From a square piece of paper, we cut out a single-piece net of an open-top, square-based prism such that the cutting and folding lines are parallel to the diagonals of the sheet. Determine the dimensions of the net so that we can fold it into a cube. Prove that the maximum volume is obtained with a half-cube. Construct the net for both cases!
Let the side of the paper square be $a$, the base edge of the prism be $b$, and its height be $m$. The net consists of a $b$-sided square and four rectangles with sides $b$ and $m$. For a single-piece construction, the sheets can be arranged in 8 different ways. Let's consider the simplest and most aesthetically pleasing type, which is the one where all four base edges are folded and all four side edges are glued (this has four rotational symmetries and four symmetry axes), allowing $b$ and $m$ to be maximized. Such a net can be constructed by inscribing two rectangles with sides $b$ and $b+2m$ in the square sheet, with the sides parallel to the diagonals, and their common part being the $b$-sided square.

![](https://cdn.mathpix.com/cropped/2024_05_02_8255dda6274040e707b5g-1.jpg?height=404&width=330&top_left_y=432&top_left_x=884)

Figure 1

We get a cube when $m=b$ (Figure 1); in this case, the sides of the inscribed rectangles are $b$ and $3b$. The sum of their projections on one side of the sheet equals the side of the sheet, so $4b / \sqrt{2} = a$ and thus $b = a / 2\sqrt{2}$. Therefore, the free-standing vertices of the net are at the first and third quarters of the sides of the sheet; the volume is: $V_k = b^3 = a^3 \sqrt{2} / 32 = a^3 / 16\sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8255dda6274040e707b5g-1.jpg?height=395&width=376&top_left_y=1109&top_left_x=861)

Figure 2

In general, $b / \sqrt{2} + (2m + b) / \sqrt{2} = a$ (Figure 2), and from this, $b + m = a / \sqrt{2}$ is constant, meaning that as $b$ increases, $m$ decreases and vice versa. The volume $V = b^2 m$ can be transformed into a product where the sum of the factors is constant:

$$
V = b^2 m = b \cdot b \cdot m = 4 \frac{b}{2} \cdot \frac{b}{2} \cdot m
$$

and for $V / 4$ the factors are:

$$
\frac{b}{2} + \frac{b}{2} + m = b + m = \frac{a}{\sqrt{2}}
$$

Thus, using the known inequality between the arithmetic and geometric means for three numbers, $V / 4$ is maximized when the factors are equal: $b / 2 = m$, and thus $b + m = 3m = a / \sqrt{2}$, $m = a / 3\sqrt{2}$, $b = a\sqrt{2} / 3$, and $V_{\max} = a^3 \sqrt{2} / 27 (> V_k)$. The resulting solid, with $m = b / 2$, can be called a half-cube. The vertices of the net on the sides of the square sheet are at a distance of $b / \sqrt{2} = a / 3$ from the corners of the sheet, dividing the sides into 3 equal parts.

Gyula Tomcsányi (Budapest, Toldy F. g. III. o. t)

Remarks. 1. There is no need to use the inequality between the arithmetic and geometric means to prove the previously stated assertion that the maximum volume is achieved with a half-cube. Let's choose the vertices of the net on the sides of the sheet at a distance of $a / 3 - z$ from the corners of the sheet. Then $b = (a / 3 - z) \sqrt{2}$, $m = a / \sqrt{2} - b = (a / 6 + z) \sqrt{2}$, and the modified volume is

$$
V(z) = b^2 m = 2 \sqrt{2} \left( \frac{a}{3} - z \right)^2 \left( \frac{a}{6} + z \right) = \sqrt{2} \left( \frac{a^3}{27} - a z^2 + 2 z^3 \right)
$$

and the difference from the volume $V_0 = a^3 \sqrt{2} / 27$ of the half-cube is:

$$
V(z) - V_0 = \sqrt{2} z^2 (2z - a)
$$

$V(z)$ would be greater than $V_0$ if this difference were positive, which would only occur if $z > a / 2$. However, this is excluded because $z$, the distance to the corners of the sheet, cannot reach the value $a / 3$, so the difference cannot be positive. - Our examination also includes the possibility of moving the vertices of the net towards the midpoints of the sides of the sheet, in which case $z$ is negative, and the difference is also negative, so $V_0$ is greater than $V(z)$ as long as $z$ is not 0.

2. Ten papers mention or examine some or all of the other possible arrangements of the net's sheets. The authors of these received 1 bonus point each. The other papers, with one exception, examined the net of the type presented.

Similar calculations can show that for each of the other 7 possible arrangements, the maximum volume achievable is less than $V_0$. - However, it is a known problem that if the folding and cutting lines of the net are parallel to the sides of the sheet, the maximum volume achievable is $2a^3 / 27 = V_0 \sqrt{2}$, in which case the height of the prism is one quarter of the base edge.