At the base of a quadrilateral pyramid lies a rhombus $A B C D$, in which $\angle B A D=60^{\circ}$. It is known that $S A=S C$, $S D=S B=A B$. A point $E$ is taken on the edge $D C$ such that the area of triangle $B S E$ is the smallest among the areas of all sections of the pyramid containing the segment $B S$ and intersecting the segment $D C$. Find the ratio $D E: E C$.
The lateral edges $S A$ and $S C$ of the given pyramid are equal, so the base $O$ of the height of the pyramid is equidistant from points $A$ and $C$. Therefore, point $O$ lies on the perpendicular bisector of segment $A C$, i.e., on line $B D$. Similarly, point $O$ lies on line $A C$. Therefore, the height of the pyramid passes through the center of the rhombus.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-13.jpg?height=46&width=258&top_left_y=1209&top_left_x=14)

Let $S D = S B = A B = a$, $E C = x$. Then

$$
B E = \sqrt{C B^{2} + C E^{2} - 2 C B \cdot C E \cos 60^{\circ}} = \sqrt{a^{2} + x^{2} - a x}
$$

Let $P$ be the foot of the perpendicular dropped from point $O$ to $B E$. By the theorem of three perpendiculars, $S P \perp B E$, i.e., $S P$ is the height of triangle $B S E$. Drop a perpendicular $D K$ from point $D$ to line $B E$. Then $O P$ is the midline of triangle $D B K$. Let $M$ be the midpoint of $C D$. Then, by writing the area of triangle $D B E$ in two ways, we get that

$$
O P = \frac{1}{2} D K = \frac{1}{2} \cdot \frac{D E \cdot B M}{B E} = \frac{\frac{1}{2}(a - x) \cdot a \sqrt{3}}{\sqrt{a^{2} + x^{2} - a x}} = \frac{a(a - x) \sqrt{3}}{\sqrt{a^{2} + x^{2} - a x}}
$$

By the Pythagorean theorem from right triangles $S O D$ and $S O P$, we find that

$$
\begin{aligned}
& S O^2 = S D^2 - O D^2 = a^2 - \frac{a^2}{2} = \frac{3 a^2}{2},
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-13.jpg?height=82&width=669&top_left_y=1897&top_left_x=700)

$$
\begin{aligned}
& = \frac{3 a^2 \left(a^2 + x^2 - a x + a^2 + x^2 - 2 a x\right)}{16 \left(a^2 + x^2 - a x\right)} = \frac{3 a^2 \left(5 a^2 - 6 a x + 5 x^2\right)}{16 \left(a^2 + x^2 - a x\right)}, \\
& S^2_{\triangle B S E} = \frac{1}{2} B E^2 \cdot S P^2 = \frac{1}{2} \frac{\left(a^2 + x^2 - a x\right) \cdot \left(a^2 \left(5 a^2 - 6 a x + 5 x^2\right)\right)}{16 \left(a^2 + x^2 - a x\right)} =
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-13.jpg?height=75&width=272&top_left_y=2395&top_left_x=893)

Since the area is a positive quantity, it reaches its minimum value when its square is minimized. The square of the area is a quadratic polynomial in $x$. Its minimum value is achieved at $x = \frac{3 a}{5}$. In this case,

$$
E C = \frac{3 a}{5}, D E = a - x = \frac{2 a}{5}
$$

Therefore, $\frac{D E}{E C} = \frac{2}{3}$.