On the lateral edge $A B$ of the pyramid, points $K$ and $M$ are taken such that $A K=B M$. Through these points, sections parallel to the base of the pyramid are drawn. It is known that the sum of the areas of these sections is $\frac{2}{3}$ of the area of the base of the pyramid. Find the ratio $K M: A B$.
Let $S$ be the area of the base of the pyramid, $S_1$ and $S_2$ be the areas of the sections through points $K$ and $M$ respectively. These sections are polygons similar to the polygon of the base of the pyramid with coefficients $\frac{AK}{AB}$ and $\frac{AM}{AB}$ respectively. Let $KM = x$ and $AB = y$. Then,

\[
\begin{gathered}
AK = BM = \frac{y - x}{2}, \quad AM = AK + KM = \frac{y - x}{2} + x = \frac{y + x}{2}, \\
S_1 = \left(\frac{AK}{AB}\right)^2 \cdot S = \left(\frac{y - x}{2y}\right)^2 \cdot S, \quad S_2 = \left(\frac{AM}{AB}\right)^2 \cdot S = \left(\frac{y + x}{2y}\right)^2 \cdot S.
\end{gathered}
\]

Since $S_1 + S_2 = \frac{2}{3} S$, we have

\[
\left(\frac{y - x}{2y}\right)^2 \cdot S + \left(\frac{y + x}{2y}\right)^2 \cdot S = \frac{2}{3} S, \quad \left(\frac{y - x}{2y}\right)^2 + \left(\frac{y + x}{2y}\right)^2 = \frac{2}{3},
\]

from which we find that

\[
\frac{KM}{AB} = \frac{x}{y} = \frac{\sqrt{3}}{3}
\]

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-36.jpg?height=100&width=1006&top_left_y=190&top_left_x=0)

The base of a triangular pyramid is an equilateral triangle. The height of the pyramid is $h$. All lateral faces are inclined to the base plane at an angle $\alpha$. Find the area of the base. (List all possibilities.)

## Solution

Since the lateral faces of the pyramid form equal dihedral angles with the base plane, the height of the pyramid passes either through the center of the inscribed circle or through the center of one of the exscribed circles of the base triangle. Let the height of the pyramid pass through the center $O$ of the inscribed circle of the base triangle $ABC$ of the triangular pyramid $ABCD$, and let $M$ be the midpoint of $BC$ (Fig. 1). Denote $AB = BC = AC = a$. Since $OM \perp BC$, by the theorem of three perpendiculars, $DM \perp BC$, so $DMO$ is the linear angle of the dihedral angle formed by the lateral face $DBC$ with the base plane $ABC$. By the problem's condition, $\angle DMO = \alpha$, and $DO = h$. From the right triangle $DMO$, we find that

\[
OM = DO \operatorname{ctg} \angle DMO = h \operatorname{ctg} \alpha.
\]

On the other hand, since $OM$ is the radius of the inscribed circle of an equilateral triangle with side $a$, we have $OM = \frac{a \sqrt{3}}{6}$. From the equation $\frac{a \sqrt{3}}{6} = h \operatorname{ctg} \alpha$, we find that $a = 2h \sqrt{3} \operatorname{ctg} \alpha$. Therefore,

\[
S_{\triangle ABC} = \frac{a^2 \sqrt{3}}{4} = (2h \sqrt{3} \operatorname{ctg} \alpha)^2 \cdot \frac{\sqrt{3}}{4} = 3h^2 \sqrt{3} \operatorname{ctg}^2 \alpha.
\]

Let the height of the pyramid pass through the center $O_1$ of the exscribed circle touching the side $BC$ of the base triangle $ABC$ of the pyramid $ABCD$ (Fig. 2). Similarly, we find that

\[
O_1M = DO \operatorname{ctg} \angle DMO = h \operatorname{ctg} \alpha.
\]

On the other hand, since $O_1M$ is the radius of the exscribed circle of an equilateral triangle with side $a$, we have $O_1M = \frac{a \sqrt{3}}{2}$. From the equation $\frac{a \sqrt{3}}{2} = h \operatorname{ctg} \alpha$, we find that $a = \frac{2}{3} h \sqrt{3} \operatorname{ctg} \alpha$. Therefore,

\[
S_{\triangle ABC} = \frac{a^2 \sqrt{3}}{4} = \left(\frac{2}{3} h \sqrt{3} \operatorname{ctg} \alpha\right)^2 \cdot \frac{\sqrt{3}}{4} = \frac{1}{3} h^2 \sqrt{3} \operatorname{ctg}^2 \alpha.
\]

## Answer

$3h^2 \sqrt{3} \operatorname{ctg}^2 \alpha; \frac{1}{3} h^2 \sqrt{3} \operatorname{ctg}^2 \alpha$.