A weighted triangle is not uniform on its conditional slices

A random vector $(X,Y)$ has joint density $$ f_{X,Y}(x,y)= \begin{cases} C(x+y),&0<x<y<1,\\ 0,&\text{otherwise}. \end{cases} $$ Find $C$. Derive $f_X(x)$, $f_Y(y)$, $f_{X\mid Y}(x\mid y)$, and $f_{Y\mid X}(y\mid x)$, stating every support and the conditioning values for which each conditional density is defined. Verify directly that both conditional densities integrate to one. Compute exactly $$ \mathbb P(X<Y/2\mid Y=y) $$ for $0<y<1$, and $$ \mathbb P\!\left( Y<\frac{1+X}{2}\,\middle|\,X=x \right) $$ for $0<x<1$. Explain why neither calculation may assume a uniform conditional distribution merely because the support slice is an interval. Find $\mathbb E[X\mid Y=y]$ and verify $\mathbb E[\mathbb E[X\mid Y]]=\mathbb E[X]$ by explicit integration. Then compute $\mathbb E[Y]$, $\mathbb E[XY]$, and $\operatorname{Cov}(X,Y)$. Use either a conditional law or the covariance, together with the nonrectangular support, to give two valid dependence diagnostics. Density values on boundary sets may be chosen arbitrarily.

Normalize with $\int_0^1\int_0^y C(x+y)\,dx\,dy$. For fixed $x$, remember that $y$ starts at $x$, not at zero.