## Task A-2.6.

Trapezoid $ABCD$ with bases $\overline{AB}$ and $\overline{CD}$ has a circumscribed circle $k$. Its diagonals are perpendicular to each other and intersect at point $S$. Determine the ratio of the area of the circle bounded by circle $k$ to the sum of the areas of triangles $ABS$ and $CDS$.
## Second Solution.

Let $|A B|=a$ and $|C D|=c$. Let $M$ and $N$ be the feet of the altitudes from vertex $S$ in triangles $A B S$ and $C D S$, respectively. Denote $|S M|=h_{a}$ and $|S N|=h_{c}$.

Since it is inscribed in a circle, trapezoid $A B C D$ is isosceles.

Due to symmetry, triangles $A B S$ and $C D S$ are isosceles.

Since right triangle $A B S$ is isosceles, it follows that $|A M|=|S M|=h_{a}$, i.e., $h_{a}=\frac{a}{2}$. Similarly, it follows that $h_{c}=\frac{c}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-10.jpg?height=709&width=716&top_left_y=979&top_left_x=590)

On the other hand, let $O$ be the center and $R$ the radius of the circle $k$. Further, let $v_{a}=|O M|$ be the length of the altitude from $O$ to $\overline{A B}$, and $v_{c}=|O N|$ be the length of the altitude from $O$ to $\overline{C D}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-10.jpg?height=714&width=720&top_left_y=1876&top_left_x=591)

From the right triangles $A M O$ and $C N O$, we have

$$
\begin{aligned}
& \left(\frac{a}{2}\right)^{2}+v_{a}^{2}=R^{2} \\
& \left(\frac{c}{2}\right)^{2}+v_{c}^{2}=R^{2}
\end{aligned}
$$

Subtracting these equations, we get

$$
v_{a}^{2}-v_{c}^{2}+\left(\frac{a}{2}\right)^{2}-\left(\frac{c}{2}\right)^{2}=0
$$

or

$$
\left(v_{a}-v_{c}\right)\left(v_{a}+v_{c}\right)=\frac{a+c}{2} \cdot \frac{c-a}{2} .
$$

Notice that $v_{a}+v_{c}$ is the height of the trapezoid, so $v_{a}+v_{c}=h_{a}+h_{c}=\frac{a+c}{2}$.

Substituting this into the above equation and dividing by $\frac{a+c}{2}$, we get $v_{a}-v_{c}=\frac{c-a}{2}$.

Solving the system of equations

$$
\begin{aligned}
& v_{a}-v_{c}=\frac{c-a}{2} \\
& v_{a}+v_{c}=\frac{a+c}{2}
\end{aligned}
$$

we obtain $v_{a}=\frac{c}{2}$ and $v_{c}=\frac{a}{2}$.

Now we have $R^{2}=\left(\frac{a}{2}\right)^{2}+v_{a}^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}$.

The sum of the areas of triangles $A B S$ and $C D S$ is

$$
\frac{a \cdot h_{a}}{2}+\frac{c \cdot h_{c}}{2}=\frac{a^{2}}{4}+\frac{c^{2}}{4}=R^{2}
$$

so the desired ratio is $\pi$.