14. (6 points) The figure is composed of 5 identical squares, where points $B$, $P$, and $C$ are collinear, and points $B$, $N$, and $F$ are collinear. If the area of the shaded region to the left of line $B F$ is twice the area of the shaded region to the right of line $B F$, then $M N: N P=$ . $\qquad$
【Solution】Solution: According to the analysis, let the side length of the square be $a$. As shown in the figure, draw $P D \perp B D$ intersecting $B D$ at $D$,
$$
\because O F=A B, P E=D P, \therefore S_{\triangle O N F}=S_{\triangle A B N}, S_{\triangle P E C}=S_{\triangle B D P} \text {, }
$$

The area of the left shaded part $=S_{\triangle O N F}+S_{\text {quadrilateral } B N M G}=S_{\text {quadrilateral } A B G M}$;
The area of the right shaded part $=S_{\triangle A B P}+S_{\triangle P E C}=S_{\text {rectangle } A P D B}$,
According to the problem, the area of the left shaded part $=2 \times$ the area of the right shaded part,
$$
\therefore(A M \times A B):(A P \times A B)=2: 1 \Rightarrow A M: A P=2: 1
$$

Therefore, $A P=\frac{1}{2} A M=E C, F C=E F+E C=2.5 a$, and since $N P=\frac{1}{2} F C=\frac{5}{4}$ a,
Thus, $M N=M P-N P=1.5 a-\frac{5}{4} \mathrm{a}=\frac{1}{4} a$,
$M N: N P=\frac{1}{4} a: \frac{5}{4} a=1: 5$,
Therefore, the answer is: $1: 5$.