Problem 4. Point $O$ is the center of the circumcircle of triangle $ABC$ with sides $AB=8$ and $AC=5$. Find the length of side $BC$ if the length of the vector $\overrightarrow{OA}+3 \overrightarrow{OB}-4 \overrightarrow{OC}$ is 10.
Answer: 4.

Answer: Let the radius of the circumscribed circle be denoted by $R$. Then for any numbers $x, y, z$, the following equality holds:

$$
\begin{aligned}
& (x \overrightarrow{O A}+y \overrightarrow{O B}+z \overrightarrow{O C})^{2}= \\
& =x^{2} O A^{2}+y^{2} O B^{2}+z^{2} O C^{2}+2 x y(\overrightarrow{O A} \cdot \overrightarrow{O B})+2 y z(\overrightarrow{O B} \cdot \overrightarrow{O C})+2 x z(\overrightarrow{O A} \cdot \overrightarrow{O C})= \\
& =\left(x^{2}+y^{2}+z^{2}\right) R^{2}+x y\left(O A^{2}+O B^{2}-A B^{2}\right)+y z\left(O B^{2}+O C^{2}-B C^{2}\right)+x z\left(O A^{2}+O C^{2}-A C^{2}\right)= \\
& =\left(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right) R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}= \\
& =(x+y+z)^{2} R^{2}-x y A B^{2}-y z B C^{2}-x z A C^{2}
\end{aligned}
$$

Then, for $x=1, y=3, z=-4$, we obtain the equality $10^{2}=0 \cdot R^{2}-3 A B^{2}+12 B C^{2}+4 A C^{2}$, from which $B C^{2}=\frac{1}{12}\left(100+3 A B^{2}-4 A C^{2}\right)=16$, i.e., $B C=4$.