<s>[INST] How do I find the lengths of the sides of a triangle that is inscribed in the unit circle that has the greatest perimeter? I know how to find the sides of the triangle when seeking to maximize the area of ​​the triangle, since the following relationship exists $Area= \frac{abc}{4R}$ where $a, b,c$ are the sides and $R$ the radius.
But I don't know what to occupy to maximize the perimeter.
 [/INST] Recall that the Law of Sines tells you that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$
where $R$ is the circumradius.  Then the perimeter is
$$2R(\sin A + \sin B + \sin C).$$
Note that
$$\sin A + \sin B = 2\cos(\tfrac{A - B}{2})\sin(\tfrac{A + B}{2}) \le 2\sin(\tfrac{A + B}{2})$$
with equality only when $A = B$.  Thus, the perimeter is maximized when $A = B$.  Using similar logic, you can show that $B = C$ as well when the perimeter is maximized.  So, the perimeter is maximized when $A = B = C = 60^\circ$ with a value of
$$2R(\sin 60^\circ + \sin 60^\circ + \sin 60^\circ) = 3\sqrt3 R.$$
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