3. Let $A, B, C$ be three distinct points on a circle, $P$ be the area of triangle $ABC$, and $P_{1}$ be the area of the triangle determined by the tangents to the circle at points $A, B, C$. Determine the limit of the ratio $P_{1}: P$ as point $A$ is fixed and points $B$ and $C$ approach $A$ along the circle such that $B \neq C$.
Solution. Let $A_{1} B_{1} C_{1}$ be the triangle defined by the tangents to the circle at points $A$, $B$, and $C$, let $B$ be on the shorter arc $A C$, $O$ be the center of the given circle, and $r=O A, \alpha=\measuredangle A O B$, $\beta=\measuredangle B O C$, as shown in the diagram on the right. Then

$$
\begin{gathered}
C_{1} A=C_{1} B=r \operatorname{tg} \frac{\alpha}{2} \text { and } \\
A_{1} C=A_{1} B=r \operatorname{tg} \frac{\beta}{2},
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_34f3dbec3bea085bdc82g-08.jpg?height=430&width=665&top_left_y=958&top_left_x=802)

it follows that

$$
A_{1} C_{1}=A_{1} B+B C_{1}=r\left(\operatorname{tg} \frac{\alpha}{2}+\operatorname{tg} \frac{\beta}{2}\right)=r \frac{\sin \frac{\alpha+\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}}
$$

Since $A B=2 r \sin \frac{\alpha}{2}, B C=2 r \sin \frac{\beta}{2}$ and $\measuredangle A B C=\pi-\frac{\alpha+\beta}{2}$, we get

$$
P=\frac{1}{2} A B \cdot B C \sin \measuredangle A B C=2 r^{2} \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\alpha+\beta}{2}
$$

Furthermore, $\measuredangle A_{1} C_{1} B_{1}=\measuredangle B O A=\alpha$ and $\measuredangle C_{1} A_{1} B_{1}=\measuredangle B O C=\beta$ (as angles with perpendicular sides) and $\measuredangle A_{1} B_{1} C_{1}=\pi-\alpha-\beta$, so

$$
\begin{aligned}
& A_{1} B_{1}=A_{1} C_{1} \frac{\sin \alpha}{\sin (\alpha+\beta)}, \quad B_{1} C_{1}=A_{1} C_{1} \frac{\sin \beta}{\sin (\alpha+\beta)}, \\
& P_{1}=\frac{1}{2} A_{1} B_{1} \cdot B_{1} C_{1} \sin \measuredangle A_{1} B_{1} C_{1}=r^{2} \operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2} \operatorname{tg} \frac{\alpha+\beta}{2}
\end{aligned}
$$

Finally, we get

$$
\frac{P_{1}}{P}=\frac{1}{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\alpha+\beta}{2}}
$$

If $B \rightarrow A$ and $C \rightarrow A$, then $\alpha \rightarrow 0, \beta \rightarrow 0$, so $\frac{P_{1}}{P}=\frac{1}{2}$.