3. Over the trapezoid $A B C D$, a circle is described with center on the base $A B$ and radius $R$. Express the length of the leg $A D$ in terms of $R$, if

$$
\overline{A D}^{2}+3 \overline{C D}^{2}=2 R^{2}
$$
Solution. Since a circle is circumscribed around the trapezoid with its center on the base, it follows that the trapezoid is isosceles. Let $O$ be the center of the circumscribed circle (see diagram). Triangle $COD$ is isosceles with legs $R$ and base $CD$. Let $h$ be the height of the trapezoid. Then

$$
\left(\frac{\overline{CD}}{2}\right)^{2}=R^{2}-h^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6fcf99234988cbfe775eg-3.jpg?height=485&width=548&top_left_y=1762&top_left_x=935)

From the similarity of triangles $CMB$ and $BDA$, it follows that $h=\frac{\overline{AD} \cdot \overline{BD}}{2}$. Therefore, $\overline{CD}^{2}=4R^{2}-\frac{\overline{AD}^{2}-\overline{BD}^{2}}{R^{2}}$, from which, together with the given equation $\overline{AD}^{2}+3\overline{CD}^{2}=2R^{2}$, it follows that

$$
3 \overline{CD}^{2}=12R^{2}-\frac{3 \overline{AD}^{2} \cdot \overline{BD}^{2}}{R^{2}}=2R^{2}-\overline{AD}^{2}
$$

Since triangle $ABD$ is a right triangle, $\overline{BD}^{2}=4R^{2}-\overline{AD}^{2}$, from which, by substitution into (1), we get:

$$
12R^{4}-12 \overline{AD}^{2} R^{2}+3 \overline{AD}^{4}=2R^{4}-\overline{AD}^{2} R^{2}
$$

i.e.,

$$
10R^{4}-11 \overline{AD}^{2} R^{2}+3 \overline{AD}^{4}=0
$$

Solving the above equation for $\overline{AD}^{2}$, we get: $\overline{AD}^{2}=\frac{11R^{2} \pm R^{2}}{6}$. Since $ABCD$ is a trapezoid (not a triangle), we get $\overline{AD}^{2}=\frac{5R^{2}}{3}$, i.e., $\overline{AD}=R \frac{\sqrt{15}}{3}$.