Question:

Step 1: Use the given condition $2\sin 2A + \sin(2B+C) = \sin C$ with $C = \pi/3$. Expand $\sin(2B+C) = \sin(2B+C)$. Since $A+B+C=\pi$, we have $2B+C = \pi + B - A$, so $\sin(2B+C) = -\sin(B-A)$. Alternatively, expand directly: $$2\sin 2A + \sin(2B+C) = \sin C$$ $$2\sin 2A + \sin 2B\cos C + \cos 2B\sin C = \sin C$$ $$2\sin 2A + \sin 2B\cos C = \sin C(1 - \cos 2B)$$ $$2\sin 2A + \sin 2B\cos C = 2\sin C\sin^2 B$$ Step 2: Substitute $C = \pi/3$, so $\cos C = 1/2$, $\sin C = \sqrt{3}/2$: $$2\sin 2A + \frac{1}{2}\sin 2B = 2\cdot\frac{\sqrt{3}}{2}\sin^2 B = \sqrt{3}\sin^2 B$$ $$2\sin 2A + \frac{1}{2}\cdot 2\sin B\cos B = \sqrt{3}\sin^2 B$$ $$2\sin 2A + \sin B\cos B = \sqrt{3}\sin^2 B$$ $$2\sin 2A = \sin B(\sqrt{3}\sin B - \cos B) = 2\sin B\sin\left(B - \frac{\pi}{6}\right)\cdot\frac{1}{\sin(\pi/6)}$$ Actually: $\sqrt{3}\sin B - \cos B = 2\sin\left(B - \frac{\pi}{6}\right)$ So: $2\sin 2A = 2\sin B\sin\left(B-\frac{\pi}{6}\right)$ $$\sin 2A = \sin B\sin\left(B-\frac{\pi}{6}\right)$$ Step 3: Use $A + B = \pi - C = 2\pi/3$, so $A = 2\pi/3 - B$: $$\sin 2A = \sin\left(\frac{4\pi}{3} - 2B\right) = -\sin\left(2B - \frac{4\pi}{3}\right)$$ Alternatively, $\sin 2A = \sin(4\pi/3 - 2B)$. Let's try: $2A = \pi - 2B - 2C + 2A$... use $A = 2\pi/3 - B$: $$\sin 2A = \sin\left(\frac{4\pi}{3} - 2B\right)$$ Setting equal: $\sin\left(\frac{4\pi}{3}-2B\right) = \sin B\sin\left(B-\frac{\pi}{6}\right)$ Step 4: Try $A = \pi/6$, then $B = 2\pi/3 - \pi/6 = \pi/2$. Check: $\sin 2A = \sin(\pi/3) = \sqrt{3}/2$. $\sin B\sin(B-\pi/6) = \sin(\pi/2)\sin(\pi/2-\pi/6) = 1\cdot\sin(\pi/3) = \sqrt{3}/2$. ✓ So $A = \pi/6$, $B = \pi/2$, $C = \pi/3$. Step 5: Find sides using sine rule. $c = 2$, $C = \pi/3$: $$\frac{c}{\sin C} = \frac{2}{\sqrt{3}/2} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$$ Circumradius $R = \dfrac{c}{2\sin C} = \dfrac{2}{2\cdot(\sqrt{3}/2)} = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}$. ✓ (option 3) Step 6: Find sides: $$a = R\cdot 2\sin A = \frac{4\sqrt{3}}{3}\cdot\sin\frac{\pi}{6} = \frac{4\sqrt{3}}{3}\cdot\frac{1}{2} = \frac{2\sqrt{3}}{3}$$ $$b = R\cdot 2\sin B = \frac{4\sqrt{3}}{3}\cdot 1 = \frac{4\sqrt{3}}{3}$$ Step 7: Compute area: $$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}\cdot\frac{2\sqrt{3}}{3}\cdot\frac{4\sqrt{3}}{3}\cdot\frac{\sqrt{3}}{2} = \frac{1}{2}\cdot\frac{8\cdot 3}{9}\cdot\frac{\sqrt{3}}{2} = \frac{1}{2}\cdot\frac{8}{3}\cdot\frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{3}$$ ✓ (option 4) Step 8: Compute inradius $r = \Delta/s$ where $s = (a+b+c)/2$: $$s = \frac{\frac{2\sqrt{3}}{3} + \frac{4\sqrt{3}}{3} + 2}{2} = \frac{\frac{6\sqrt{3}}{3}+2}{2} = \frac{2\sqrt{3}+2}{2} = \sqrt{3}+1$$ $$r = \frac{2\sqrt{3}/3}{\sqrt{3}+1} = \frac{2\sqrt{3}}{3(\sqrt{3}+1)}$$ ✓ (option 2) Step 9: Check option (1): $\sin 2A + \sin 2B + \sin 2C$: $$= \sin\frac{\pi}{3} + \sin\pi + \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} + 0 + \frac{\sqrt{3}}{2} = \sqrt{3} \neq \sqrt{2}$$ Option (1) is incorrect.