Half-disc: Area ratios

Adding to the diagram, we have the following:

With the altitudes of the two new triangles being perpendicular to the line $y=0$ . The angle defining the sector within $A$ is $180\degree - \theta-(\theta-90\degree)=90\degree$ .

$A$ comprises a quarter circle of area $1^2\cdot\pi\times\dfrac{1}{4}=\dfrac{\pi}{4}$ and two congruent triangles of area $\dfrac{\sin\theta\cos\theta}{2}$ each.

This is because on the edge of the disc we have $\sin^2\theta+y^2=1\impliedby y=\cos\theta$ for the right-hand triangle, since $\sin^2 x+\cos^2 x\equiv 1$ . For the left-hand triangle we have $(-\cos\theta)^2=1\impliedby y=\sin\theta$ .

The area of $A=\dfrac{\pi}{4}+2\cdot\dfrac{\sin\theta\cos\theta}{2}=\dfrac{\pi}{4}+\sin\theta\cos\theta$ .

Evidently, since the half-disk has radius 1, area of $A$ $+$ area of $B=\dfrac{\pi}{2}$ , so the area of $B=\dfrac{\pi}{2}-\dfrac{\pi}{4}-\sin\theta\cos\theta=\dfrac{\pi}{4}-\sin\theta\cos\theta$ .

$\begin{aligned}\frac{\textnormal{Area of }A}{\textnormal{Area of }B}&=\frac{\frac{\pi}{4}+\sin\theta\cos\theta}{\frac{\pi}{4}-\sin\theta\cos\theta}\end{aligned}$

If we put $\dfrac{\pi}{4}:=k$ and $\sin\theta\cos\theta:=x$ then $\dfrac{k+x}{k-x}$ is a (strictly) increasing function for $|x|<k$ , so the maximum value of the ratio that we require is at the maximum value of $x=\sin\theta\cos\theta$ .

We have that $x=\dfrac{\sin 2\theta}{2}\leq \dfrac{1}{2}$ .

Alternatively, note that $r^2\geq 0\ \forall\ r\in\mathbb{R}\implies (\sin\theta-\cos\theta)^2\geq 0$ , so $1-2\sin\theta\cos\theta\geq 0$ which gives the same result.

The maximum value of $\begin{aligned}\frac{\textnormal{Area of }A}{\textnormal{Area of }B}&=\frac{\frac{\pi}{4}+\frac{1}{2}}{\frac{\pi}{4}-\frac{1}{2}}\\&=\frac{\pi+2}{\pi-2}\\&=\color{#20A900}{\boxed{4.50}\textnormal{ (3 s.f.)}}\end{aligned}$

Let the area of region marked $A$ be $A$ and that marked $B$ be $B$ . Then $B$ is the area of semicircle with radius $1$ less $A$ or $B = \dfrac \pi 2 - A$ . Then the ratio $\dfrac AB = \dfrac A{\frac \pi 2 - A} = \dfrac 1{\frac \pi{AA}-1}$ . Then $\dfrac AB$ is maximum, when $A$ is maximum.

We can see from the figure that $A$ is equal to a quarter circle with radius $1$ plus a rectangle with side lengths $\sin \theta$ and $\cos \theta$ or $A = \dfrac \pi 4 + \sin \theta \cos \theta = \dfrac \pi 4 + \dfrac {\sin (2\theta)}2$ . $\implies \max(A) = \dfrac \pi 4 + \dfrac {\max(\sin (2\theta))}2 = \dfrac \pi 4 + \dfrac 12$ .

Therefore $\max \left(\dfrac AB\right) = \dfrac {A_{\max}}{\frac \pi 2 - A_{\max}} = \dfrac {\frac \pi 4+\frac 12}{\frac \pi 4-\frac 12} = \dfrac {\pi + 2}{\pi -2} \approx \boxed{4.50}$ .