Problem 7. On the diagonals $A C$ and $C E$ of a regular hexagon $A B C D E F$, points $M$ and $K$ are taken such that $A M: A C = C K: C E = n$ (Fig. 6). Points $B, M$, and $K$ lie on the same straight line. Find $n$.
Solution. As point $K$ moves along the diagonal $EC$ (from $E$ to $C$), the ratio $CK: CE$ decreases from 1 to 0, while the ratio $AM: AC$ increases from 0.5 ( $AM_1$: $AC=0.5$ ) to 1. Therefore, there exists a unique point $K$ on diagonal $CE$ such that

$$
AM: AC = CK: CE
$$

By trial, we hypothesize that equality (1) holds if $AM = AB$. Let's try to prove this hypothesis.

Let $AM = AB = 1$. Then $AC = 2 \sin 60^\circ = \sqrt{3}$ and $MC = \sqrt{3} - 1$. From the isosceles triangle $ABM$, we get: $\angle ABM = \angle AMB = 75^\circ$. Therefore, $\angle CMK = 75^\circ$, $\angle MCK = 60^\circ$, and $\angle MKC = 45^\circ$.

From triangle $MCK$, using the Law of Sines, we find:

$$
MC: \sin 45^\circ = CK: \sin 75^\circ
$$

Thus,
$CK = MC \cdot \sin 75^\circ: \sin 45^\circ = (\sqrt{3} - 1) \sin (45^\circ + 30^\circ): \sin 45^\circ = (\sqrt{3} - 1) \frac{\sqrt{3} + 1}{2} = 1$.

Therefore, $n = \sqrt{3}: 3$.