I have been attempting to solve the problem here which is:
Given three concentric circles of radii 1, 2, and 3, respectively, find the maximum area of a triangle that has one vertex on each of the three circles.
A solution is here which is (I have edited it a little for clarity. Note that $A=1$, $B=2$ and $C=3$):
Let radii A,B, and C be at angles a,b, and c respectively. Position radius A on the positive x-axis at angle $a=0$ (no loss in generality). From the equation for triangle area
(1) area = $\frac12 BC \sin(b-c) + \frac12 CA \sin(c) + \frac12 AB \sin (2\pi -b)$.
Take the total partial of area w.r.t $b$ and $c$ and set equal to $0$. This gives
(2) $C \cos c = B \cos b$.
Also, from condition (2) extended radii are perpendicular to the triangle side. Next, the value of angle $b$ is determined. A little fancy geometry shows that $b$ is $225^o$ from $A$ ($-45^o$ in the third quadrant). From (2) angle $c$ is obtained.
I am happy with the expression for the triangle's area, and also with the differentiation and derivation of $C\cos c = B\cos b$.
But I don't see why the extended radii are perpendicular to the triangle's sides, which makes the centre of the concentric circles the orthocentre of the triangle. And I'm also not seeing the "fancy geometry" that gives the angle $b$, nor, indeed, why angle $b$ is constant.
Could someone please explain what's happening here?