E = DH \cap AC$, $F = AH \cap BC$ and $M = AD \cap O_{1}O_{2}$.
We have $\widehat{CEH} = \widehat{CFH} = 90^{\circ} \Rightarrow CEHF$ is cyclic, hence $\widehat{ACD} = \widehat{AHD}$.
But $\widehat{ACD} = \operatorname{arc} AD / 2 = \widehat{DO_{1}O_{2}}$, so $\widehat{AHD} = \widehat{DO_{1}O_{2}}$. We know that $AH = O_{1}O_{2}$.
We also have $\widehat{DAH} = \widehat{O_{1}O_{2}D}$ since $AO_{2}FM$ is cyclic with $\widehat{AMO_{2}} = \widehat{AFO_{2}}$.
Therefore $\triangle HDA \equiv \triangle O_{1}DO_{2} \Rightarrow DA = DO_{2} = BD/2$, so in right-angled $\triangle ABD$ we have $m(\widehat{ABD}) = 30^{\circ}$.
Then we get $\widehat{ACB} = 45^{\circ}$ and $\widehat{BAC} = 105^{\circ}$.