The unit-edged cube $A B C D E F G H$ has the midpoint of edge $A E$ as $P$, and the center of face $B C G F$ as $R$.

![](https://cdn.mathpix.com/cropped/2024_05_02_827ab25d39816d2e520cg-1.jpg?height=293&width=312&top_left_y=222&top_left_x=901)

a) What is the area of the plane figure in which the plane passing through points $P, B, R$ intersects the cube?

b) The above plane cuts the cube into two solids. What is the ratio of the volumes of the two parts?
Solution. Let's attach another instance of the cube $K$, denoted as $K^{\prime}$, to the face $A D H E$ as shown in Figure 1. The resulting prism's plane through points $M, B, G$ clearly passes through points $P$ and $R$ as well, thus it is identical to the cutting plane. This plane intersects the surface of the prism along the diagonals $M G, M B, B G$. In the cube $K$, this plane section is a quadrilateral whose vertices $P$ and $Q$ bisect the segments $M B$ and $M G$, respectively. Therefore, $P Q$ is the midline of the isosceles triangle $M G B$, making the plane section $B P Q G$ a symmetric trapezoid. The longer base, $B G$, is the face diagonal of the cube, while the shorter base, $P Q$, is half of it. The lengths of the legs, $Q G$ and $P B$, are half the length of the face diagonal $M G$ of the prism.

![](https://cdn.mathpix.com/cropped/2024_05_02_092c2645ecae3bd70aebg-1.jpg?height=415&width=610&top_left_y=432&top_left_x=736)

Figure 1

a) The height of the trapezoid is half the height $M R$ of the isosceles triangle $M G B$. By the Pythagorean theorem,

$$
M R^{2}=M G^{2}-G R^{2}=M F^{2}+F G^{2}-G R^{2}=4+1-\frac{1}{2}=\frac{9}{2},
$$

thus the height of the trapezoid is $h=\frac{3 \sqrt{2}}{4}$. The area of the trapezoid is

$$
t=\frac{h}{2}(P Q+B G)=\frac{3 \sqrt{2}}{8} \cdot \frac{3 \sqrt{2}}{2}=\frac{9}{8}
$$

Remarks. 1. The fact that the plane section is a symmetric trapezoid also follows from the symmetry of the cube and the cutting plane with respect to the diagonal plane $E F C D$ of the cube.

2. If we use the spatial extension of the Pythagorean theorem, which states that if a plane figure is projected onto three mutually perpendicular planes, then the sum of the squares of the areas of the projections is the square of the area of the figure, we can reach the goal more quickly. In Figure 2, the projections of the trapezoid are visible on three faces of the cube. The area of the two congruent projections is $\frac{3}{4}$ of the area of the cube face, and the third is half of that, $\frac{3}{8}$. The square of the area of the trapezoid is thus

$$
t^{2}=2 \cdot\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{8}\right)^{2}=\frac{9}{8}+\frac{9}{64}=\frac{81}{64}
$$

from which we again obtain the previous result.

![](https://cdn.mathpix.com/cropped/2024_05_02_092c2645ecae3bd70aebg-1.jpg?height=304&width=634&top_left_y=1905&top_left_x=732)

b) The cutting plane divides the cube $K$ into two solids. One of them is a frustum of a pyramid, whose bases are isosceles right triangles. In the prism formed by the two cubes, the pyramid $Q E P M$ completes this frustum to the pyramid $G F B M$. The two pyramids are centrally similar, with a similarity ratio of $1: 2$. The ratio of their volumes is thus the cube of this ratio, $1: 8$, so the volume of the frustum in $K$ is $\frac{7}{8}$ of the volume of the pyramid $G F B M$, which is one-third of the volume of the cube. (The area of the face $G F B$ is half the area of the cube face, and the corresponding height $M F$ is twice the length of the cube edge.) The volume of the frustum is thus $\frac{7}{24}$ of the volume of the cube, so the ratio of the volumes of the parts is $7: 17$.

Remark. Of course, we could have reasoned using the known volume formula for the frustum as well.