Task B-2.4. Dundo and Maro were competing by giving each other unusual math problems. Thus, Maro was given the task to calculate the exact value of the cosine of an angle of $36^{\circ}$. He was not allowed to use any tools except paper and a pencil for writing and calculating.

It's impossible, Maro thought, until Dundo drew a few isosceles triangles, at least one of whose angles measured $36^{\circ}$. Maro observed one of the triangles and quickly calculated $\cos 36^{\circ}$. What value did Maro get and how did he do it?
## Solution.

We distinguish two cases - the triangle can have one or two angles of $36^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_30_38fd7d226823545bc90fg-09.jpg?height=618&width=1388&top_left_y=641&top_left_x=301)

First case. Suppose Maro chose an isosceles triangle with the angle opposite the base measuring $36^{\circ}$ and labeled its vertices as $A, B, C$. Let the angle of $36^{\circ}$ be denoted by $x$.

If we draw the angle bisector of the angle at vertex $A$ to the base, we get a triangle $\triangle A B D$ that is similar to the given triangle $\triangle A B C$ (by the AA similarity criterion).

We write $b: a=a:(b-a)$, and from this we have

$$
a^{2}=b^{2}-a b
$$

(This equality can also be obtained using the angle bisector theorem.)

From the right triangle $\triangle C D E$, we get

$$
\cos x=\frac{\frac{b}{2}}{a}, \text{ or } b=2 a \cos x
$$

Now we have

$$
a^{2}=(2 a \cos x)^{2}-a(2 a \cos x)
$$

This gives us the quadratic equation $4 \cos ^{2} x-2 \cos x-1=0$. Its positive solution is

$$
\cos x=\frac{1+\sqrt{5}}{4}
$$

Second case. Maro could have also chosen an isosceles triangle $ABC$ with two angles of $36^{\circ}$.

In this case, the angle opposite the base is divided into three angles of $36^{\circ}$ by the segments $\overline{C E}$ and $\overline{C D}$. Consider the triangle $\triangle A E C$. This is an isosceles triangle with one angle of $36^{\circ}$, so we have reduced the problem to the previous case.