Given are the following proportional pairs:

$$
\frac{a}{b}=\frac{c}{d} \text { and } \frac{a^{\prime}}{b^{\prime}}=\frac{c^{\prime}}{d^{\prime}}
$$

What is the condition for the numbers $a+a^{\prime}, b+b^{\prime}, c+c^{\prime}$, and $d+d^{\prime}$ to also form a proportional pair?
According to the condition,

$$
\frac{a}{b}=\frac{c}{d}=k \text { and } \frac{a^{\prime}}{b^{\prime}}=\frac{c^{\prime}}{d^{\prime}}=k^{\prime}
$$

thus

$$
a=b k, c=d k, a^{\prime}=b^{\prime} k^{\prime}, c^{\prime}=d^{\prime} k^{\prime}
$$

If the numbers $a+a^{\prime}, b+b^{\prime}, c+c^{\prime}$ and $d+d^{\prime}$ form a geometric proportion, then

$$
\left(a+a^{\prime}\right)\left(d+d^{\prime}\right)=\left(b+b^{\prime}\right)\left(c+c^{\prime}\right)
$$

or

$$
a d+a d^{\prime}+a^{\prime} d+a^{\prime} d^{\prime}=b c+b c^{\prime}+b^{\prime} c+b^{\prime} c^{\prime}
$$

Substituting the values of $a, a^{\prime}, c, c^{\prime}$ from (1), we get:

$$
b d k+b d^{\prime} k+b^{\prime} d k^{\prime}+b^{\prime} d^{\prime} k^{\prime}=b d k+b d^{\prime} k^{\prime}+b^{\prime} d k+b^{\prime} d^{\prime} k^{\prime}
$$

or

$$
k\left(b d^{\prime}-b^{\prime} d\right) 7=k^{\prime}\left(b d^{\prime}-b^{\prime} d\right)
$$

or

$$
\left(b d^{\prime}-b^{\prime} d\right)\left(k-k^{\prime}\right)=0
$$

The left side is 0 if

$$
b d^{\prime}-b^{\prime} d=0, \quad \text { or } \quad k-k^{\prime}=0
$$

from which the required condition is:

$$
\frac{b}{b^{\prime}}=\frac{d}{d^{\prime}}, \frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}, \frac{c}{c^{\prime}}=\frac{d}{d^{\prime}}
$$

or

$$
\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=\frac{d}{d^{\prime}}
$$

(Hermann Szóllös, Esztergom.)