What angle should two forces enclose so that their resultant is as large as their geometric mean? What is the condition for this angle to be a minimum, and what is the size of this minimum?
Without loss of generality, we can assume that the point of attack of the two forces is common. Let these two forces be $p_{1}, p_{2}$, the angle between them $\alpha$, and their resultant $p$. The problem requires that

$$
p^{2}=p_{1}^{2}+p_{2}^{2}+2 p_{1} p_{2} \cos \alpha=p_{1} p_{2}
$$

Thus,

$$
\cos \alpha=\frac{1}{2}\left[1-\left(\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}\right)\right]
$$

Since $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}>1, \quad \cos \alpha<0, \quad 90^{\circ}<\alpha<270^{\circ}$.

However, in the case of $180^{\circ}<\alpha<270^{\circ}$, the angle between the two forces can also be considered a negative rotation-generated obtuse angle, so we can say: $\alpha$ is an obtuse angle.

It is known that if $a, b$ are positive numbers, then

$$
\frac{a}{b}+\frac{b}{a} \geq 2
$$

Accordingly,

$$
\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}} \geq 2
$$

and

$$
\cos \alpha \leq-\frac{1}{2}, \quad 180^{\circ} \geq \alpha \geq 120^{\circ}
$$

Thus, the maximum value of $\cos \alpha$ is $-\frac{1}{2}$, which occurs when $p_{1}=p_{2}$ and in this case $\alpha=120^{\circ}$ is the smallest value of $\alpha$.

Since $\cos \alpha \geq-1$ must hold, therefore

$$
\begin{gathered}
\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}} \leq 3, \text { or } p_{1}^{2}+p_{2}^{2} \leq 3 p_{1} p_{2} \\
\quad \text { or }\left(\frac{p_{1}}{p_{2}}\right)^{2}-3\left(\frac{p_{1}}{p_{2}}\right)+1 \leq 0
\end{gathered}
$$

From this:

$$
\frac{3-\sqrt{5}}{2} \leq \frac{p_{1}}{p_{2}} \leq \frac{3+\sqrt{5}}{2}
$$

This is the condition for the resultant of two forces to be equal to their geometric mean.

George Egri (Ferenc Kölcsey g. VIII. o. Bp.)

Note. The requirement $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}} \leq 3$

$$
p_{1}^{2}-2 p_{1} p_{2}+p_{2}^{2} \leq p_{1} p_{2}, \quad \text { or } \quad\left|p_{1}-p_{2}\right| \leq \sqrt{p_{1} p_{2}}
$$

can also be written. $\left|p_{1}-p_{2}\right|$ is the resultant of the two forces, regardless of direction, if their angle is $180^{\circ}$; thus, in this case, $\left|p_{1}-p_{2}\right|$ must be equal to the geometric mean of the two forces.