Problem source

Two points are chosen independently at random on the perimeter (including the diameter) of a semicircle of unit radius. The area of the triangle whose vertices are these two points and the midpoint of the diameter is denoted by the random variable $A$. 
Show that the expected value of $A$ is $(2+\pi)^{-1}$.

Solution source

There are $3$ possible numbers of points on the curved part of the perimeter.

$0$: The area is $0$

$1$:


\begin{center}
    \begin{tikzpicture}
        \draw (-3,0) -- (3,0);
        \draw[domain=0:180, samples=101]
            plot({3*cos(\x)}, {3*sin(\x)});

        \coordinate (O) at (0,0);
        \coordinate (A) at (1.5, 0);
        \coordinate (B) at ({3*cos(70)}, {3*sin(70)});
        
        \filldraw (O) circle (1.5pt);
        \filldraw (A) circle (1.5pt) node[below] {$x$};
        \filldraw (B) circle (1.5pt);

        \draw (A) -- (O) -- (B) -- cycle;

        \pic [draw, angle radius=.5cm, angle eccentricity=1.3, "$\theta$"] {angle = A--O--B}; 
        
    \end{tikzpicture}
\end{center}

The area of the triangle is $\frac12 |x| \sin \theta$

Where $X$ is the point along the diameter which is $U[-1,1]$ and $\theta \sim U(0, \pi)$

Therefore 
\begin{align*}
    \mathbb{E}(A|\text{one on diameter}) &= \int_{0}^\pi \frac{1}{\pi} \int_{-1}^1\frac{1}{2}\frac12 |x| \sin \theta \d x \d \theta \\
    &= \frac{1}{2\pi}\frac12 \int_{0}^\pi  \sin \theta  \d \theta \cdot 2\int_{0}^1 x\d x  \\
    &=\frac{1}{2\pi}\cdot 2 \cdot \frac12 = \frac{1}{2\pi}
\end{align*}

$2$: If both are on the curved section


\begin{center}
    \begin{tikzpicture}
        \draw (-3,0) -- (3,0);
        \draw[domain=0:180, samples=101]
            plot({3*cos(\x)}, {3*sin(\x)});

        \coordinate (O) at (0,0);
        \coordinate (A) at ({3*cos(40)}, {3*sin(40)});
        \coordinate (B) at ({3*cos(100)}, {3*sin(100)});
        
        \filldraw (O) circle (1.5pt);
        \filldraw (A) circle (1.5pt);
        \filldraw (B) circle (1.5pt);

        \draw (A) -- (O) -- (B) -- cycle;

        \pic [draw, angle radius=.5cm, angle eccentricity=1.3, "$\theta$"] {angle = A--O--B}; 
        
    \end{tikzpicture}
\end{center}

Then the area is $\frac12 \sin \theta$ where $\theta = |\theta_1 - \theta_2|$ and $\theta_i \sim U[0, \pi]$

Therefore the area is

\begin{align*}
    \mathbb{E}(A|\text{none on diameter}) &= \int_{0}^\pi\frac{1}{\pi} \int_{0}^\pi\frac{1}{\pi} \frac12 \sin |\theta_1 - \theta_2| \d \theta_1 \d \theta_2 \\
    &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left (\int_{0}^{\theta_2}  \sin (\theta_2 - \theta_1) \d \theta_1-\int_{\theta_2}^{\pi}  \sin (\theta_2 - \theta_1) \d \theta_1 \right)\d \theta_2 \\
    &= \frac{1}{\pi^2}\frac12 \int_{0}^\pi \left [2\cos(\theta_2 - \theta_2)-\cos(\theta_2 - 0)-\cos(\theta_2 - \pi) \right]\d \theta_2 \\
    &= \frac{1}{\pi}
\end{align*}

Therefore the expected area is:

\begin{align*}
    \mathbb{E}(A ) &= \mathbb{E}(A|\text{one on diameter})\cdot \mathbb{P}(\text{one on diameter}) + \mathbb{E}(A|\text{none on diameter})\cdot \mathbb{P}(\text{none on diameter}) \\
    &= \frac{1}{2\pi}\mathbb{P}(\text{one on diameter}) + \frac{1}{\pi}\cdot \mathbb{P}(\text{none on diameter}) \\
    &= \frac{1}{2\pi} \cdot 2 \cdot \frac{\pi}{\pi + 2} \cdot \frac{2}{\pi + 2} + \frac1{\pi} \cdot \frac{\pi}{\pi + 2} \cdot \frac{\pi}{\pi+2} \\
    &= \frac{2 + \pi}{(\pi+2)^2} \\
    &= \frac{1}{\pi+2}
\end{align*}