In triangle \(ABC\), point \(M\) lies on side \(AC\) and point \(N\) lies on side \(BC\), such that \(MN\) is parallel to \(AB\) (so triangle \(CMN\) is the smaller triangle near vertex \(C\), and \(ABNM\) is the trapezium formed below it). If the area of trapezium \(ABNM\) is twice the area of triangle \(CMN\), what is the ratio \(CM : AM\)?