In a quadrilateral $A B C D$, a circle can be inscribed and a circle can be circumscribed around it. The diagonals of this quadrilateral are perpendicular to each other. Find its area if the radius of the circumscribed circle is $R$ and $A B=2 B C$.

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1. In a cyclic quadrilateral with mutually perpendicular diagonals, the distance from the center of the circumscribed circle to one of the sides is half of the opposite side.
2. Since a circle can be inscribed in the quadrilateral, the sums of its opposite sides are equal, and since the diagonals of the quadrilateral are perpendicular, the sums of the squares of its opposite sides are also equal.

## Solution

## First Method.

Let $BC = x$. Then $AB = 2x$. Let $O$ be the center of the circumscribed circle, and $K$ and $M$ be the midpoints of $AD$ and $DC$. Then

$$
\begin{gathered}
O K = \frac{1}{2} B C = \frac{x}{2}, O M = \frac{1}{2} A B = x \\
A D = 2 D K = \sqrt{R^{2} - \frac{x^{2}}{4}}, C D = 2 M D = 2 \sqrt{R^{2} - x^{2}}
\end{gathered}
$$

Since $AB + CD = BC + AD$, then

$$
2 x + 2 \sqrt{R^{2} - x^{2}} = x + 2 \sqrt{R^{2} - \frac{x^{2}}{4}}
$$

From this, we find that

$$
x = \frac{2 R}{\sqrt{5}}, AB = \frac{4 R}{\sqrt{5}}, AD = \frac{4 R}{\sqrt{5}}, BC = \frac{2 R}{\sqrt{5}}, CD = \frac{2 R}{\sqrt{5}}
$$

Therefore, $AC$ is the bisector of angle $BCD$ and $\angle ABC = 90^\circ$. Consequently, the desired area is twice the area of triangle $ABC$, i.e., $\frac{8}{5} R^{2}$.

## Second Method.

Let $BC = x, AB = 2x, AD = y, CD = z$. Since a circle can be inscribed in the quadrilateral, the sums of its opposite sides are equal, and since the diagonals of the quadrilateral are perpendicular, the sums of the squares of its opposite sides are also equal. Then

$$
\left\{\begin{array} { l } 
{ x + y = 2 x + z } \\
{ x ^ { 2 } + y ^ { 2 } = 4 x ^ { 2 } + z ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x = y - z } \\
{ 3 x ^ { 2 } = y ^ { 2 } - z ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x = y - z \\
3 x^{2} = (y - z)(y + z)
\end{array} \Leftrightarrow\right.\right.\right.
$$

$$
\Leftrightarrow\left\{\begin{array} { l } 
{ x = y - z } \\
{ 3 x ^ { 2 } = x ( y + z ) }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x = y - z } \\
{ 3 x = y + z }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y = 2 x \\
z = x
\end{array}\right.\right.\right.
$$

Therefore, $BC = DC$ and $BA = DA$, i.e., points $A$ and $C$ are equidistant from the endpoints of segment $BD$. Thus, line $AC$ is the perpendicular bisector of chord $BD$ of the circumscribed circle of quadrilateral $ABCD$. Consequently, $AC$ is the diameter of this circle, $\angle ABC = \angle ADC = 90^\circ$.

By the Pythagorean theorem

$$
BC^{2} + AB^{2} = AC^{2}, \text { or } x^{2} + 4 x^{2} = 4 R^{2},
$$

from which $x^{2} = \frac{4}{5} R^{2}$. Therefore,

$$
S_{\mathrm{ABCD}} = 2 S_{\triangle \mathrm{ABC}} = 2 \cdot \frac{1}{2} BC \cdot AB = 2 x^{2} = \frac{8}{5} R^{2}
$$

First Method.

Let $BC = x$. Then $AB = 2x$. Let $O$ be the center of the circumscribed circle, and $K$ and