We inscribe a straight circular cone in a sphere such that the center of the sphere divides the height of the cone according to the golden ratio. What is the ratio of the volumes of the two bodies to each other?

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Let the radius of the sphere be $r$, the radius of the base of the cone be $r_{1}$, and its height be $r+x$. According to the problem,

$$
x: r=r: r+x
$$

from which

$$
r^{2}-x^{2}=r x
$$

If the volume of the sphere is $K$, and that of the cone is $K_{1}$, then

$$
K: K_{1}=\frac{4}{3} r^{3} \pi: \frac{1}{3} r_{1}^{2} \pi(r+x)
$$

but

$$
r_{1}^{2}=r^{2}-x^{2}=r x
$$

so

$$
K: K_{1}=4 r^{3}: r x(r+x)
$$

but according to (1)

$$
x(r+x)=r^{2}
$$

thus

$$
K: K_{1}=4 r^{3}: r^{3}
$$

or

$$
K: K_{1}=4: 1
$$

(Béla Bayer, Losoncz.)

The problem was also solved by: Zs. Benedek, K. Burján, K. Czank, J. Demeter, N. Fekete, J. Gellért, T. Kerekes, Gy. Krisztián, A. Kürth, Gy. Lukhaub, Gy. Lupsa, G. Messik, V. Messik, I. Mikuleczky, Gy. Perl, Gy. Póka, Á. Rosenberg, G. Sasvári, J. Sasvári, K. Smodits, H. Spitzer.