Add an integer-coefficient polynomial whose zero is $\cos 18^{\circ}$!
I. solution. Calculate the value of $\cos 18^{\circ}$: consider the isosceles triangle $ABC$ where the legs are of length 1 and the included angle is $36^{\circ}$.

The angle bisector from $B$ intersects $AC$ at $D$. Then the triangles $ABC$ and $DAB$ are similar, so $CB: DB = AB: DA$, i.e., $\frac{1}{a} = \frac{a}{1-a}$.

From this, $a = \frac{\sqrt{5}-1}{2}$. The altitude from $C$ bisects the $36^{\circ}$ angle and the segment $AB$, so $\sin 18^{\circ} = \frac{a}{2} = \frac{\sqrt{5}-1}{4}$, $\cos 18^{\circ} = \sqrt{1 - \left(\frac{\sqrt{5}-1}{4}\right)^2} = \frac{\sqrt{10+2\sqrt{5}}}{4}$.

Let $\frac{\sqrt{10+2\sqrt{5}}}{4} = x$, where $x > 0$. Then $16x^2 = 10 + 2\sqrt{5}$, or $\sqrt{5} = 8x^2 - 5$. Squaring this, we get: $64x^4 - 80x^2 + 20 = 0$. Dividing by 4, the polynomial that meets the problem's requirement is $16x^4 - 20x^2 + 5$.

II. solution. Let $\cos 18^{\circ} = x (0 < x < 1)$, then $\sin 18^{\circ} = \sqrt{1 - x^2}$. Using these:

$$
\sin 36^{\circ} = 2x\sqrt{1 - x^2}, \quad \cos 36^{\circ} = x^2 - (1 - x^2) = 2x^2 - 1
$$

$\sin 72^{\circ} = 4x\sqrt{1 - x^2}(2x^2 - 1)$. Since $\cos 18^{\circ} = \sin 72^{\circ}$, we have $4x\sqrt{1 - x^2}(2x^2 - 1) = x$. We can divide by $x$ since it is not 0. Thus, the polynomial that meets the problem's requirement is $16(1 - x^2)(2x^2 - 1)^2 - 1 = -64x^6 + 128x^4 - 80x^2 + 15$.

Attila Dénes (Békéscsaba, Rózsa F. Gymnasium, 12th grade)

Remarks. 1. The polynomial $-64x^6 + 128x^4 - 80x^2 + 15$ obtained in the II. solution can be written in product form as: $(3 - 4x^2)(16x^4 - 20x^2 + 5)$. The second factor is identical to the polynomial given in the I. solution.

2. Balázs Nyul (Debrecen, Fazekas M. Gymnasium, 9th grade) started from the fact that $5 \cdot 18^{\circ} = 90^{\circ}$, so the equation $\cos 5\alpha = 0$ should be written in terms of powers of $\cos \alpha$, where $\alpha = 18^{\circ}$.

$\cos 5\alpha$ can be expressed using the addition theorems of trigonometric functions with some calculation:

$$
\cos 5\alpha = 64\cos^5\alpha - 80\cos^3\alpha + 20\cos\alpha
$$

Thus, $\cos 18^{\circ}$ is a root of the polynomial $64x^5 - 80x^3 + 20x$. This is incidentally the $4x$ multiple of the polynomial obtained in the I. solution.

For any integer $n$, there exists an integer-coefficient polynomial $T_n$ such that $T_n(\cos \alpha) = \cos n\alpha$. The polynomial $T_n$ is called the $n$-th Chebyshev polynomial.

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