Overlapping $\chi^2$ random variables I have 3 independent random variables that follow $\chi^2$ laws, with $m$ and $n$ the degrees of freedom:
\begin{align}A&\sim\chi^2_m\\B&\sim\chi^2_n\\C&\sim\chi^2_m\end{align}
I am interested to know the conditional probability distribution of $B+C$, knowing that $A+B=x$. In equation, this means:
$$f_{B+C|A+B}(y|x)=\frac{f_{A+B,B+C}(x,y)}{f_{A+B}(x)}$$
I can compute the denominator as it's simply a $\chi^2_{m+n}$ distribution, given that $A$ and $B$ are independent. However, I don't know how to compute the term in the numerator. It is not two independent $\chi^2$ distributions as the $B$ overlaps.

Allowing for some slight abuse of notation, you can find the numerator directly by solving the following integral.
\begin{align*}
f(A+B=x, C+B=y) &= f(A= x-B, C+y-B) \\[1.5ex]
&= \int_0^\infty f(A=x-b, C=y-b|B=b)f(B=b) db \\[1.5ex]
&= \int_0^\infty f_A(x-b)f_C(y-b)f_B(b) db \\[1.5ex]
&= \frac{1}{2^{m/2}\Gamma(m/2)}\frac{1}{2^{m/2}\Gamma(m/2)}\frac{1}{2^{n/2}\Gamma(n/2)} \times \\
&\quad\int_0^{\min\{x, y\}}(x-b)^{m/2-1}(y-b)^{m/2-1}b^{n/2-1}\exp\left(-\frac{1}{2}\left(x+y-b\right)\right) db \\[1.5ex]
&= c\int_0^{\min\{x, y\}}\left[(x-b)(y-b)\right]^{m/2-1}b^{n/2}\exp\left(\frac{b}{2}\right)db
\end{align*}
where $c = \left(2^{m+2/n}\Gamma(m/2)^2\Gamma(n/2)\right)^{-1}\exp(-(x+y)/2)$
I will revisit this later today if I have the time, but if @whuber cannot obtain an analytical solution then I doubt one exists.
