12.312. The side of the base of a regular triangular pyramid is equal to $a$, and the dihedral angle at the base is $\alpha$. The pyramid is intersected by a plane parallel to the base. The area of the section is equal to the lateral surface area of the resulting frustum of the pyramid. Find the distance from the cutting plane to the base of the pyramid.
Solution.

Let $DO$ be the height of the regular pyramid $DABC$ (Fig. 12.174), $A_1B_1C_1$ be the section of the pyramid, $E$ be the midpoint of $AB$, $E_1$ be the midpoint of $A_1B_1$, and $O_1$ be the center of $A_1B_1C_1$.

Then $EE_1$ is the apothem of the regular truncated pyramid $ABCA_1B_1C_1$, $OO_1$ is the distance between the parallel planes $ABC$ and $A_1B_1C_1$; $\angle E_1EO = \alpha$.

Let $A_1B_1 = b, EE_1 = d$.

Draw the perpendicular $E_1F$ to the plane $ABC$.

Then the point $F$ is on $OE$, $E_1F = OO_1$, $EF = \frac{(a-b) \sqrt{3}}{6}$, $d = EE_1$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1067.jpg?height=722&width=777&top_left_y=77&top_left_x=306)

Fig. 12.174

\[
\begin{aligned}
= & \frac{(a-b) \sqrt{3}}{6 \cos \alpha} \cdot \text{ The lateral surface of the truncated pyramid } S = 3 \cdot \frac{a+b}{2} d = \\
= & 3 \cdot \frac{a+b}{2} \cdot \frac{(a-b) \sqrt{3}}{6 \cos \alpha} = \frac{(a^2 - b^2) \sqrt{3}}{4 \cos \alpha}. \\
& \text{ Since } S = S_{\triangle A_1B_1C_1}, \text{ then } \frac{(a^2 - b^2) \sqrt{3}}{4 \cos \alpha} = \frac{b^2 \sqrt{3}}{4} \Leftrightarrow a^2 - b^2 = b^2 \cos \alpha \Rightarrow \\
\Rightarrow & b = \frac{a}{\sqrt{1 + \cos \alpha}} = \frac{a \sqrt{2}}{2 \cos \frac{\alpha}{2}} \Rightarrow E_1F = EF \tan \angle E_1EF = \frac{(a-b) \sqrt{3}}{6} \tan \alpha = \\
= & \frac{\sqrt{3}}{6} \tan \alpha \left(a - \frac{a \sqrt{2}}{2 \cos \frac{\alpha}{2}}\right) = \frac{a \sqrt{3} \tan \alpha}{6 \cos \frac{\alpha}{2}} \left(\cos \frac{\alpha}{2} - \frac{\sqrt{2}}{2}\right) = \frac{a \sqrt{3} \tan \alpha}{6 \cos \frac{\alpha}{2}} \left(\cos \frac{\alpha}{2} - \cos \frac{\pi}{4}\right) = \\
= & \frac{a \sqrt{3} \tan \alpha}{3 \cos \frac{\alpha}{2}} \sin \left(\frac{\pi}{8} - \frac{\alpha}{4}\right) \sin \left(\frac{\pi}{8} + \frac{\alpha}{4}\right)
\end{aligned}
\]

\[
\text{ Answer: } \frac{a \sqrt{3} \tan \alpha}{3 \cos \frac{\alpha}{2}} \sin \left(\frac{\pi}{8} - \frac{\alpha}{4}\right) \sin \left(\frac{\pi}{8} + \frac{\alpha}{4}\right).
\]