[ Rhombuses. Properties and Characteristics ] [ Two tangents drawn from one point ]

A rhombus and a triangle are circumscribed around a circle with a radius of 1, two sides of which are parallel to the diagonals of the rhombus, and the third side is parallel to one of the sides of the rhombus and is equal to 5. Find the side of the rhombus.
Let the side $KM$ of the triangle $KLM$ be parallel to the diagonal $BD$ of the rhombus $ABCD$ and touch the given circle at point $P$, the side $KL$ of the triangle be parallel to the diagonal $AC$ of the rhombus and touch the circle at point $Q$, and the side $LM$ of the triangle be parallel to the side $AB$ of the rhombus and touch the circle at point $T$. Since the sides $KM$ and $KL$ are parallel to the diagonals of the rhombus, and the diagonals of the rhombus are perpendicular, the triangle $KLM$ is a right triangle, with $\angle LKM = 90^\circ$. If $O$ is the center of the circle, then the quadrilateral $KPOQ$ is a square with a side equal to the radius of the circle, i.e., 1. Let $MP = MD = x$. Then

$$
MT = MP = x, LQ = LT = ML - MT = 5 - x, KL = KQ + LQ = 1 + (5 - x) = 6 - x.
$$

By the Pythagorean theorem, $ML^2 = KM^2 + KL^2$, or

$$
25 = (1 + x)^2 + (6 - x)^2, x^2 - 5x + 6 = 0.
$$

Assume that $KM < KL$. Then we take the smaller root of the obtained equation: $x = 2$. Thus, $KM = 1 + x = 1 + 2 = 3, KL = 6 - x = 6 - 2 = 4$. Let $\angle KLM = \alpha$. Then

$$
\sin \alpha = \frac{KM}{ML} = \frac{3}{5}, \cos \alpha = \frac{4}{5}, \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}.
$$

Let $H$ be the projection of the vertex $B$ of the rhombus onto its side $AD$. Then $BH$ is the height of the rhombus, so $BH$ is twice the radius of the circle inscribed in the rhombus, i.e., $BH = 2$. From the right triangle $AHB$, we find that

$$
AB = \frac{BH}{\sin \angle BAD} = \frac{BH}{\sin 2 \angle BAC} = \frac{BC}{\sin 2 \angle KLM} = \frac{2}{\sin 2 \alpha} = \frac{2}{\frac{24}{25}} = \frac{25}{12}.
$$

## Answer

$\frac{25}{12}$.