Apologies if this is a duplicate, the problem is very hard to word in a search
The bivariate distribution of random variables X and Y is uniform over the triangle with vertices (1,0), (1,1), (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 $ABC$ is obtuse if and only if $$y>\frac{x^2-2x+2}{2-x}$$ Deduce that the probability ABC is obtuse is $3-4 \ln 2$
This question regards the "Deduce that" part, and I will quote the obtuse result
I'm going to (WLOG?) roll X first and say x=t. With this paramaterisation, I can write $$P(y>\frac{x^2-2x+2}{2-x})=\int_0 ^1{P(X=t)P(Y>\frac{t^2-2t+2}{2-t}|X=t)}$$ Since $X\sim U[0,1]$, then $P(X=t)=t dt$ and if you draw a sketch of the triangle, it is clear $Y\sim U[1-X,1]$
therefore $P(Y\le u|X=t)=\frac{u-(1-t)}{t}$ for $1-t\le u \le 1$, so $P(Y>\frac{t^2-2t+2}{2-t}|X=t)=1-\frac{1}{2-t}$ after taking the compliment and letting u equal the rational function.
Putting these into the integral I get $P(A)=\int^1_0{t(1-\frac{1}{2-t})}dt=\frac{1}{2}(3-4\ln2)$
I can't see at all where this factor of one half came in. It feels hard to believe that my method was wrong since my answer is very similar to the correct answer. I think I'm missing one last bit and I can't see what it is.
Sketch for the solution: an uniform distribution over some set means that the probability of some point to be chosen from some subset is proportional to the area or volume of this subset, hence in this case if $A$ is a subset of the triangle $T$ then $$ \Pr [(X,Y)\in A]=\frac{\text{ area of }A}{\text{ area of }T}=2 \cdot \text{area of }A\tag1 $$ as the area of the triangle is one-half. Now, we have that $$ \begin{align*} A:=&\left\{(x,y)\in T: y>\frac{x^2-2x+2}{2-x}\right\}\\ =&T\cap \left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\}\\ =&\left\{(x,y)\in \mathbb{R}^2: 0\leqslant x\leqslant 1 \,\land\, 1-x\leqslant y\leqslant 1\right\}\cap \left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\} \end{align*}\tag2 $$ Therefore setting $B:=\left\{(x,y)\in \mathbb{R}^2: y>\frac{x^2-2x+2}{2-x}\right\}$ we find that $$ \Pr [(X,Y)\in A]=2\int_{0}^1\int_{1-x}^1 \mathbf{1}_{B}(x,y)dydx\tag3 $$
The value of the inner integral is just the length of the interval $$ I_x:=\left[\max\left\{1-x,\frac{x^2-2x+2}{2-x}\right\},1\right]\tag4 $$ that is $$ \Pr [(X,Y)\in A]=2\int_{0}^1 \ell (I_x)dx\tag5 $$ where $\ell (I_x)$ is the length of $I_x$. So, depending of the value of $x\in[0,1]$ it length will be $1-(1-x)=x$ or $1-\frac{x^2-2x+2}{2-x}=\frac{x -x^2}{2-x}$, then the integral on (5) can be separated in (maybe just two) integrals where it integrand will be $x$ or $\frac{x-x^2}{2-x}$.