A circle can be inscribed in a symmetric trapezoid, and the center of this circle bisects the radius of the circumscribed circle that passes through it, with $R=1$. What is the radius of the inscribed circle?
In the trapezoid $ABCD$, $AB \| CD, AB=2a \geq 2b=CD$, the center of the circumscribed circle is $O$, and the center of the inscribed circle is $K$. Since there is an inscribed circle, $AD + BC = 2AD = 2(a + b)$, and the diameter of the inscribed circle is the height of the trapezoid:

$$
2 \varrho = \sqrt{(a+b)^{2} - (a-b)^{2}} = 2 \sqrt{ab}, \quad \text{or} \quad \varrho^{2} = ab
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_cde052ccccbc7a85aa2ag-1.jpg?height=935&width=1412&top_left_y=394&top_left_x=346)

On the other hand, based on $OK = \frac{1}{2}$, the distances from the two bases to $O$ are $\frac{1}{2} \mp \varrho$, thus

$$
a^{2} = R^{2} - \left(\frac{R}{2} - \varrho\right)^{2} = \frac{3}{4} - \varrho^{2} + \varrho, \quad b^{2} = \frac{3}{4} - \varrho^{2} - \varrho
$$

Substituting these,

$$
\varrho^{4} = a^{2} b^{2} = \left(\frac{3}{4} - \varrho^{2}\right)^{2} - \varrho^{2}, \quad \varrho = \sqrt{\frac{9}{40}} = 0.4743 \text{ units. }
$$

Remarks. 1. Our problem, based on the required axial symmetry, is a special case of a bicentric quadrilateral, whose Euler-Fuss relation we published in the January issue in the 3rd remark of problem 2161: $2 \varrho^{2}(R^{2} + d^{2}) = (R^{2} - d^{2})^{2}$. From this,

$$
\varrho^{2} = \frac{(R^{2} - d^{2})^{2}}{2(R^{2} + d^{2})} = \left(\frac{3}{4}\right)^{2} : \frac{10}{4} = \frac{9}{40}
$$

But let the agreement not distract us from the essence! Taking any tangent to the inner circle, and then drawing the second tangents from the points $P, Q$ on the outer circle, the chord connecting the points $T$ and $S$ "closes the tangential quadrilateral", touching the circle.

2. The relationship $\varrho^{2} = ab$ also follows from the fact that, through the angle bisectors, the triangle $ADK$ is right-angled, and its legs appear as hypotenuses in those right-angled triangles where the legs are $a$ and $\varrho$, and $b$ and $\varrho$, respectively.