, 2024

Solution:

Claim 1. If $a, b, c, d$ are roots of $f(g(x))$, then one can permute them so that $a+b=c+d$.

Proof. Let $v$ be the point for which $g(v)$ is the local minimum or maximum. Note that if $g(x)=g(y)$, then $x$ and $y$ are symmetric around $v$, or $x+y=2v$. Moreover, if $a, b, c, d$ are roots of $f(g(x))$, then we can permute them so that $g(a)$ and $g(b)$ are equal to one root of $f$ and $g(c)$ and $g(d)$ are equal to another root of $f$. This means that $a+b=c+d=2v$.

In the case of our problem, if three roots are $r, s$, and $t$, then the fourth can be $r+s-t$, $r+t-s$, or $s+t-r$, with sum $r+s+t$. Using the given values, we get that the answer is $112+131+146=389$. Observe that these are all possible; indeed, if we let $g(x)=x^{2}-(r+s)x$, then $g(r)=g(s)$ and $g(t)=g(r+s-t)$. Now let $f(x)=(x-g(r))(x-g(t))$; then $f(g(x))$ has roots $r, s, t$, and $r+s-t$. The other two values are similarly achievable.