11.1 In triangle $\mathrm{ABC}$, angle $\mathrm{A}$ is the largest, and angle $\mathrm{C}$ is the smallest. Which of the vertices A, B, C is closest to the center of the inscribed circle?
Solution: See Fig.

$$
\begin{aligned}
& \frac{r}{A O}=\sin \frac{A}{2} \\
& A O=\frac{r}{\sin \frac{A}{2}}
\end{aligned}
$$

Similarly, $B O=\frac{r}{\sin \frac{B}{2}}, \quad C O=\frac{r}{\sin \frac{C}{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_29b6123696d1dc639d6bg-1.jpg?height=429&width=508&top_left_y=468&top_left_x=1274)

Since the angles $\frac{A}{2}, \frac{B}{2}, \frac{C}{2}$ lie in the first quadrant, where the sine function is increasing, we have $\sin \frac{A}{2}>\sin \frac{B}{2}>\sin \frac{C}{2}$. Therefore, $\frac{r}{\sin \frac{A}{2}}<\frac{r}{\sin \frac{B}{2}}<\frac{r}{\sin \frac{C}{2}}$, i.e., $A O<B O<C O$, i.e., vertex $\mathrm{A}$ is the closest to point $\mathrm{O}$.