SuperHero Circles

Area of the sector is given by $A = \dfrac {\theta r^2}2$ and its perimeter $p = \theta r + 2r$ . When $A=p$ , we have:

$\begin{aligned} \frac {\theta r^2}2 & = \theta r + 2r & \small \blue{\text{Multiply both sides by }\frac 2{\theta r}} \\ \implies r & = 2 + \frac 4 \theta \end{aligned}$

For $\theta \in \left[\dfrac \pi 2 , \dfrac {3\pi}2 \right]$ , integer values of $\theta$ are $2$ , $3$ , and $4$ . For $r$ to be an integer, $\theta$ can only be $2$ and $4$ , then $r = 4$ and $3$ respectively. Therefore there are only $\boxed 2$ solutions.

Assuming a constraint $r > 0$ to exclude the degenerate case...

The area of the sector is $A = \frac{\theta r^2}{2}$ .

The perimeter is $P = 2r + r\theta$ .

Set $A = P$ : $\frac{\theta r^2}{2} = 2r + r\theta$

Divide both sides by (non-zero) $r$ and multiply by $2$ :

$r \theta = 4 + 2 \theta$ , so $r = \frac{4}{\theta} + 2.$

We are told that $r$ is an integer, thus, $\frac{4}{\theta}$ must be an integer.
The only integral values of $\frac{4}{\theta}$ in the range from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$ are found when $\theta = 2,4$ .

Thus there are exactly two solutions. $(\theta,r) = (2,4)$ and $(\theta,r) = (4,3)$

If r is the radius and theta is the central angle, we have Perimeter = 2 r + r (theta). Also, Area = (.5) r r (theta). Setting these equal and solving for theta gives: theta = 4/(r-2). The restriction on theta means (Pi) / 2 < 4/(r-2) < 3 Pi/2, or Pi < 8/(r-2) < 3*Pi. Substituting integral values of r into the middle expression shows that r = 3 and r = 4 both satisfy the inequality. The fact that the middle expression is decreasing with increasing r means that these can be the only two solutions.

@Mahdi Raza , I guess you should mention that r > 0 because at r = 0 (integer), both area and perimeter are 0 so there can be 3 solutions