A compact cube has each edge 1 dm. On each of its faces, we draw a smaller square whose center coincides with the center of the face and whose sides are parallel to the edges of the face. The side of the drawn square is $a$ on the base and top faces, $b$ on the front and back faces, and $c$ on the left and right faces, where $c<b<a<1$. We sequentially remove the square prisms defined by the pairs of squares with sides $c$, $b$, and $a$ (in the second and third cases, removing the material still present in the prisms' volumes). Determine the values of $c$, $b$, and $a$ so that the volumes of the sequentially removed parts and the remaining body are equal.
The volumes of the 3 removed parts and the remaining part are equal, so each is $1 / 4 \mathrm{dm}^{3}$. The first removed part is a solid square column, its volume is

$$
V_{1}=c^{2} \cdot 1=1 / 4, \quad \text { hence } \quad c=1 / 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f5a71d3c5b2a367db63ag-1.jpg?height=520&width=1184&top_left_y=344&top_left_x=450)

The second part to be removed is a square column with side $b$ and height 1, but part of the previously drilled square column also falls into it. Therefore, if we want to remove $1 / 4 \mathrm{dm}^{3}$ of material again, we need to drill a square column with a side greater than $1 / 4 = c$ into the cube. The part that falls into the already empty space is a square column with side $c$ and height $b$. The volume of the removed material is therefore - which again must be a quarter of the cube -

$$
V_{2}=b^{2}-c^{2} b=b^{2}-\frac{b}{4}=\frac{1}{4}
$$

the positive root of which is

$$
b=\frac{1+\sqrt{17}}{8} \approx 0.640
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f5a71d3c5b2a367db63ag-1.jpg?height=736&width=1234&top_left_y=1450&top_left_x=432)

It is obviously clear that $a > b$. The third removed part already has two square column-shaped holes, their axes are perpendicular to each other. Individually, their volumes are $c^{2} a$ and $b^{2} a$, but if we calculate $V_{3}$ by subtracting these from $a_{2}$, we would subtract the volume of the hole in the second part twice, so we need to add it back once:

$$
V_{3}=a^{2}-c^{2} a-b^{2} a+c^{2} b=\frac{1}{4}
$$

Substituting the found values of $c$ and $b$,

$$
32 a^{2}-(17+\sqrt{17}) a-(7-\sqrt{17})=0
$$

and the positive root is:

$$
a=\frac{17+\sqrt{17}+\sqrt{1202-94 \sqrt{17}}}{64} \approx 0.776
$$

Kármán Antal (Jászberény, Lehel Vezér g. IV. o. t.)

Remarks. 1. Our diagrams represent the (intended to be removed in one piece) parts, but they can also be considered as images of the cavities formed in the original body. In this sense, the "middle" part of the 3rd part can also fit the somewhat whimsical phrase "a hole in a hole in a hole".
2. Equation (1) can also be obtained from the volume of the remaining body. This is what remains after removing the $a^{2}$ volume square column and 4 "window areas" from the cube. The areas of the windows are $c^{2}$ and $b^{2}$, and their "depth" is $(1-a) / 2$. Thus,

$$
V_{4}=1-a^{2}-\left(c^{2}+b^{2}\right)(1-a)=1 / 4
$$