11.169. A plane is drawn through the vertex of the base and the midpoints of the two lateral edges of a regular triangular pyramid. Find the ratio of the lateral surface area of the pyramid to the area of its base, given that the intersecting plane is perpendicular to the lateral face.
Solution.

Let $K O$ be the height of the regular pyramid $K A B C$ (Fig. 11.63), $M$ be the midpoint of $A K$, and $N$ be the midpoint of $C K$. The plane $M B N$ is perpendicular to the plane $A K C$. $B M$ and $B N$ are medians of the equal isosceles $\triangle A K B$ and $\triangle C K B$, drawn to their lateral sides. Therefore, $B M = B N$. $Q$ is the point of intersection of the midline $M N$ and the median $K E$ of $\triangle A K C$. Consequently, $Q$ is the midpoint of $K E$ and $M N$, and since $B M = B N$, $B Q$ is the height and median of $\triangle M B N$. Since the planes $M B N$ and $A K C$ are perpendicular, the segment $B Q$ is perpendicular to the plane $A K C$ and, therefore, $B Q \perp K E$. Thus, $B Q$ is the median and height of $\triangle K B E$. It follows that $B K = B E$. Let $B K = B E = 1$. Then $B O = \frac{2}{3}$, $O E = \frac{1}{3}$. From the right triangle $B O K \left(\angle B O K = 90^{\circ}\right)$: $K O^{2} = B K^{2} - B O^{2} = 1 - \frac{4}{9} = \frac{5}{9}$; further from $\triangle E O K \left(\angle E O K = 90^{\circ}\right)$ we find $K E = \sqrt{K O^{2} + O E^{2}} = \sqrt{\frac{5}{9} + \frac{1}{9}} = \frac{\sqrt{6}}{3}$. Since the pyramid is regular, $S_{6} = \frac{S_{\text{base}}}{\cos \angle K E O}$. Then

$$
\frac{S_{6}}{S_{\text{base}}} = \frac{1}{\cos \angle K E O} = \frac{K E}{E O} = \frac{\sqrt{6}}{3} : \frac{1}{3} = \sqrt{6}.
$$

Answer: $\sqrt{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0873.jpg?height=629&width=618&top_left_y=77&top_left_x=93)

Fig. 11.64