\section*{Problem 2 - 131212}

From a straight circular frustum, a cone is to be cut out whose apex is the center of the (larger) base of the frustum and whose base coincides with the top face of the frustum.

Determine the values of the ratio of the radius of the base to the radius of the top face of the frustum, for which the volume of the remaining body is six times as large as that of the cut-out cone.
Let \(r_{1}\) and \(r_{2}\) be the radii of the base and top of a frustum of a cone \(\left(r_{1} \geq r_{2}>0\right)\) and \(h\) the height of this body. Then the volume \(V\) of the considered body is given by

\[
V_{1}=\frac{\pi}{3} h\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right)
\]

Furthermore, the volume \(v_{2}\) of the cut-out cone is given by

\[
V_{2}=\frac{\pi}{3} h r_{2}^{2}
\]

and therefore the volume \(V\) of the remaining body is

\[
V=\frac{\pi}{3} h\left(r_{1}^{2}+r_{1} r_{2}\right)
\]

Due to (1) and (2), \(r_{1}: r_{2}\) satisfies the conditions of the problem if and only if \(6 r_{2}^{2}=r_{1}^{2}+r_{1} r_{2}\) and \(r_{1} \geq r_{2}>0\), or equivalently,

\[
\left(\frac{r_{1}}{r_{2}}\right)^{2}+\frac{r_{1}}{r_{2}}-6=0 \quad \text { and } \quad \frac{r_{1}}{r_{2}} \geq 1
\]

Thus, only

\[
\frac{r_{1}}{r_{2}}=-\frac{1}{2}+\sqrt{\frac{1}{4}+6}=2
\]

can satisfy the conditions of the problem. By substituting into (3) and reasoning backwards, it is found that for \(\frac{r_{1}}{r_{2}}=2\), indeed \(V=6 V_{2}\).

Adapted from \([5]\)