A mere square

First note that as $\cos(x)$ is an even function, $\arcsin(\cos(x))$ will be even as well. Also note that $\arcsin(\cos(0)) = \arcsin(1) = \frac{\pi}{2}$ and that $\arcsin(\cos(\frac{\pi}{2})) = \arcsin(0) = 0$ . Further, on the interval $0 \le x \le \frac{\pi}{2}$ we have that $\arcsin(\cos(x)) = \arcsin(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$ .

Thus the square in question will have its upper right corner touching the curve $y = (\frac{\pi}{2} - x)^{2}$ , and by the symmetry of $f(x)$ this corner will have coordinates $(a,2a)$ where $a$ is such that

$2a = (\frac{\pi}{2} - a)^{2} \Longrightarrow 8a = (\pi - 2a)^{2} = \pi^{2} - 4\pi a + 4a^{2} \Longrightarrow 4a^{2}- (4\pi + 8)a + \pi^{2} = 0$

$\Longrightarrow a = \dfrac{4\pi + 8 \pm \sqrt{(4\pi + 8)^{2} - 16\pi^{2}}}{8} = \dfrac{4\pi + 8 \pm 4\sqrt{(\pi + 2)^{2} - \pi^{2}}}{8} = \dfrac{\pi}{2} + 1 \pm \sqrt{\pi + 1}$ .

Now as we require that $a \le \dfrac{\pi}{2}$ we take the negative root, i.e., $a = \dfrac{\pi}{2} + 1 - \sqrt{\pi + 1}$ .

The area of the desired square is then $(2a)^{2} = (\pi + 2 - 2\sqrt{\pi + 1})^{2} =$

$(\pi + 2)^{2} + 4(\pi + 1) - 4(\pi + 2)\sqrt{\pi + 1} = \pi^{2} + 8\pi + 8 - 4(\pi + 2)\sqrt{\pi + 1}$ .

Thus $\dfrac{AC}{CD} = \dfrac{2*8}{4*1} = \boxed{4}$ .

We note that the vertex of the square in the positive quadrant must be of the form $a, 2a$ and $0 < a < \frac{\pi}{2}$ .

$\begin{aligned} f(a) & = 2a \\ \Rightarrow \left( \sin^{-1} (\cos a) \right)^2 & = 2a \\ \sin^{-1} (\cos a) & = \sqrt{2a} \\ \cos a & = \sin (\sqrt{2a}) \\ \Rightarrow \sin a & = \sqrt{1-\sin^2 (\sqrt{2a})} = \cos (\sqrt{2a}) \\ \Rightarrow \cot a & = \tan (\sqrt{2a}) \\ \Rightarrow \frac{\pi}{2} - a & = \sqrt{2a} \\ a^2 - (\pi + 2)a + \frac{\pi^2}{4} & = 0 \\ \Rightarrow a & = \frac{\pi + 2 \pm \sqrt{(\pi+2)^2-\pi^2}}{2} \\ \Rightarrow a & = \frac{\pi}{2} +1 - \sqrt{\pi+1} \quad \quad \small \color{#3D99F6}{\text{The other solution is }> \frac{\pi}{2} \text{ and rejected.}} \end{aligned}$

The area of the square is $4a^2 = \pi^2 + 8\pi + 8 -4(\pi+2)\sqrt{\pi + 1}$ . $\quad \Rightarrow \dfrac{AB}{CD} = \dfrac{2 \times 8}{4 \times 1} = \boxed{4}$

The rightmost part of the function $f(x)= (\textrm{arcsin} \; \cos x)^2$ for $x \ge 0$ is equal to $g(x)= \left ( x-\frac{\pi}{2} \right )^2$ on the interval $[0, \pi]$ .

Let's look at these four points: $(0, \; 0), \: \: (a, \; 0), \: \: (a, \; g(a)), \: \:$ and $(0, \; g(a) )$ . When connecting the points we see a rectangle. The height of the rectangle is $g(a)$ and the length is $a$ . We want some value of $a$ such that the horizontal distance $D_1$ from $(0, \; g(a) )$ to $(a, \; g(a))$ is equal to half the vertical distance $D_2$ from $(a, \; 0)$ to $(a, \; g(a) )$ .

We can readily see that $D_1=a$ and $D_2=g(a)$ , and so we want to solve the equation

$D_1=\frac{D_2}{2} \: \Rightarrow \: a=\frac{1}{2} \left (a-\frac{\pi}{2} \right )^2.$

This is simply a quadratic with a solution of

$a= 1+\frac{\pi}{2} \pm \sqrt{1+\pi}$

However, the only feasible solution here is $1+\frac{\pi}{2} - \sqrt{1+\pi}$ , and this is the value of the side of the square. Now we want to find the area of this square and it is

$a^2= \frac{\pi^2}{4}+2+2 \pi-(2+\pi) \sqrt{1+\pi}$ , so $A=B=2,$ and $C=D=1$ and $\frac{AB}{CD}=\boxed{4}$

---

This is where I'm stuck. My solution is of the form $\frac{\pi^A}{4}+B \pi+B-C(\pi+A) \sqrt{\pi+D}$ and the OP's form is ${\pi^A}+B \pi+B-C(\pi+A) \sqrt{\pi+D}$ .

I don't understand why I differ from the OP's form. Can anyone explain?