[Theorem of Three Perpendiculars]

At the base of a quadrilateral pyramid lies a rhombus $ABCD$, in which $\angle BAD=60^{\circ}$. It is known that $SD=SB$, $SA=SC=AB$. A point $E$ is taken on the edge $DC$ such that the area of triangle $BSE$ is the smallest among the areas of all sections of the pyramid containing the segment $BS$ and intersecting the segment $DC$. Find the ratio $DE: EC$.
Denote $S D=S B=A B=a, E C=x$. Then the square of the area of triangle $B S E$ can be represented as a quadratic trinomial in terms of $x$.

## Solution

The lateral edges $S D$ and $S B$ of the given pyramid are equal, so the base $O$ of the height of the pyramid is equidistant from points $D$ and $B$. Therefore, point $O$ lies on the perpendicular bisector of segment $D B$, i.e., on the line $A C$. Similarly, we can prove that point $O$ lies on the line $B D$. Therefore, the height of the pyramid passes through the center of the rhombus.

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

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

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 the line $B E$. Then $O P$ is the midline of triangle $D B K$. Let $M$ be the midpoint of $C D$. Then, by calculating the area of triangle $D B E$ in two ways, we get that

$$
\begin{gathered}
O P=\frac{1}{2} D K=\frac{1}{2} D E \cdot B M / B E=\frac{1}{2}(a-x) \cdot\left(a \sqrt{a^{2}+x^{2}-a x}=\right. \\
\left.=a \sqrt{a^{2}+x^{2}-a x}\right)
\end{gathered}
$$

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

$$
\begin{gathered}
S O^{2}=S D^{2}-O D^{2}=a^{2}-3 a^{2} / 4=a^{2} / 4, \\
S P^{2}=S O^{2}+O P^{2}=a^{2} / 4+3 a^{2}(a-x)^{2} /\left(16\left(a^{2}+x^{2}-a x\right)\right)= \\
=a^{2}\left(4 a^{2}+4 x^{2}-4 a x+3 a^{2}+3 x^{2}-6 a x\right) /\left(16\left(a^{2}+x^{2}-a x\right)\right)= \\
=a^{2}\left(7 x^{2}-10 a x+7 a^{2}\right) /\left(16\left(a^{2}+x^{2}-a x\right)\right) \\
S^{2}(B S E)=\frac{1}{4} B E^{2} \cdot S P^{2}= \\
=\frac{1}{4}\left(a^{2}+x^{2}-a x\right) \cdot a^{2}\left(7 x^{2}-10 a x+7 a^{2}\right) /\left(16\left(a^{2}+x^{2}-a x\right)\right)= \\
=a^{2}\left(7 x^{2}-10 a x+7 a^{2}\right) / 64
\end{gathered}
$$

Since the area is a positive quantity, it attains its minimum value when and only when its square is minimized. The square of the area is a quadratic trinomial in $x$. Its minimum value is attained at $x=5 a / 7$. In this case,

$$
E C=5 a / 7, D E=a-x=2 a / 7
$$

Therefore, $D E / E C=2 / 5$.

## Answer

$$
2: 5
$$

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