A non-zero vector $\mathbf{p}$ and its rotation by $2 \alpha$ angle, when added together, equal the vector $\mathbf{p}$ rotated by $\alpha$ angle. What is the measure of the angle $\alpha$?
I. solution. If we denote the $\alpha$ and $2\alpha$ angle rotated versions of $\mathbf{p}$ by $\mathbf{p}'$ and $\mathbf{p}''$, respectively, then according to the condition, $\mathbf{p} + \mathbf{p}'' = \mathbf{p}'$, which means

$$
\mathbf{p} - \mathbf{p}' + \mathbf{p}'' = \mathbf{0}
$$

The three vectors, $\mathbf{p}$, $-\mathbf{p}'$, and $\mathbf{p}''$, have the same magnitude and are non-zero. If their sum is $\mathbf{0}$, then connecting them in sequence forms a closed broken line consisting of three equal segments, which means that the three-vector closed chain forms an equilateral triangle.

This can happen in two ways: if the $\mathbf{p}$ vector is followed by the $-\mathbf{p}'$ vector in the chain in a positive direction (Figure 1), or if the sequence is $\mathbf{p}$, $\mathbf{p}''$, and $-\mathbf{p}'$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_92f7449dca66110a4bf2g-1.jpg?height=271&width=247&top_left_y=512&top_left_x=511)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_92f7449dca66110a4bf2g-1.jpg?height=174&width=252&top_left_y=859&top_left_x=1378)

Figure 2

In the first case, the $\mathbf{p}'$ vector is the $-60^\circ$ rotation of the $\mathbf{p}$ vector, and in the second case, due to the reverse order, the $\mathbf{p}'$ vector is the $60^\circ$ rotation of the $\mathbf{p}$ vector.

Thus, the angle $\alpha$ is either $-60^\circ + n \cdot 360^\circ$ or $60^\circ + k \cdot 360^\circ$, where $n$ and $k$ are integers.

Note. It is clear that the condition of the problem is satisfied for $\alpha$ if and only if $\cos \alpha = \frac{1}{2}$.

II. solution. If for any angle $\varphi$ and vector $\mathbf{p}$, $\varphi(\mathbf{p})$ denotes the $\varphi$-angle rotated version of the $\mathbf{p}$ vector, then according to the condition,

$$
\mathbf{p} + 2\alpha(\mathbf{p}) = \alpha(\mathbf{p})
$$

Now, let's rotate the vectors in (1) by the angle $\alpha$. Since the sum of vectors can be rotated term by term, we have

$$
\alpha(\mathbf{p} + 2\alpha(\mathbf{p})) = \alpha(\mathbf{p}) + \alpha(2\alpha(\mathbf{p})) = \alpha(\alpha(\mathbf{p}))
$$

When a rotated vector is further rotated, the resulting rotation angle is the sum of the rotation angles, so $\alpha(2\alpha(\mathbf{p})) = 3\alpha(\mathbf{p})$ and $\alpha(\alpha(\mathbf{p})) = 2\alpha(\mathbf{p})$. Thus,

$$
\alpha(\mathbf{p}) + 3\alpha(\mathbf{p}) = 2\alpha(\mathbf{p})
$$

Considering the sum of (1) and (2),

$$
\mathbf{p} + \alpha(\mathbf{p}) + 2\alpha(\mathbf{p}) + 3\alpha(\mathbf{p}) = \alpha(\mathbf{p}) + 2\alpha(\mathbf{p})
$$

which simplifies to

$$
\mathbf{p} + 3\alpha(\mathbf{p}) = \mathbf{0}
$$

If (1) holds, then the $3\alpha$-angle rotated version of the $\mathbf{p}$ vector is the negative of the $\mathbf{p}$ vector, meaning that the $3\alpha$ rotation rotates the $\mathbf{p}$ vector by $180^\circ$.

Thus, $3\alpha = 180^\circ + k \cdot 360^\circ$, from which $\alpha_1 = 60^\circ$, $\alpha_2 = 180^\circ$, and $\alpha_3 = 300^\circ$. Since we used the sum of (1) and (2), the found values are not necessarily solutions to the original problem; indeed, $\alpha_2 = 180^\circ$ is clearly not a solution, as in this case $\mathbf{p} + 360^\circ(\mathbf{p}) = 2\mathbf{p}$, which is not equal to $180^\circ(\mathbf{p}) = -\mathbf{p}$ if $\mathbf{p} \neq \mathbf{0}$.

The other two values, $\alpha_1 = 60^\circ$ and $\alpha_2 = 300^\circ$, are clearly solutions (Figure 3).

Notes. 1. In the above solution, the notation $2\alpha(\mathbf{p})$ requires some caution: it would be more correct to write $(2\alpha)\mathbf{p}$, since it is not the double of the $\alpha(\mathbf{p})$ vector but the $\mathbf{p}$ vector rotated by $2\alpha$.

2. The meaning of the problem and the algebraic nature of the solution can be easily demonstrated on the complex plane. If the $\alpha$-angle rotation is performed by multiplying by a complex number $a$ with unit magnitude and argument $\alpha$, then according to the condition, for the non-zero complex number $p$,

$$
p + a^2 p = a p
$$

Dividing by $p$, we get the quadratic equation

$$
a^2 - a + 1 = 0
$$

Multiplying by $(a + 1)$, we get the equation $a^3 + 1 = 0$. The complex roots of this equation are the negatives of the complex cube roots of unity, which are solutions to equation (3): $a_1 = \cos 60^\circ + i \sin 60^\circ$ and $a_2 = \cos(-60^\circ) + i \sin(-60^\circ)$. From this, we also get the solutions $\alpha_1 = 60^\circ$ and $\alpha_2 = -60^\circ$, the latter being the same as the $300^\circ$ rotation.

![](https://cdn.mathpix.com/cropped/2024_05_02_92f7449dca66110a4bf2g-2.jpg?height=206&width=1042&top_left_y=472&top_left_x=520)

Figure 3

4. If arbitrary rotation angles are also possible—there is no restriction on the angles in the problem statement—then infinitely many solutions arise: $\alpha_k = 60^\circ + k \cdot 360^\circ$ and $\alpha_n = -60^\circ + n \cdot 360^\circ$, where $k$ and $n$ are arbitrary integers.