Problem 1350

I. Solution. If the sides of a closed polygon are projected onto a straight line in the same direction of rotation, the algebraic sum of the projections is zero.

Apply this theorem to the regular heptagon by projecting all sides onto the line $X^{\prime} X$ that contains one of its sides, moving in the positive direction of rotation.

!

Each angle of the regular heptagon is $\pi - \frac{2 \pi}{7}$.

The angle between the first side and the line $X^{\prime} X$ is 0; its projection is $a \cos 0$. The angle between the next side and $X^{\prime} X$ is $\frac{2 \pi}{7}$; its projection is $a \cos \frac{2 \pi}{7}$. Each subsequent side turns by an angle of $\frac{2 \pi}{7}$ from the previous one, and the angle with $X^{\prime} X$ increases by $\frac{2 \pi}{7}$ from side to side. Therefore, the sum of the projections is:

$$x = a \left( \cos 0 + \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{10 \pi}{7} + \cos \frac{12 \pi}{7} \right) = 0 \ldots$$

However,

$$\cos \frac{12 \pi}{7} = \cos \left(2 \pi - \frac{12 \pi}{7}\right) = \cos \frac{2 \pi}{7}$$

Similarly,

$$\cos \frac{10 \pi}{7} = \cos \frac{4 \pi}{7} \quad \text{and} \quad \cos \frac{8 \pi}{7} = \cos \frac{6 \pi}{7}$$

Thus, the sum inside the parentheses is:

$$\begin{gathered} 1 + 2 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) = 0 \\ \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} = -\frac{1}{2} \end{gathered}$$

II. Solution. In the previous solution, we showed that

$$\sum_{k=0}^{6} \cos \frac{2 k \pi}{7} = \cos 0 + \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{10 \pi}{7} + \cos \frac{12 \pi}{7} = 0$$

If we consider that the roots of the equation

$$x^{7} - 1 = 0$$

are

$$\cos \frac{2 k \pi}{7} + i \sin \frac{2 k \pi}{7} \quad (k=0,1,2, \ldots 6)$$

and the sum of these roots is zero, then

$$\sum_{k=0}^{6} \cos \frac{2 k \pi}{7} = 0 \quad \text{and} \quad \sum_{k=0}^{6} \sin \frac{2 k \pi}{7} = 0$$

III. Solution. Consider the more general sum

$$S = \cos \alpha + \cos 2 \alpha + \cos 3 \alpha + \ldots + \cos n \alpha$$

where the angles form an arithmetic progression; the first term of the progression is $\alpha$, and the common difference is also $\alpha$. Multiply each term of (1) by $2 \sin \frac{\alpha}{2}$. Then

$$\begin{aligned} & 2 \cos \alpha \sin \frac{\alpha}{2} = \sin \left( \alpha + \frac{\alpha}{2} \right) - \sin \left( \alpha - \frac{\alpha}{2} \right) \\ & 2 \cos 2 \alpha \sin \frac{\alpha}{2} = \sin \left( 2 \alpha + \frac{\alpha}{2} \right) - \sin \left( 2 \alpha - \frac{\alpha}{2} \right) \\ & 2 \cos 3 \alpha \sin \frac{\alpha}{2} = \sin \left( 3 \alpha + \frac{\alpha}{2} \right) - \sin \left( 3 \alpha - \frac{\alpha}{2} \right) \\ & \ldots \ldots \cdot \\ & 2 \cos n \alpha \sin \frac{\alpha}{2} = \sin \left( n \alpha + \frac{\alpha}{2} \right) - \sin \left( n \alpha - \frac{\alpha}{2} \right) \end{aligned}$$

Since $\alpha + \frac{\alpha}{2} = 2 \alpha - \frac{\alpha}{2}, \quad 2 \alpha + \frac{\alpha}{2} = 3 \alpha - \frac{\alpha}{2} \quad$ and so on, the terms on the right side cancel out during summation, except for two, so

$$2 S \sin \frac{\alpha}{2} = \sin \left( n \alpha + \frac{\alpha}{2} \right) - \sin \left( \alpha - \frac{\alpha}{2} \right) = 2 \cos \frac{(n+1) \alpha}{2} \sin \frac{n \alpha}{2}$$

In the given case, $n=3, \alpha=\frac{2 \pi}{7}$, so

$$2 S \sin \frac{\pi}{7} = 2 \cos \frac{4 \pi}{7} \sin \frac{3 \pi}{7} = -2 \cos \frac{3 \pi}{7} \sin \frac{3 \pi}{7} = -\sin \frac{6 \pi}{7}$$

However,

$$\sin \frac{6 \pi}{7} = \sin \left( \pi - \frac{6 \pi}{7} \right) = \sin \frac{\pi}{7}$$

and thus

$$S = -\frac{1}{2}$$[^0]

[^0]: ${ }^{1}$ For the calculation of such sums with any common difference, see our 4th year, page 197 (Goldziher: Goniometric Polynomials).