3. In triangle $ABC$, whose area is 20, the median $CD$ is drawn. Find the radius of the circumcircle of triangle $ABC$, given that $AC=\sqrt{41}$, and the center of the circle inscribed in triangle $ACD$ lies on the circumcircle of triangle $BCD$.

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Answer: $\frac{41}{10}$ or $\frac{41}{8}$.

Solution. Let $Q$ be the center of the circle inscribed in triangle $ACD$. Then the segments $AQ$ and $CQ$ are the bisectors of angles $BAC$ and $ACD$ respectively, by the properties of inscribed angles $\angle DBQ = \angle DCQ$. Therefore, triangles $ABQ$ and $ACQ$ are equal by side and two angles. Hence, $AB = AC$, i.e., triangle $ABC$ is isosceles.

Let $BC = 2x$, then $S_{ABC} = x \sqrt{41 - x^2}$. Therefore, taking into account the condition, we get the equation $x \sqrt{41 - x^2} = 20$, which has two roots $x = 4$ or $x = 5$. The radius of the circumscribed circle around triangle $ABC$ is

$$
R = \frac{AB \cdot BC \cdot AC}{4S} = \frac{41 \cdot 2x}{4 \cdot 20} = \frac{41x}{40}
$$

Substituting the found values of $x$ here, we get two possible answers, and both possibilities are realized.

![](https://cdn.mathpix.com/cropped/2024_05_06_84c54589ca5e85eb41a6g-1.jpg?height=331&width=465&top_left_y=2353&top_left_x=1458)

Answer to variant 2: 10 or $\frac{8}{5}$. Answer to variant 3: $\frac{17}{5}$ or $\frac{17}{3}$. Answer to variant 4: $\frac{15}{4}$ or $\frac{27}{20}$.