<s>[INST] Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b=\frac{2}{3}a,c=\left(\frac{2}{3}\right)^2a$
which gives $$\frac{\sin A}{\sin B}=\frac{\sin B}{\sin C}$$
I am stuck now,how to find the longest side $a$.
 [/INST] If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by:
$$ R=\frac{abc}{4\Delta}=\frac{\frac{8}{27}l^3}{l^2\sqrt{\left(1+\frac{2}{3}+\frac{4}{9}\right)\left(1-\frac{2}{3}+\frac{4}{9}\right)\left(1+\frac{2}{3}-\frac{4}{9}\right)\left(-1+\frac{2}{3}+\frac{4}{9}\right)}}$$
by Heron's formula, hence:
$$ R = \frac{24}{\sqrt{1463}} l $$
gives:

$$ l=\color{red}{\frac{7}{4}}.$$

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