8. a) How to choose some number of vectors from $\overline{O A}_{1}, \overline{O A}_{2}, \overline{O A}_{3}, \ldots, \overline{O A}_{25}$, where $O$ is the center of a regular 25-gon $A_{1} A_{2} A_{3} \ldots A_{25}$, so that the length of the sum of these vectors is as large as possible?

b) * What is this length if all the considered vectors are unit vectors?
8. a) Let $\overline{O S}$ be the desired sum, $X Y$ be a line passing through point $O$ perpendicular to line $O S$. If the system of selected vectors contains a vector $\overline{O A}_{i}$ (where $i=1,2,3, \ldots$, or 25), located on the line $X Y$ or on the other side of the line $X Y$ compared to vector $\overline{O S}$, then, by excluding it from the system, we subtract vector $\overline{O A_{i}}$ from the sum $\overline{O S}$, i.e., we replace vector $\overline{O S}$ with vector $\overline{O S}=\overline{O S}-\overline{O A_{i}}=\overline{A_{i} S}$; but $A_{i} S > O S$, because triangle $O A_{i} S$ is either a right triangle or an obtuse triangle: $\angle A_{i} O S \geqslant 90^{\circ}$ (Fig. $38, a$), and the length of the sum of the selected vectors increases. On the other hand, if the system of selected vectors does not contain some vector $\overline{O A}_{j}$, located on the same side of the line $X Y$ as vector $\overline{O S}$, then by adding $\overline{O A_{j}}$ to this system, we replace vector $\overrightarrow{O S}$ with vector $\overline{O S^{\prime \prime}}=\overline{O S}+\overline{O A_{j}}$, where $S S^{\prime \prime} \# O A_{j}$; but in this case $O S^{\prime \prime} > O S$, because triangle $O S S^{\prime \prime}$ is obtuse: $\angle O S S^{\prime \prime} > 90^{\circ}$ (Fig. $38, \sigma$), i.e., the length of the sum of the selected vectors increases again.

Thus, the selected system of vectors must contain all vectors lying on one side of the line $X Y$, and only these vectors. From this, it follows that this system can contain only 12 or 13 adjacent vectors, since on one side of any line $X Y$ passing through the common origin $O$ of all vectors, there are 12 vectors of the system, and on the other side - 13 vectors.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-071.jpg?height=435&width=762&top_left_y=428&top_left_x=263)

Fig. 38.

[If the line $X Y$ coincides with one of the given vectors, then on either side of it there are 12 vectors.] But it is not difficult to show that the sum of all 25 vectors is the zero vector (see the text in square brackets at the end of the solution; compare also the solution of problem b)); therefore, our system of vectors contains either 12 or 13 adjacent vectors, and in both cases the sum of all selected vectors has the same length: since

$$
\begin{aligned}
& \overline{O A}_{1}+\overline{O A}_{2}+\ldots+\overline{O A}_{25}=\overline{0}, \text { then } \overline{O A}_{1}+\overline{O A}_{2}+\ldots+\overline{O A}_{12}= \\
&=-\left(\overline{O A}_{13}+\overline{O A}_{14}+\ldots+\overline{O A}_{25}\right)
\end{aligned}
$$

and, therefore,

$$
\left|\overline{O A}_{1}+\overline{O A}_{2}+\ldots+\overline{O A}_{12}\right|=\left|\overline{O A}_{13}+\overline{O A}_{14}+\ldots+\overline{O A}_{25}\right|
$$

[Let us now prove that the sum of all 25 vectors is the zero vector. Indeed, a rotation by an angle $\varphi=\frac{360^{\circ}}{25}=14^{\circ} 24^{\prime}$ translates vector $\overline{O A}_{1}$ into vector $\overline{O A_{2}}$, vector $\overline{O A_{2}}$ into $\overline{O A_{3}}$, vector $\overline{O A}_{3}$ into $\overline{O A}_{4}, \ldots$, vector $\overline{O A}_{25}$ into $\overline{O A}_{1}$, and therefore, the entire sum $\overline{O A}_{1}+\overline{O A}_{2}+$ $+\overline{O A}_{3}+\ldots+\overline{O A_{25}}$ is translated into itself by this rotation (since each term of this sum is translated into another term). But only the zero vector can translate into itself under a rotation by an angle not a multiple of $360^{\circ}$! $]$

b) We will consider our plane as the plane of a complex variable, aligning the origin 0 of the complex numbers with point $O$, and point $A_{1}$ with the point corresponding to the number 1 (Fig. 39).

In this case, point $A_{2}$ will correspond to some complex number $z=\cos \varphi+i \sin \varphi$, where $\varphi=\frac{360^{\circ}}{25}=14^{\circ} 24^{\prime}$, and points $A_{3}, A_{4}, \ldots, A_{12}$ will correspond to complex numbers $z^{2}, z^{3}, \ldots, z^{11}$. Thus, point $S$ - the end of the vector $\overline{O S}=\overline{O A}_{1}+\overline{O A}_{2}+\ldots+\overline{O A}_{12}$ - corresponds to the complex number

$$
\begin{gathered}
Z=1+z+z^{2}+z^{3}+\ldots+z^{11}=\frac{1-z^{12}}{1-z}= \\
=\frac{1-\cos (12 \varphi)-i \sin (12 \varphi)}{1-\cos \varphi-i \sin \varphi}
\end{gathered}
$$

The modulus $|Z|$ of this number $Z$ - the length of

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-072.jpg?height=284&width=414&top_left_y=257&top_left_x=693)

Fig. 39. the vector $O S$ - is equal to

$$
\begin{aligned}
& \frac{\sqrt{[1-\cos (12 \varphi)]^{2}+\sin ^{2}(12 \varphi)}}{\sqrt{(1-\cos \varphi)^{2}+\sin ^{2} \varphi}}=\frac{\sqrt{2-2 \cos (12 \varphi)}}{\sqrt{2-2 \cos \varphi}}= \\
& =\frac{\sqrt{4 \sin ^{2} 6 \varphi}}{\sqrt{4 \sin ^{2} \frac{\varphi}{2}}}=\frac{\sin 6 \varphi}{\sin \frac{\varphi}{2}} \approx 7,97 .
\end{aligned}
$$