4. Given that the circumradius of $\triangle A B C$ is 1, and $A B, B C, \frac{4}{3} C A$ form a geometric sequence in order, then the maximum value of $B C$ is $\qquad$ .
Solve (Wang Zheng's problem) $\frac{4 \sqrt{2}}{3}$.
Using the sine rule to convert the original expression into trigonometric functions, we get $4 \sin C \sin B=3 \sin ^{2} A$. By transforming the left side into a sum and difference product, we obtain $2(\cos (B-C)+\cos A)=3 \sin ^{2} A$, thus $3 \sin ^{2} A-2 \sqrt{1-\sin ^{2} A} \in(-2,2]$. The expression is increasing with respect to $\sin A$, so the expression equals 2 when $\sin A$ is maximized, at which point $\sin A=\frac{2 \sqrt{2}}{3}$. Therefore, $BC=2 \sin A$ has a maximum value of $\frac{4 \sqrt{2}}{3}$.