Through point $M(2,1)$, a chord $AB$ is drawn of the ellipse $x^{2}+4 y^{2}=16$ such that $|A M|:|M B|=1: 2$. Find the length of the chord $AB$.

The text is translated while preserving the original text's line breaks and format.
Let the equation of the chord $AB$ be
$$
\left\{\begin{array}{l}
x=2+t \cos \alpha \\
y=1+t \sin \alpha .
\end{array}\right.
$$

Substituting into the ellipse equation and simplifying, we get
$$
\left(3 \sin ^{2} \alpha+1\right) t^{2}+4(\cos \alpha+2 \sin \alpha) t-8=0 \text {. }
$$

By Vieta's formulas, we have $t_{1}+t_{2}=-\frac{4(\cos \alpha+2 \sin \alpha)}{3 \sin ^{2} \alpha+1}$,
$$
\begin{array}{l}
t_{1} \cdot t_{2}=-\frac{8}{3 \sin ^{2} \alpha+1} . \text { Given }|A M|:|M B| \\
=1: 2 \text {, we have } t_{2}=-2 t_{1}, \\
\therefore t_{1}+t_{2}=-t_{1}, \\
t_{1} \cdot t_{2}=-2 t_{1}^{2}=-2\left(t_{1}+t_{2}\right)^{2} . \\
\therefore-\frac{8}{3 \sin ^{2} \alpha+1}=-\frac{32(\cos \alpha+2 \sin \alpha)^{2}}{\left(3 \sin ^{2} \alpha+1\right)^{2}} .
\end{array}
$$

Simplifying, we get
$$
\begin{array}{c}
12 \sin ^{2} \alpha+16 \sin \alpha \cos \alpha+3 \cos ^{2} \alpha=0 . \\
\because \cos ^{2} \alpha=0 \text { is not a solution to this equation, } \\
\therefore \text { we have } 12 \operatorname{tg}^{2} \alpha+16 \operatorname{tg} \alpha+3=0, \\
\operatorname{tg} \alpha=\frac{-4 \pm \sqrt{7}}{6} . \\
\therefore \sin ^{2} \alpha=\frac{36}{59 \pm 8 \sqrt{7}} . \\
=\sqrt{-\frac{1}{2} t_{1} t_{2}-4 t_{1} t_{2}}=\frac{6}{\sqrt{3 \sin ^{2} \alpha+1}} \\
=\frac{6}{165} \sqrt{165(59 \pm 8 \sqrt{7}) .}
\end{array}
$$