A mouse on a ring

We note that coordinates of the mouse $(x,y)$ are given by: $x = 1 - \cos t$ and $y = \sin t$ . Then, we have:

$\begin{aligned} \tan \left(f(t)\right) & = \frac{y}{x} = \frac{\sin t}{1-\cos t} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Using half-angle identities}} \\ & = \frac{2 \tan \left(\frac{t}{2}\right)/\left(1+\tan^2 \left(\frac{t}{2}\right)\right)}{1 - \left(1-\tan^2 \left(\frac{t}{2}\right)\right)/\left(1+\tan^2 \left(\frac{t}{2}\right)\right)} \\ & = \frac{2\tan \left(\frac{t}{2}\right)}{1+\tan^2 \left(\frac{t}{2}\right)-1+\tan^2 \left(\frac{t}{2}\right)} \\ & = \frac{1}{\tan \left(\frac{t}{2}\right)} = \cot \left(\frac{t}{2}\right) = \tan \left(\frac{\pi}{2}-\frac{t}{2}\right) \\ \Rightarrow f(t) & = \frac{\pi-t}{2} \\ \Rightarrow \int_0^\pi f(t) \space dt & = \int_0^\pi \frac{\pi-t}{2} dt = \frac{1}{2} \left[\pi t - \frac{t^2}{2}\right]_0^\pi = \frac{\pi^2}{4} \\ \Rightarrow A+B+C & = 1+2+4 = \boxed{7} \end{aligned}$

Let the mouse move for a time 't'.

Given velocity of mouse is 1, so in 't' time, it will cover a distance of 't' units on the circle.

Let the central angle subtended be 'x' radians.

We know $angle = \frac{arc}{radius} => x = t/1 => x = t.$

So the blue angle shown in the question is $\frac{180-t}{2} = 90 - \frac{t}{2} = f(t)$ .

So integrating it and then putting limit from $0 --> \pi$ , we get $\frac{\pi^{2}}{4}$

So, $A+B+C = 1+2+4 = \boxed7$