3.2. A quadrilateral $ABCD$ is inscribed in a circle $\omega$ such that $AB = AD$ and $BC = CD$ (a deltoid). It turns out that the radius of the inscribed circle of triangle $ABC$ is equal to the radius of the circle that touches the smaller arc $BC$ of circle $\omega$ and the side $BC$ at its midpoint (see Fig. 3). Find the ratio of the side $AB$ to the radius of the inscribed circle of triangle $ABC$.
3.2. Let $I$ be the center of the inscribed circle of triangle $ABC$, $P$ the point of tangency of this inscribed circle with side $AB$, $M$ the midpoint of side $AB$, and $Q$ the midpoint of arc $AB$ (see Fig. 5).

Note that triangle $IPB$ is isosceles and right-angled. Moreover, by the trident theorem, triangle $IQB$ is also isosceles. Therefore, $\angle QPB = \angle QPI = 135^{\circ}$, and thus $\angle MPQ = 45^{\circ}$. It is easy to see that $\angle QMP = 90^{\circ}$, so triangle $QMP$ is isosceles. Therefore, $MP = MQ$.

Thus, the ratio of side $AB$ to the radius is $2 \cdot (MP + PB) / PI = 6$.