\section*{Problem 12 - V01112}

A regular hexagon with side length \(a=10 \mathrm{~cm}\) is cut out of paper that is \(d=0.1 \mathrm{~mm}\) thick. A smaller hexagon is then glued on top, with its corners lying at the midpoints of the sides of the previous one.

On top of this, a third hexagon is glued, with its corners again lying at the midpoints of the sides of the previous one. If this process is continued, a three-dimensional structure is formed.

a) How high is this structure if it is assumed that the lower limit for cutting is \(2 \mathrm{~mm}\) and the thickness of the glue can be neglected?

b) What is the volume?

c) What would be the height and volume if there were no lower limit set for cutting?
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1662.jpg?height=400&width=459&top_left_y=1585&top_left_x=296)

A regular hexagon with side length \(a=a_{0}\) has an area of \(F=\frac{3}{2} \sqrt{3} a^{2}\). Therefore, the first hexagonal prism, with a side length of \(a=10 \mathrm{~cm}\) and a height of \(h=0.1 \mathrm{~mm}\), has a volume of

\[
V_{0}=\frac{3}{2} \sqrt{3} a^{2} \cdot h=\frac{3}{2} \sqrt{3} \mathrm{~cm}^{3}
\]

The first attached hexagon has a side length \(a_{1}=\frac{1}{2} \sqrt{3} \cdot 10 \mathrm{~cm}\) (height in the equilateral triangle with side length \(a\)). The volume of the prism becomes

\[
V_{1}=V_{0} \cdot\left(\frac{1}{2} \sqrt{3}\right)^{2}=\frac{3}{4} V_{0}
\]

Each subsequent attached prism has \(\frac{3}{4}\) of the volume of the previous prism. For \(n\) prisms attached to the bottom prism, the total volume is given by the partial sum of a geometric series:

\[
V=\sum_{i=0}^{n}\left(\frac{3}{4}\right)^{i} V_{0}=\sum_{i=0}^{n}\left\{\left(\frac{3}{4}\right)^{i} \frac{3}{2} \sqrt{3} \mathrm{~cm}^{3}\right\}=6 \sqrt{3}-\frac{9}{2} \sqrt{3}\left(\frac{3}{4}\right)^{n} \mathrm{~cm}^{3}
\]

a) The topmost prism should still have a side length of at least \(2 \mathrm{~mm}\). For the \(n\)-th prism, the side length is

\[
a_{n}=\left(\frac{1}{2} \sqrt{3}\right)^{n} \cdot 10 \mathrm{~cm}
\]
and thus

\[
\begin{aligned}
a_{n}=\left(\frac{1}{2} \sqrt{3}\right)^{n} \cdot 10 \mathrm{~cm}>0.2 \mathrm{~cm} \\
n \lg \left(\frac{1}{2} \sqrt{3}\right)>\lg 0.02 \quad n<27.1968 \ldots
\end{aligned}
\]

27 more hexagons can be attached to the base hexagon. The body will thus be 280.1 \(\mathrm{mm}=2.8 \mathrm{~cm}\) high.

b) Substituting \(n=27\) into (1) yields a volume of the body of \(\approx 10.389 \mathrm{~cm}^{3}\).

c) For the limit \(n \rightarrow \infty\), (1) becomes

\[
\lim _{n \rightarrow \infty} 6 \sqrt{3}-\frac{9}{2} \sqrt{3}\left(\frac{3}{4}\right)^{n}=6 \sqrt{3}
\]

If there were no limit to the cutting, the hexagonal tower would be infinitely high and have a volume of \(V=6 \sqrt{3} \approx 10.3923 \mathrm{~cm}^{3}\).