Problem:
$f(x) = (x-1)(x-3)(x-5)$ intersects the x axis at $A(1,0)$, $B(3,0)$ and $C(5,0)$. A point $P(t,f(t))$ is selected on the curve such that $PB+PC$ is maximum and $t \in (3,5).$ Let $PX$ be tangent to $f(x)$ at $P$, then find $∠XPC$ + $∠XPB$.
My attempt:
Other than brute force (which I'm not sure will even work out in this case), all I can notice is that $PB + PC$ can be taken to be a constant $K$. Then we can say that $B$ and $C$ are the focii of an ellipse of having $a=K/2$ and $P$ is a point on that ellipse. However, I don't know how this helps me either.
Any help with my approach/an alternative approach to the question is appreciated.
Sketch of the solution:
Consider ellipse $\cal E$ with foci $B$ and $C$ which is tangent to the graph of $f$. Observe that the point of tangency is the point $P$. Indeed, each point $Q(s, f(s))$ lies inside the ellipse $\cal E$ which means that $QB+QC\le PB+PC$.
The tangent $\ell$ to the graph of $f$ at point $P$ is tangent to the ellipse $\cal E$. But this means that $\ell$ is bisector of exterior angle $BPC$ (this is well-known fact about ellipses). Using this fact, easy angle calculation gives $\angle XPC + \angle XPB = 180^\circ$.