18. A circle is circumscribed around a right-angled triangle. Another circle of the same radius touches the legs of this triangle, with one of the points of tangency being the vertex of the triangle. Find the ratio of the area of the triangle to the area of the common part of the two given circles.
18. Let (Fig. 4) \(O_{1}\) be the center of the second circle. From the condition, its radius is equal to one of the legs; let the leg \(|C A|=R\); since \(R\) is the radius

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-067.jpg?height=415&width=385&top_left_y=933&top_left_x=167)

Fig. 4. of the circumscribed circle, then \(\widehat{C B A} = 30^{\circ}, |C B| = R \sqrt{3}\).

If \(O\) is the midpoint of \(A B\), then \(\widehat{O A O_{1}} = 30^{\circ}\). If \(A_{1}\) is the second intersection point, then \(\widehat{O A_{1} O_{1}} = \widehat{O A O_{1}} = 30^{\circ}\). Therefore, the common chord of these circles corresponds to arcs of \(150^{\circ}\), and the common part consists of two segments corresponding to arcs of \(150^{\circ}\), the area of each of which is

\[
S_{\mathrm{cer \textrm {M }}} = \frac{5}{12} \pi R^{2} - \frac{R^{2}}{4}
\]

Answer: \(\frac{3 \sqrt{3}}{5 \pi - 3}\).