Problem source

\begin{questionparts}
\item A rod of unit length is cut into pieces of length $X$ and $1-X$; the latter is then cut in half. The random variable $X$ is uniformly distributed over $[0,1].$ For some values of $X$ a triangle can be formed from the three pieces of the rod. Show that the conditional probability that, if a triangle can be formed, it will be obtuse-angled is $3-2\sqrt{2.}$
\item The bivariate distribution of the random variables $X$ and $Y$ is uniform over the triangle with vertices $(1,0),(1,1)$ and $(0,1).$ A pair of values $x,y$ is chosen at random from this distribution and a (perhaps degenerate) triangle $ABC$ is constructed with $BC=x$ and $CA=y$ and $AB=2-x-y.$ Show that the construction is always possible and that $\angle ABC$ is obtuse if and only if 
\[
y>\frac{x^{2}-2x+2}{2-x}.
\]
Deduce that the probability that $\angle ABC$ is obtuse is $3-4\ln2.$
\end{questionparts}

Solution source

\begin{questionparts}
The side lengths are $X, \frac{1-X}{2}, \frac{1-X}{2}$, so the triangle is isoceles. Since we cannot have two obtuse angles in a triangle, the only way we can have an obtuse angle is if the angle opposite the side with length $X$ is larger than $90^{\circ}$, ie

$\frac{X}{\frac{1-X}2} > \sqrt{2} \Rightarrow 2X > \sqrt{2}-\sqrt{2}X \Rightarrow (2+\sqrt{2})X > \sqrt{2} \Rightarrow X > \frac{\sqrt{2}}{2+\sqrt{2}} = \sqrt{2}-1$.

A triangle can be formed if $\frac{1-X}{2}+\frac{1-X}{2} > X \Rightarrow 1-X > X \Rightarrow X < \frac12$.

Therefore we want the probability $X > \sqrt{2}-1$, subject to $X < \frac12$, ie $\frac{\frac12-(\sqrt{2}-1)}{\frac12} = 3 - 2\sqrt{2}$


\item 
\begin{center}
    \begin{tikzpicture}[scale = 2]

        \draw[->] (-0.1, 0) -- (2, 0);    
        \draw[->] (0, -0.1) -- (0, 2);    

        \draw[thick] (1,0) -- (1,1) -- (0,1) -- cycle;
        
    \end{tikzpicture}
\end{center}

The construction is possible if $x + y > 2-x-y \Rightarrow x+y > 1$ (which is as the triangle is above the diagonal line), and $x + (2-x-y) > y \Rightarrow 1 > y$ (true as the triangle is below the horizontal line) and $y + (2-x-y) > x \Rightarrow 1 > x$ (true as the triangle is left of the vertical arrow).

By the cosine rule:

\begin{align*}
&& y^2 &= x^2 + (2-x-y)^2 - 2 x (2-x-y) \cos \angle ABC \\
\Rightarrow && \cos \angle ABC &= \frac{x^2+(2-x-y)^2 - y^2}{2x(2-x-y)} \\
&&&= \frac{4+2x^2-4x-4y+2xy}{2x(2-x-y)} \\
\underbrace{\Rightarrow}_{\cos \angle ABC < 0} && 0 &> 4+2x^2-4x-4y+2xy \\
\Rightarrow && 0 &> 2x^2-4x+4 - 2(x-2)y \\
\Rightarrow && y &> \frac{x^2-2x+2}{2-x} \\
&&&= -x + \frac{2}{2-x}
\end{align*}


\begin{center}
    \begin{tikzpicture}[scale = 2]

        \draw[->] (-0.1, 0) -- (2, 0);    
        \draw[->] (0, -0.1) -- (0, 2);    

        \draw[thick] (1,0) -- (1,1) -- (0,1) -- cycle;

        \draw[thick, blue, smooth, domain=0:1, samples=100] 
            plot (\x, {-\x + 2/(2-\x)});
        
    \end{tikzpicture}
\end{center}

Therefore the area we want is:

\begin{align*}
A &= 1 - \int_0^1 \left ( -x + \frac{2}{2-x} \right)\d x \\
&= 1 - \left [-\frac12 x^2 - 2 \ln(2-x) \right]_0^1 \\
&= 1 + \frac12 -2 \ln 2 \\
&= \frac32 - 2 \ln 2
\end{align*}

Therefore the relative area is: $\frac{\frac32 - 2 \ln 2}{1/2} = 3 - 4 \ln 2$

\end{questionparts}