Given a circle with center $O$ and radius 1. From point $A$, tangents $A B$ and $A C$ are drawn to the circle. Point $M$, lying on the circle, is such that the quadrilaterals $O B M C$ and $A B M C$ have equal areas. Find $M A$.
Note that point $M$ lies on the smaller arc $BC$ (otherwise, $OBMC$ is either inside $ABMC$ or is not a quadrilateral at all). Then

$S_{OBMC} - S_{ABMC} = S_{OBC} + 2S_{MBC} - S_{ABC}$. Therefore, the geometric locus of points for which $S_{OBMC} = S_{ABMC}$ is the perpendicular bisector of segment $OA$. Hence, $AM = OM = 1$.

## Answer

1.