Question 99, In $\triangle ABC$, the perimeter is 20, the area is $10 \sqrt{3}$, and $2 \angle A = \angle B + \angle C$.
(1) Find the length of $BC$;
(2) If point $P$ is a point on the incircle $\odot I$ of $\triangle ABC$, and $\overrightarrow{A I} = x \overrightarrow{A B} + y \overrightarrow{A C}$,
(1) Find the value of $x + y$;
(2) Find the range of $|\overrightarrow{A P} + \overrightarrow{B P} + \overrightarrow{C P}|$.
Question 99, Answer: Let $\triangle \mathrm{ABC}$ have three interior angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ with corresponding side lengths $\mathrm{a}, \mathrm{b}, \mathrm{c}$. Given that $\angle A=60^{\circ}$, we have $\left\{\begin{array}{c}a+b+c=20 \\ \frac{b c \cdot \sin 60^{\circ}}{2}=10 \sqrt{3} \\ b^{2}+c^{2}-2 b c \cdot \cos A=a^{2}\end{array} \Rightarrow\left\{\begin{array}{c}a=7 \\ \{b, c\}=\{5,8\}^{\circ}\end{array}\right.\right.$ Therefore, the length of BC is 7.

Assume $b=5, c=8$. Let the inradius of $\triangle \mathrm{ABC}$ be $\mathrm{r}$. According to the given conditions, $\mathrm{r}=\frac{2 S_{\triangle A B C}}{a+b+c}=\sqrt{3}$, so $A I=2 \sqrt{3}$. Construct $I M / / A C$ intersecting $A B$ at $M$, and construct $I N / / A B$ intersecting $A C$ at $N$. Then $A M=A N=2$, so $x+y=$ $\frac{A M}{A B}+\frac{A N}{A C}=\frac{2}{8}+\frac{2}{5}=\frac{13}{20}$.

Let the centroid of $\triangle \mathrm{ABC}$ be $\mathrm{G}$. Then $\overrightarrow{\mathrm{AP}}+\overrightarrow{\mathrm{BP}}+\overrightarrow{\mathrm{CP}}=3 \overrightarrow{\mathrm{GP}}=3(\overrightarrow{\mathrm{GI}}+\overrightarrow{\mathrm{IP}})$, so the range of $|\overrightarrow{\mathrm{AP}}+\overrightarrow{\mathrm{BP}}+\overrightarrow{\mathrm{CP}}|$ is
$$
[3(|I P|-|I G|), 3(|I P|-|I G|)]=[3(\sqrt{3}-|I G|), 3(\sqrt{3}+|I G|)]
$$

Thus, we only need to find $|G|$.
As shown in the figure, establish a Cartesian coordinate system with $A$ as the origin and $\overline{A C}$ as the positive direction of the $x$-axis, then $A(0,0), B(4,4 \sqrt{3})$, $\mathrm{C}(5,0)$. Therefore, $\mathrm{G}\left(3, \frac{4 \sqrt{3}}{3}\right)$. Also, $\mathrm{I}(3, \sqrt{3})$, so $|\mathrm{IG}|=\frac{\sqrt{3}}{3}$. Substituting into $(*)$, the range of $|\overrightarrow{\mathrm{AP}}+\overrightarrow{\mathrm{BP}}+\overrightarrow{\mathrm{CP}}|$ is $[2 \sqrt{3}, 4 \sqrt{3}]$.