Let's calculate the sides of an isosceles triangle if the radius of the inscribed circle and the radius of the circumscribed circle are given. Let's also construct the triangle.

Given:
- The radius of the inscribed circle (r)
- The radius of the circumscribed circle (R)

We need to find the sides of the isosceles triangle and construct it.
I. Solution. Let $ABC$ be the sought triangle $(AC=BC)$. Let the radius of the circumscribed circle and the inscribed circle be denoted by $r$ and $\varrho$, respectively. Let the midpoint of $AB$ be $D$, the point opposite to $C$ on the circumscribed circle be $C'$, and the point of tangency of the inscribed circle on the side $AC$ be $T$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_4c5cb35f15cbf7c16719g-1.jpg?height=531&width=570&top_left_y=298&top_left_x=772)

We can express the distances appearing in the relationship

$$
AC = AT + CT = AD + CT
$$

using the height $CD = m$, and thus we obtain an equation for $m$.

The triangle $ACC'$ is right-angled at $A$, so by the geometric mean property of the leg and the altitude,

$$
\begin{gathered}
AC^2 = CC' \cdot CD = 2rm \\
AD^2 = CD \cdot DC' = m(2r - m)
\end{gathered}
$$

Finally, the following relationship holds between the tangent $CT$ of the inscribed circle and the secant $CD$ of the circle, since $CD$ (the axis of symmetry of the isosceles triangle) passes through the center of the circle:

$$
CT^2 = CD \cdot (CD - 2\varrho) = m(m - 2\varrho)
$$

Substituting these into (1),

$$
\sqrt{2rm} = \sqrt{m(2r - m)} + \sqrt{m(m - 2\varrho)}
$$

We can simplify by $\sqrt{m}$, since it cannot be 0. Then, by squaring the square roots and rearranging in the usual way, we arrive at the equation

$$
m^2 - 2(r + \varrho)m + \varrho(\varrho + 4r) = 0
$$

This equation has solutions only if $r \geq 2\varrho$, and in that case, the solutions are:

$$
m_{1,2} = r + \varrho \pm \sqrt{r(r - 2\varrho)}
$$

From (2) and (3),

$$
\begin{aligned}
a_{1,2} &= b_{1,2} = AC = \sqrt{2r(r + \varrho \pm \sqrt{r(r - 2\varrho)})} \text{ and } \\
c_{1,2} &= 2AD = 2\sqrt{(r + \varrho \pm \sqrt{r(r - 2\varrho)})(r - \varrho \mp \sqrt{r(r - 2\varrho)})} = \\
&= 2\sqrt{\varrho(2r - \varrho \mp 2\sqrt{r(r - 2\varrho)})}
\end{aligned}
$$

Here,

$$
\sqrt{r(r - 2\varrho)} = \sqrt{r^2 - 2r\varrho} \quad r + \varrho > m_2