The base of the regular, unit volume pyramid $A B C D E$ is the rectangle $A B C D$. The pyramid is cut by a plane passing through the edge $B C$, and the intersection domain intersects the edges $E A, E D$ at points $A_{1}$ and $D_{1}$, respectively. The intersection of the lines $B A_{1}$ and $C D_{1}$ is point $M$. What is the volume of the tetrahedron $E M A_{1} D_{1}$, if $E M=2 A B$?
Draw a line through point $E$ parallel to $AB$, let this be $e$; then of course it is also parallel to $CD$. The line $e$ is thus in both the planes $ABE$ and $CDE$, or rather, it is precisely the intersection line of these planes. $BA_1$ lies in the plane $ABE$, and $CD_1$ lies in the plane $CDE$, so their intersection point $M$ is on the intersection line of the planes $ABE$ and $CDE$, which is the line $EM$. Therefore, $EM \parallel AB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_133cf927f3433a080dbdg-1.jpg?height=756&width=1629&top_left_y=365&top_left_x=237)

Let $B'$ be the reflection of point $B$ with respect to $A$, and $C'$ be the reflection of point $C$ with respect to $D$. Then $B'B = C'C = 2AB = EM$, so the planes $B'C'M$ and $BCE$ are parallel. This implies that the volumes of the tetrahedra $MBCE$ and $B'BCE$ are equal:

$$
V_{MBCE} = V_{B'BCE}
$$

It is clear that the area of triangle $B'BC$ is equal to the area of rectangle $ABCD$, so the volume of tetrahedron $B'BCE$ is equal to the volume of pyramid $ABCDE$:

$$
V_{B'BCE} = V_{ABCDE} = 1
$$

The triangles $ABA_1$ and $EMA_1$ are similar because the corresponding sides are parallel or coincide. This implies that $A_1M = 2A_1B$ (since $EM = 2AB$ according to the problem), or $A_1M = \frac{2}{3}MB$. Similarly, we find that $MD_1 = \frac{2}{3}MC$. Therefore, the triangles $MA_1D_1$ and $MBC$ are similar, and $MA_1D_1$ can be obtained by a $\frac{2}{3}$-scale reduction of $MBC$ centered at $M$. The ratio of the areas is the square of the ratio of the linear dimensions:

$$
T_{MA_1D_1} = \frac{4}{9} T_{MBC}
$$

From this, it immediately follows that the volume of tetrahedron $EMA_1D_1$ is $\frac{4}{9}$ of the volume of tetrahedron $EMBC$:

$$
V_{EMA_1D_1} = \frac{4}{9} V_{EMBC}
$$

Comparing equations (1), (2), and (3), we find that the volume of tetrahedron $EMA_1D_1$ is $\frac{4}{9}$ cubic units. András Réz (Bp., Fazekas M. Gyak. Gimn., IV. o. t.)

Remark: We did not use the fact that quadrilateral $ABCD$ is a right angle, only that it is a parallelogram, and even that not the fact that $EA = EC$ and $EB = ED$.