Question:

Step 1: Define the region and its properties. The region is formed by points $P(z)$ lying inside the incircle of a regular hexagon centered at the origin. Let the radius of this incircle be $R$. The region is further constrained by the argument condition $\frac{\pi}{3} \leq \arg(z) \leq \frac{5\pi}{6}$. This describes a sector of the incircle. Step 2: Determine the central angle of the sector. The central angle $\theta$ of the sector is the difference between the upper and lower bounds of the argument: $$ \theta = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} $$ Step 3: Calculate the area of the sector. The area of a sector of a circle with radius $R$ and central angle $\theta$ is given by the formula $A = \frac{1}{2} R^2 \theta$. Substituting the calculated angle: $$ A = \frac{1}{2} R^2 \left(\frac{\pi}{2}\right) = \frac{R^2 \pi}{4} $$ Step 4: Determine the radius of the incircle. The problem states that the area is $\frac{m\pi}{n}$, where $m$ and $n$ are relatively prime natural numbers. For the sum $m+n$ to be 7, with $m$ and $n$ relatively prime, the only possible values are $m=3$ and $n=4$. Therefore, we set the calculated area equal to $\frac{3\pi}{4}$: $$ \frac{R^2 \pi}{4} = \frac{3\pi}{4} $$ Dividing both sides by $\frac{\pi}{4}$ yields: $$ R^2 = 3 $$ Thus, the radius of the incircle is $R = \sqrt{3}$. Step 5: State the final area and calculate $m+n$. With $R=\sqrt{3}$, the area of the region is: $$ A = \frac{(\sqrt{3})^2 \pi}{4} = \frac{3\pi}{4} $$ Comparing this to the form $\frac{m\pi}{n}$, we have $m=3$ and $n=4$. These are relatively prime natural numbers. The sum $m+n$ is: $$ m+n = 3+4 = 7 $$