A plane that bisects the volume of a right pyramid with a square base, passing through one of the base edges and forming a 30-degree angle with the base plane. What is the height of the pyramid if the base edge is $2 \mathrm{~cm}$?
Let the vertices of the pyramid be $A, B, C, D$ and $E$; the plane intersecting the edge $AB$ cuts the face $DCE$ at the segment $FG$. According to problem F. 3021 (see the solution in the 97th page of the 1995/2 issue of our journal), if $\frac{DC}{FG}=\frac{\sqrt{5}+1}{2}$, then the plane $ABFG$ bisects the volume of the pyramid. Clearly, there is exactly one volume-bisecting plane intersecting $AB$, so it is also true for the plane in our problem, which forms a $30^{\circ}$ angle with the plane $ABCD$, that $\frac{DC}{FG}=\frac{\sqrt{5}+1}{2}=\frac{2}{\sqrt{5}-1}$.

Let the midpoints of segments $AB$, $CD$, and $FG$ be denoted by $X$, $Y$, and $Z$, respectively, and the orthogonal projections of $E$ and $Z$ onto the plane $ABCD$ by $M$ and $T$, respectively. Since $ABCDE$ is a right pyramid, the points $M$, $T$, and $Z$ lie in the plane $EXY$, and the angle $ZXY$ is equal to the angle between the planes $ABCD$ and $ABFG$, which is $30^{\circ}$ (see the 2nd figure). Clearly, $XY=2$ and $MX=MY=1$. By the theorem of parallel chords,

$$
\frac{MT}{MY}=\frac{EZ}{EY}=\frac{FZ}{DY}=\frac{FG}{DC}=\frac{\sqrt{5}-1}{2}
$$

From this, $MT=\frac{\sqrt{5}-1}{2}$ follows.

The triangle $ZTX$ is half of an equilateral triangle, so $ZT \cdot \sqrt{3}=TX=XM+MT=\frac{\sqrt{5}+1}{2}$, from which $ZT=\frac{\sqrt{15}+\sqrt{3}}{6}$. Finally, using the theorem of parallel chords again:

$$
\frac{EM}{MY}=\frac{ZT}{TY}=\frac{ZT}{MY-MT}=\frac{\frac{\sqrt{15}+\sqrt{3}}{6}}{1-\frac{\sqrt{5}-1}{2}}
$$

From this, a simple calculation yields $EM=\frac{\sqrt{15}+2\sqrt{3}}{3}$.

Thus, the height of the pyramid is $\frac{\sqrt{15}+2\sqrt{3}}{3} \approx 2.45 \text{ cm}$.

Based on the solutions of Péter Frenkel (Fazekas M. Főv. Gimn., II. class) and Zoltán Kocsis (Fonyód, Mátyás Király Gimn., III. class)
![](https://cdn.mathpix.com/cropped/2024_05_02_1b7d70693ab23b3ad123g-1.jpg?height=1366&width=626&top_left_y=1334&top_left_x=716)