Which is the geometric mean pair, where the sum of the outer terms is 24, the sum of the inner terms is 16, and the sum of the squares of all terms is 580?
If $x, y, 16-y, 24-x$ are the terms of a geometric sequence, then

$$
x: y=(16-y):(24-x), \text { or } 24 x-x^{2}=16 y-y^{2}
$$

and

$$
x^{2}+y^{2}+(16-y)^{2}+(24-x)^{2}=580
$$

or

$$
\left(16 y-y^{2}\right)+\left(24 x-x^{2}\right)=126
$$

From $1^{\circ}$ and $2^{\circ}$, we get

$$
2\left(24 x-x^{2}\right)=126
$$

or

$$
x^{2}-24 x+63=0,
$$

from which

$$
x_{1}=21, x_{2}=3
$$

Furthermore,

$$
\begin{aligned}
& 2\left(16 y-y^{2}\right)=126 \\
& y^{2}-16 y+63=0
\end{aligned}
$$

from which

$$
y_{1}=9, y_{2}=7
$$

Thus, $21: 7=9: 3$ is the sought ratio.