Random Triangle

Clearly, the locus of $C$ would be a circle. Consider the figure shown. The probability that $\angle AOC = \theta = \frac{d \theta}{2 \pi}$ .

As median divides a triangle in two equal parts in terms of area.

Required area = $A = 2ar(\triangle AOC) = 2 \times \frac{1}{2} R \times R |\sin \theta| = R^2 | \sin \theta|$

Hence, expected area = $\displaystyle \int A dP = \displaystyle \int_{0}^{2 \pi} R^2 |\sin \theta| \frac{d \theta}{2 \pi}$

= $\frac{2R^2}{\pi} = \frac{d^2}{2 \pi}$

Here, $d^2 = 2^2 + 4^2 = 20$

Hence, expected area = $\frac{20}{2 \pi} = \frac{10}{\pi}$

Without loss of generality, we can consider $A \big(-\frac{\sqrt{20}}{2},0 \big)$ and $B\big(\frac{\sqrt{20}}{2}, 0\big)$ . Being $C(x,y)$ and $AC \perp AB$ , we have that

$\displaystyle \bigg(x-\frac{\sqrt{20}}{2}, y\bigg)\cdot \bigg(x+\frac{\sqrt{20}}{2},y \bigg)=0 \hspace{5pt} \Longrightarrow \hspace{5pt} x^2+y^2=5$

that is, clearly, a circumference. It can be also written as

$\displaystyle c(\theta)=\bigg( \frac{\sqrt{20}}{2}\cos{\theta}, \frac{\sqrt{20}}{2}\sin{\theta}\bigg), \hspace{5pt} \theta \in [0, 2\pi]$ .

The PDF is simply $\displaystyle f(\theta)=\frac{1}{2\pi}$ . The area $A$ of the triangle is

$\displaystyle A(\theta)=\frac{1}{2} \cdot \sqrt{20} \cdot \frac{\sqrt{20}}{2} |\sin{\theta}|=5|\sin{\theta}|$

where

$\displaystyle |\sin{\theta}|=\begin{cases} \sin{\theta}, & \text{for } 0 \leq \theta \leq \pi \\[3pt] -\sin{\theta}, & \text{for } \pi < \theta \leq 2\pi \end{cases}$ .

Expected value can be calculated

$\displaystyle \mathbb{E}[A(\theta)]=\int_{0}^{2\pi} A(\theta) f(\theta) d\theta =\int_{0}^{2\pi} \frac{5}{2\pi} |\sin{\theta}| d\theta =\int_{0}^{\pi} \frac{5}{2\pi} \sin{\theta} d\theta + \int_{\pi}^{2\pi} -\frac{5}{2\pi} \sin{\theta} d\theta=\frac{10}{\pi}$

Eventually

$a=10$ , $b=1$ , $a+b=\boxed{11}$