10.206. What integers represent the sides of an isosceles triangle if the radius of the inscribed circle is $3 / 2 \mathrm{~cm}$, and the radius of the circumscribed circle is $25 / 8$ cm
Solution.

In $\triangle ABC$, $AB = BC$, $BD$ is the altitude, $O_1$ is the center of the inscribed circle, and $O_2$ is the center of the circumscribed circle (see Fig. 10.16). The radii of these circles are $r = \frac{3}{2}$ cm and $R = \frac{25}{8}$ cm, respectively. $E$ is the point of intersection of the ray $BD$ and the circumscribed circle. $AO_1$ is the bisector of $\angle BAC$. Let $\angle BAO_1 = \angle O_1AD = \alpha$. Then $\angle AO_1D = 90^\circ - \alpha$. Since $BE$ is the diameter of the circumscribed circle, $\angle BAE = 90^\circ$, $\angle O_1AE = 90^\circ - \alpha$, and $BE = 2R = \frac{25}{4}$ cm. Therefore, $AE = O_1E$. Let $DE = x$ cm. Then $AE = O_1E = O_1D + DE = r + x = \frac{3}{2} + x$. From $\triangle BAE$ ($\angle BAE = 90^\circ$, $AD$ is the altitude): $AE^2 = DE \cdot BE$; $\left(\frac{3}{2} + x\right)^2 = x \cdot \frac{25}{4}$; $x = \frac{9}{4}$ or $x = 1$. From $\triangle ADE$ ($\angle ADE = 90^\circ$): $AD = \sqrt{AE^2 - DE^2}$. When $x = \frac{9}{4}$: $AE = \frac{15}{4}$ cm, $AD = 3$ cm, $AC = 6$ cm. When $x = 1$: $AE = \frac{5}{2}$ cm, $AD = \frac{\sqrt{21}}{2}$ cm, and $AC$ is expressed as an irrational number. Therefore, $AC = 6$ cm, $AB = BC = \sqrt{BE^2 - AE^2} = 5$ cm.

Answer: 5 cm, 5 cm, 6 cm.