56. In triangle $ABC$, $R$ and $r$ are the radii of the circumcircle and the incircle, respectively. Let $A_{1}, B_{1}$, $C_{1}$ be the points of intersection of the angle bisectors of triangle $ABC$ with the circumcircle. Find the ratio of the areas of triangles $ABC$ and $A_{1} B_{1} C_{1}$.
56. Let's use the following formula for the area of a triangle: $S=2 R^{2} \sin \hat{A} \sin \hat{B} \sin \hat{C}$, where $\hat{A}, \hat{B}$, and $\hat{C}$ are the angles of the triangle. Then the area of triangle $A_{1} B_{1} C_{1}$, where $A_{1}$, $B_{1}$, and $C_{1}$ are the points of intersection of the angle bisectors of $\triangle A B C$ with the circumcircle, will be

$$
S_{1}=2 R^{2} \sin \frac{\hat{A}+\hat{B}}{2} \sin \frac{\hat{B}+\hat{C}}{2} \sin \frac{\hat{C}+\hat{A}}{2}=2 R^{2} \cos \frac{\hat{C}}{2} \cos \frac{\hat{A}}{2} \cos \frac{\hat{B}}{2}
$$

and

$$
\frac{S}{S_{1}}=8 \sin \frac{\hat{A}}{2} \sin \frac{\hat{B}}{2} \sin \frac{\hat{C}}{2}
$$

On the other hand (Fig. 10) we have: $|B C|=2 R \sin \hat{A}$,

$$
\begin{aligned}
& \left(\operatorname{ctg} \frac{\hat{B}}{2}+\operatorname{ctg} \frac{\hat{C}}{2}\right)=2 R \sin \hat{A}, r \frac{\sin \frac{\hat{B}+\hat{C}}{2}}{\sin \frac{\hat{B}}{2} \sin \frac{\hat{C}}{2}}=2 R \cdot 2 \sin \frac{\hat{A}}{2} \cos \frac{\hat{A}}{2} \\
& \sin \frac{\hat{A}}{2} \sin \frac{\hat{B}}{2} \sin \frac{\hat{C}}{2}=\frac{r}{4 R} ; \text { thus, } \frac{S}{S_{1}}=\frac{2 r}{R}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-078.jpg?height=396&width=386&top_left_y=445&top_left_x=425)

Fig. 10.