A regular pentagon with side $a$ has a square inscribed in it such that two sides of the square are parallel to one side of the pentagon. Calculate the length $x$ of the side of the square.

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In a pentagon, one side of the inscribed square is parallel to a side of the pentagon, cutting out an isosceles triangle from the pentagon. Since the angle of a regular pentagon is 108 degrees, the two equal angles of the isosceles triangle are 36 degrees. Thus, the angles of the triangle \( \triangle AFE \) are 108, 54, and 18 degrees.

![](https://cdn.mathpix.com/cropped/2024_05_02_6aa6f22e4fa1372698c9g-1.jpg?height=558&width=525&top_left_y=309&top_left_x=800)

From the two mentioned triangles, using the sine rule:

$$
\frac{AF}{x}=\frac{\sin 18^{\circ}}{\sin 108^{\circ}}, \quad \frac{FB}{x}=\frac{\sin 36^{\circ}}{\sin 108^{\circ}}
$$

Adding the two equations (noting that \( AF + FB \) is exactly the length of one side of the pentagon):

$$
\frac{a}{x}=\frac{\sin 18^{\circ}+\sin 36^{\circ}}{\sin 108^{\circ}}
$$

From this, the desired length \( x \) is:

$$
x=\frac{a \sin 108^{\circ}}{\sin 18^{\circ}+\sin 36^{\circ}}=a \cdot \frac{\cos 18^{\circ}}{\sin 18^{\circ}+2 \sin 18^{\circ} \cos 18^{\circ}}=a \cdot \frac{1}{\operatorname{tg} 18^{\circ}+2 \sin 18^{\circ}}
$$

The approximate values of the trigonometric functions can be found in tables, and thus we get the following approximate value for \( x \):

$$
x \approx 1.06 a
$$

Károly Füle (Bp. V., Apáczai Csere g. III. o. t.)