Marinada '23

[lang=hr] Na stranici $AB$ trokuta $ABC$ izabrana je točka $D$ takva da vrijedi $4 \cdot AD=BD$. Točke $O_1$ i $O_2$ su središta kružnica opisanih oko trokuta $ACD$ i $BCD$. Pretpostavimo da je $O_1O_2 \parallel BC$. Pravac $O_1O_2$ dijeli trokut $ABC$ na dva dijela površina $P_1$ i $P_2$. Ako je $P_1 > P_2$ odredi $P_1 : P_2$. Odgovor zaokruži na $3$ decimale. [/lang] [lang=en] On side $AB$ of the triangle $ABC$ choose point $D$ such that $4 \cdot AD=BD$. Points $O_1$ and $O_2$ are centers of circumscribed circles of triangles $ACD$ and $BCD$ respectively. Suppose $O_1O_2 \parallel BC$. Line $O_1O_2$ divides triangle $ABC$ in two parts with areas $P_1$ and $P_2$. Suppose $P_1 > P_2$. Find $P_1 : P_2$. Round your answer to three decimal places. [/lang]