Geometry - Inadequate Information #1

Let the angle and the radius of the sector be $\theta$ and $r$ respectively, then $l = 2r + \theta r$ , $S=\dfrac {\theta r^2}2$ and

$\begin{aligned} \frac S{l^2} & = \frac {\frac 12 \theta r^2}{r^2(2+\theta)^2} = \frac \theta{2(2+\theta)^2} = \frac {\theta+2-2}{2(2+\theta)^2} \\ & = \frac 12\left( \frac 1{2+\theta} - \frac 2{(2+\theta)^2}\right) \\ & = - \left(\left(\frac 1{2+\theta} \right)^2 - \frac 12 \left(\frac 1{2+\theta} \right) \right)\\ & = \frac 1{16} - \color{#3D99F6} \left(\frac 1{2+\theta} - \frac 14 \right)^2 & \small \color{#3D99F6} \text{Note that } \left(\frac 1{2+\theta} - \frac 14 \right)^2 \ge 0 \\ & {\color{#3D99F6} \le} \frac 1{16} = \boxed{0.0625} \end{aligned}$

Maximum occurs when $\dfrac 1{2+\theta} - \dfrac 14 = 0$ or $\theta = 2$ radian.

Let the central angle of the given sector be $x$ , and the radius of the sector $r$ . Then $0<x<2\pi$ .

$l=(x+2)r$ , and $S=\dfrac{1}{2}r^2x$ .

$\dfrac{S}{l^2}=\dfrac{x}{2(x+2)^2}$ .

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Note that

$\dfrac{x}{x+2}+\dfrac{2}{x+2}\ge 2\sqrt{\dfrac{2x}{(x+2)^2}}$ .

Left side of the inequality is $1$ . Now divide both sides by $4$ .

$\dfrac{1}{4}\ge \sqrt{\dfrac{x}{2(x+2)^2}}$

$\dfrac{x}{2(x+2)^2}\le \boxed{\dfrac{1}{16}}$

with equality achieved when $\dfrac{x}{x+2}=\dfrac{2}{x+2}$ , namely $x=2$ .

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To sum up, we proved these facts below.

Fact 1: If the area of a sector is constant, the perimeter of the sector achieves its minimum when the central angle is $2$ radian.

Fact 2: If the perimeter of a sector is constant, the area of the sector achieves its maximum when the central angle is $2$ radian.