\section*{Problem 4 - 111034}

A straight circular cone with radius \(R=6\) and height \(h\) is drilled through cylindrically such that the axis of the cone coincides with that of the drill hole.

How large must the radius \(r\) ( \(R, h, r\) measured in cm) of the drill hole be chosen so that the volume of the remaining body is half as large as that of the cone?
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The drilled partial body can be decomposed into a cylinder with radius \(r\) and height \(h_{T}:=\frac{R-r}{R} \cdot h\) as well as a cone with radius \(r\) and height \(h-h_{T}=\frac{r}{R} \cdot h\), since the drill hole (viewed from the base) pierces the mantle surface of the given conical body at a height of \(h_{T}\).

For the volume \(V\) of the original cone, we get \(V=\frac{1}{3} \cdot \pi R^{2} h\) and for the partial body

\[
V_{T}=\pi r^{2} \cdot h_{t}+\frac{1}{3} \cdot \pi r^{2} \cdot\left(h-h_{T}\right)=\pi r^{2} \cdot \frac{1}{3} \cdot\left(3 \frac{R-r}{R}+\frac{r}{R}\right) \cdot h
\]

From \(V=2 V_{T}\), we get the equation

\[
R^{2}=2 r^{2} \cdot \frac{3 R-2 r}{R} \quad \text { or } \quad 6 r^{2} R-4 r^{3}-R^{3}=0
\]

With \(R=6\), we get \(4 r^{3}-36 r^{2}+216=0\) or \(r^{3}-9 r^{2}+54=0\). It is \(r^{3}-9 r^{2}+54=(r-3) \cdot\left(r^{2}-6 r+18\right)\), which has \(r=3\) as its only root. Therefore, the radius \(r\) of the drill hole must be \(3 \mathrm{~cm}\).

\section*{Alternative Solution:}

Obviously, the larger the radius \(r\) of the drill hole, the larger the volume \(V_{T}\) of the drilled partial body. If \(r=0\), then \(V_{T}=0\) and for \(r=R\), \(V_{T}=V\) is the entire original body. By varying \(r\) in the interval \([0 ; R]\), there is exactly one value for which \(V_{T}=\frac{1}{2} V\) holds.

We choose \(r=\frac{1}{2} R\) and decompose the drilled partial body into a straight circular cylinder and a cone. Both have a base area of \(\pi r^{2}=\frac{1}{4} \pi R^{2}\) and a height of \(\frac{1}{2} h\), so the volume of the partial body is

\[
V_{T}=\frac{1}{4} \pi R^{2} \cdot \frac{1}{2} h+\frac{1}{3} \cdot \frac{1}{4} \pi R^{2} \cdot \frac{1}{2} h=\left(\frac{1}{8}+\frac{1}{24}\right) \cdot \pi \cdot R^{2} h=\frac{1}{6} \cdot \pi \cdot R^{2} h=\frac{1}{2} V
\]

Thus, the choice of \(r=\frac{1}{2} R=3 \mathrm{~cm}\) is the only one that results in the remaining body having exactly half the volume of the original body.