## Πέμπτη, 12 Οκτωβρίου 2017

### 2002 JBMO problem 2 (CYP)

proposed by Cyprus

Two circles with centers ${O_1}$  and ${O_2}$  meet at two points ${A}$  and ${B}$  such that the centers of the circles are on opposite sides of the line ${AB}$. The lines ${BO_1}$  and ${BO_2}$  meet their respective circles again at ${B_1}$  and ${B_2}$. Let ${M}$  be the midpoint of ${B_1B_2 }$. Let ${M_1, M_2}$  be points on the circles of centers ${O_1}$  and ${O_2}$  respectively, such that ${\angle AO_1M_1 =\angle AO_2M_2}$, and ${B_1}$  lies on the minor arc ${AM_1}$  while ${B }$ lies on the minor arc ${AM_2}$. Show that ${\angle MM_1B = \angle MM_2B}$.

posted in aops