XXXIII OM - I - Problem 1

A regular 25-gon $ A_1, A_2, \ldots, A_{25} $ is inscribed in a circle with center $ O $ and radius of length $ r $. What is the maximum length of the vector that is the sum of some of the vectors $ \overrightarrow{OA_1}, \overrightarrow{OA_2}, \ldots, \overrightarrow{OA_{25}} $?
Consider a certain subset $X$ of the set of data vectors, for which the vector $\overrightarrow{a}$ will be equal to the sum of all vectors in the subset $X$. If the vector $\overrightarrow{OA_i}$ does not belong to $X$ and forms an acute angle with the vector $\overrightarrow{a}$, then the sum of all vectors in the set $X$ and the vector $\overrightarrow{OA_i}$ has a length greater than $a$. Assuming that $\overrightarrow{OB} = \overrightarrow{a}$, then $\overrightarrow{OC} = \overrightarrow{a} + \overrightarrow{OA_i}$ and $\overrightarrow{OC}^2 = \overrightarrow{OA_i}^2 + \overrightarrow{A_iC}^2 - 2OA_i A_i C \cos| \measuredangle OA_iC| > OA_i^2 + A_iC^2$ (since the angle $OA_iC$ is obtuse). Therefore, $OC_i^2 > a^2$, which means $OC > a$.
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On the other hand, since the sum of all data vectors $\overrightarrow{OA_1} + \ldots + \overrightarrow{OA_{25}}$ is the zero vector, the sum of vectors of any fixed subset $\{\overrightarrow{OA_{i_2}}, \ldots, \overrightarrow{OA_{i_k}}\}$ is the vector opposite to the sum of the remaining $25-k$ vectors, and thus both these sums are vectors of equal length. It follows that the maximum sum length can be obtained by adding no more than $12$ vectors. We will show that the maximum sum is obtained by adding $12$ consecutive vectors. Let the vector $\overrightarrow{a} = \overrightarrow{OB}$ denote the sum of vectors of the chosen subset $X$. Without loss of generality, we can assume that the point $B$ lies in the angle $A_6OA_7$. If, therefore, any of the vectors $\overrightarrow{OA_i}$ for $i = 1,2,\ldots,12$ did not belong to $X$, then by the initial remark, adding this vector to $X$ would yield a subset of vectors whose sum would be longer than $a$. Ultimately, then, the maximum length is the sum of $12$ consecutive vectors (the sum of $13$ consecutive vectors has the same length). It will be more convenient to calculate the length of the sum of $13$ consecutive vectors. Since consecutive vectors form an angle of measure $\frac{2\pi}{25}$, the sum of the lengths of such $13$ vectors $\{\overrightarrow{OA_1}, \ldots, \overrightarrow{OA_{13}}\}$ is equal to the sum of the lengths of the projections of these vectors onto the ray $OA_7$. It is

since the vectors $\overrightarrow{OA_1}$ and $\overrightarrow{OA_{13}}$ form an angle of measure $6 \cdot \frac{2\pi}{25}$ with $\overrightarrow{OA_7}$, the vectors $\overrightarrow{OA_1}$ and $\overrightarrow{OA_{13}}$ form angles of measure $5 \cdot \frac{2\pi}{25}, \ldots,$ the vectors $\overrightarrow{OA_6}$ and $\overrightarrow{OA_{8}}$ form angles of measure $\frac{2\pi}{25}$ with $\overrightarrow{OA_7}$.
To calculate the value (*), we will use complex numbers. The number

is the real part of the complex number

Taking $z = \cos \frac{2\pi}{25} + i \sin \frac{2\pi}{25}$, by de Moivre's formula,

It follows that

thus

The real part of this number that we are interested in is

We have therefore calculated that

Therefore,

Ultimately, the maximum length of the vector being the sum of some of the vectors $\overrightarrow{OA_1}, \ldots, \overrightarrow{OA_{25}}$ is