Maximum area of a triangle [Edit:I had a couple of links to the original probelm but they have gone the way of all things.]
I have been attempting to solve this problem:

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.


Here is a partial solution (not my own) which 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?
 A: Fix points $A$ and $B$ and maximise the area by varying point $C$. This means maximising the height of the triangle with base $AB$. 
With $C$ at the maximum point draw a line parallel to $AB$ through $C$. If it is not a tangent then there is a point on the outer circle which gives a greater area - contradiction. It is therefore a tangent and the perpendicular to $AB$ through $C$ is perpendicular to the tangent at the point it touches the circle (tangent is parallel to $AB$), and therefore passes through the centre $O$.
A: Some algebraic computation
The accepted answer by robjohn explains nicely why the origin has to be the orthocenter. I want to suggest an alternative approach to the computation that follows. To do so, I want to use coordinates for the points in question:
$$O=(0,0)\qquad A=(1,0)\qquad B=(x,b)\qquad C=(x,c)$$
The use of the same $x$ coordinate for both $B$ and $C$ already ensures that $OA$ is perpendicular to $BC$. You have three conditions to determine these three variables:
\begin{align*}
\lVert B\rVert&=2 & x^2+b^2 &= 2^2 \\
\lVert C\rVert&=3 & x^2+c^2 &= 3^2 \\
\langle C-A,B\rangle&=0 & x^2+bc-x &= 0
\end{align*}
The scalar product expresses the condition $(C-A)\perp B$, while the third condition $(C-B)\perp A$ is already implied by these conditions (since the altitudes of any triangle meet in a single point).
You can feed this to Wolfram Alpha to obtain numeric approximations of the relevant coordinates. But you can also eliminate variables, e.g. using resultants, to obtain polynomials which describe your solutions:
\begin{align*}
x^3 - 7x^2 + 18 &= 0 \\
b^6 + 37b^4 - 92b^2 + 36 &= 0 \\
c^6 + 22c^4 - 387c^2 + 1296 &= 0
\end{align*}
No ruler and compass solution possible
These are irreducible cubic polynomials in $x,b^2,c^2$ resp. so they cannot be the result of a ruler and compass construction. Every ruler and compass construction which starts from integer coordinates (like the points $O,A$ and integer radii) can only result in coordinates which are sums, differences, products, quotients or square roots of other constructible coordinates. Cubic roots are not possible.
So the main result of this investigation is this:
The triangle of maximal area cannot be constructed using ruler and compass.
Anyone looking for a “fancy geometry” construction to achieve this can stop now, unless he is knowingly looking for a good approximation only.
The value of the maximal area
Using a little bit of computation on the above coorinates using algebraic numbers in sage shows that the area $a$ satisfies the equation
$$16a^6-392a^4+133a^2+900=0$$
so that again is a cubic polynomial in $a^2$. Feeding that polynomial into WA (surprisingly only when omitting commands like “solve”) I get an ugly but exact expression for the solution in question:
$$a=\frac12\sqrt{\frac13\left(98+
\frac{9205}{\sqrt[3]{833939+7974i\sqrt{1329}}}+
\sqrt[3]{833939+7974i\sqrt{1329}}\right)}$$
This is interesting because the post you referred to claims a much simpler formula, but that was based on the incorrect $45°$ assumption robjohn already disproved. The last post by ferret, however, results in the same area I computed myself. So they, too, eventually found an exact solution.
A: Let the center be $O$. Let the foot of perpendicular from $B,C$ to $AO$ be $P,Q$. Then $B \cos b = B \cos \angle AOB = - B \cos \angle POB = - BP$, similarly $C \cos c = -CQ$ which means that $P=Q$, and hence $AO \perp BC$. (Note that this proof doesn't hold if P and Q lie on opposite sides of O, but I think that case shouldn't be too hard)
A: Here is an approach, probably far from optimal.
Let $O$ be the centre of the three circles.
First it is easy to see that if $C'$ is so that $COC'$ is a diameter, from the triangle $ABC$ and $ABC'$, the one which contains $O$ inside has more area. So WLOG we can assume $O$ is inside $ABC$. 
Then
\begin{equation}
\begin{split}
Area(ABC)&=Area(ABO)+Area(BOC)+Area(AOC)\\
&=\frac{1}{2} AO \cdot BO \cdot \sin(AOB)+ \frac{1}{2} AO \cdot CO \cdot \sin(AOC)+\frac{1}{2} BO \cdot CO \cdot \sin(BOC)\\
&=   \sin(AOB)+\frac{3}{2}  \sin(AOC)+ 3 \sin(BOC)
\end{split}
\end{equation}
Thus, the problem can be reduced to finding the maximum of the function $f(x,y,z)=\sin(x)+\frac{3}{2} \sin(y)+3 \sin (z)$, when $x,y,z >0$ and $x+y+z=2\pi$. Lagrange multipliers should finish it easely.
