What angle must $p_{1}$ and $p_{2}$ forces enclose so that their resultant is equal to their arithmetic mean? What limits must the $\frac{p_{1}}{p_{2}}$ (or $\cdot \frac{p_{2}}{p_{1}}$) ratios fall within for the problem to have a solution? What limits must the angle between the two forces fall within?
If the forces $p_{1}$ and $p_{2}$ enclose an angle $\alpha$, their resultant is: $\left(p_{1}^{2}+p_{2}^{2}+2 p_{1} p_{2} \cos \alpha\right)^{\frac{1}{2}}$. The problem's requirement is:

$$
\left(p_{1}^{2}+p_{2}^{2}+2 p_{1} p_{2} \cos \alpha\right)^{\frac{1}{2}}=\frac{1}{2}\left(p_{1}+p_{2}\right) \ldots
$$

From this,

$$
\cos \alpha=\frac{1}{8}\left[2-3\left(\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}\right)\right] \ldots
$$

Since $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}} \geq 2$, it follows that $\cos \alpha<0$, thus $90^{\circ}<\alpha<180^{\circ}$. 1

Furthermore,

$$
-1 \leq \cos \alpha \leq-\frac{1}{2} \cdots
$$

If we substitute the smallest value 2 for $\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}$, we get the largest value of $\cos \alpha$. According to (3),

$$
180^{\circ} \geq \alpha \geq 120^{\circ}
$$

From the fact that $\cos \alpha \geq-1$, it follows that

$$
\frac{1}{8}\left[2-3\left(\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}\right)\right] \geq-1, \text { i.e., } \frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}} \leq \frac{10}{3}
$$

Let

$$
\frac{p_{1}}{p_{2}}=x
$$

The

$$
x+\frac{1}{x} \leq \frac{10}{3}, \text { i.e., } x^{2}-\frac{10}{3} x+1 \leq 0
$$

The solution to the inequality is: $\frac{1}{3} \leq x \leq 3$, thus

$$
\frac{1}{3} \leq \frac{p_{1}}{p_{2}} \leq 3{ }^{2}
$$

Let's examine the boundary conditions.

$$
\frac{p_{1}}{p_{2}}+\frac{p_{2}}{p_{1}}=2, \text { if } p_{1}^{2}+p_{2}^{2}-2 p_{1} p_{2}=0, \text { i.e., } p_{1}=p_{2}=p
$$

In this case, $\alpha=120^{\circ}$ and the resultant is: $\left(p^{2}+p^{2}-p^{2}\right)^{\frac{1}{2}}=p$. Indeed:

$$
p=\frac{1}{2}\left(p_{1}+p_{2}\right)
$$

For

$$
\begin{gathered}
\frac{p_{1}}{p_{2}}=3, \text { i.e., } p_{1}=3 p_{2}, \text { then } \\
\cos \alpha=\frac{1}{8}\left(2-3 \cdot \frac{10}{3}\right)=-1, \alpha=180^{\circ}
\end{gathered}
$$

The resultant is $\left(p_{1}^{2}+p_{2}^{2}-2 p_{1} p_{2}\right)^{\frac{1}{2}}=\left|p_{1}-p_{2}\right|=2 p_{2}=\frac{1}{2}\left(3 p_{2}+p_{2}\right)$.

Miklós Hajnal (izr. g. VIII. o. Bp.)[^0]


[^0]:    ${ }^{1}$ We can also say that $90^{\circ}<\alpha<270^{\circ}$. However, if $180^{\circ}<\alpha<270^{\circ}$, then essentially the same situations arise as when $90^{\circ}<\alpha<180^{\circ}$, just after a reverse rotation.

    ${ }^{2}$ At the same time, $\frac{1}{3} \leq \frac{p_{2}}{p_{1}} \leq 3$