11.121. A regular triangular pyramid is intersected by a plane passing through the vertex of the base and the midpoints of two lateral edges. Find the ratio of the lateral surface area of the pyramid to the area of the base, given that the intersecting plane is perpendicular to one of the lateral faces (specify which one exactly).
## Solution.

Let $DO$ be the height of the regular pyramid $DABC$, $M$ be the midpoint of $DB$, $N$ be the midpoint of $DC$, and $\triangle MAN$ be the section of the pyramid mentioned in the problem (Fig. 11.18). Suppose the plane $MAN$ is perpendicular to the plane of the face $ADB$. $\triangle ADM = \triangle ADN$ (by three sides). Then their heights, dropped to the sides $AM$ and $AN$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0834.jpg?height=519&width=439&top_left_y=128&top_left_x=160)

Fig. 11.17

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0834.jpg?height=626&width=522&top_left_y=18&top_left_x=720)

Fig. 11.18

are equal. But the height of $\triangle ADM$ is perpendicular to the plane $MAN$, and the height of $\triangle ADN$ is an oblique to this plane. Then the height of $\triangle ADN$ is also perpendicular to the plane $MAN$ and, consequently, the specified heights coincide with the lateral edge $DA$ of the given pyramid, and $\angle DAB > \angle DAM = 90^\circ$, which is impossible since the given pyramid is regular. Similarly, we obtain a contradiction if we assume that the plane $AMN$ is perpendicular to the plane of the face $ADC$. Therefore, the plane $MAN$ is perpendicular to the face $BDC$. $E$ is the midpoint of $BC$. Then the point $K$ of intersection of $MN$ and $DE$ is the midpoint of these segments. Consequently, the median $AK$ of the isosceles $\triangle MAN$ is its height and, since the planes $MAN$ and $BDC$ are perpendicular, $AK$ is perpendicular to the plane $BDC$ and, therefore, $AK \perp DE$. Hence, $AK$ is the median and height of $\triangle DAE$.

Then $DA = AE$. Let $OE = a$. Then $OA = 2a$, $AD = AE = 3a$.

From $\triangle AOD (\angle AOD = 90^\circ): DO^2 = AD^2 - AO^2 = 5a^2$.

From $\triangle DOE (\angle DOE = 90^\circ): DE = \sqrt{DO^2 + OE^2} = a\sqrt{6}$. Since the pyramid is regular, the ratio of the lateral surface area to the base area $\frac{S_6}{S_{\text{base}}} = \frac{1}{\cos \angle DEO} = \frac{DE}{OE} = \sqrt{6}$.

Answer: $\sqrt{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0835.jpg?height=519&width=557&top_left_y=80&top_left_x=91)

Fig. 11.19

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0835.jpg?height=533&width=551&top_left_y=67&top_left_x=675)

Fig. 11.20