There are two situations for this question due to different rotation directions, as shown in Figure 1 and Figure 2 respectively, so there should be two answers. Trigonometric function is to establish rectangular coordinate system and complete it with analytic geometry.

Let c be the origin (0,0), EB be the positive direction of X axis, and CA be the positive direction of Y axis.

Figure 1

Rotate 60 clockwise, and then ∠ ECE' = ∠ DCD' = 60. Therefore, the d' coordinate is (6cos30,6s in30) = (3 √ 3,3), and the e' coordinate is (6cos 120, 6s in120) = (-3,3 √ 3).

Because the coordinate of b is (10,0), the equation of straight line BE' is 3√3x+ 13y-30√3=0. Because CN⊥BE', | cn | = | 3 √ 3 * 0+13 * 0-30 √ 3 |/√ [(3 √ 3) 2+132] =15 √ 3/7.

The slope of the straight line CM⊥BE' is 13√3/9, so the equation of the straight line CM is y= 13√3x/9. The equation of straight line ad' is 7√3x+9y-90=0. The coordinate of point M can be obtained by simultaneous expression as (3√3/2, 13/2). So | cm | = √ [(3 √ 3/2) 2+(13/2) 2] = 7.

Finally, we get |MN|=|CM|-|CN|=7- 15√3/7.

Figure 2

Rotate 60 counterclockwise, which also applies to ∠ ECE' = ∠ DCD' = 60. Don't forget next, or you'll be in trouble!

By observing Figure 1 and Figure 2, we can see that E, E', D and D' are all on a circle with the center of C and the radius of 6. Point E' in Figure 1 and Figure 2 is symmetrical about X axis, and point B is on X axis, so |CN|= 15√3/7.

Two BEs are symmetrical about X, then two CM perpendicular to them are symmetrical about Y, that is, 1 is the same as ∠ACM in Figure 2. There are two graphs whose d' is symmetric about y, that is, the two ∠ Macs are the same, so the two △ AMCs are exactly the same. Get |CM|=7。

Finally, we get | Mn | =| cm |+cn | = 7+ 15 √ 3/7.

To sum up: | Mn | = 7 15 √ 3/7.