Two circles inscribed in a semicircle Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)

 A: 
It's not easy to believe, but
by pure luck, the solution is found as:
\begin{align}
R&=\frac67\sqrt{187\sqrt2}\approx 13.939
,\\
r_1&=\frac13\sqrt{187\sqrt2}\approx 5.42
,\\
r_2&=\frac38\sqrt{187\sqrt2}\approx 6.0983
.
\end{align}
Edit
$\require{begingroup} \begingroup$
$\def\i{\cdot\mathbf{i}}$
I just started to build the construction from
two radii of inscribed circles,
and it was a pure stroke of luck: a pair
$r_1=8,\ r_2=9$ looks nice to start with.
When the inradii are known,
the radius of semicircle $R$ can be easily found as
\begin{align}
R&=
\frac{2r_1 r_2(r_1+r_2+2\sqrt{2r_1 r_2})}{6r_1 r_2-r_1^2-r_2^2}
.
\end{align}
Using complex numbers to express coordinates of the points, we have:
\begin{align}
O&=0
\\
,\quad
O_1&=
O-\sqrt{(R-r_1)^2-r_1^2}+r_1\i
=-\tfrac{48}7\sqrt2+8\i
,\\
O_2&=
O+\sqrt{(R-r_2)^2-r_2^2}+r_2\i
=\frac{36}7\sqrt2+9\i
,\\
D&=O_{1x}=-\frac{48}7\sqrt2
,\\
E&=O_{2x}=\frac{36}7\sqrt2
,\\
F&=O_1+r_1\cdot\frac{O_1-O}{|O_1-O|}
=-\frac{864}{77}\sqrt2+\frac{144}{11}\i
,\\
G&=O_2+r_2\cdot\frac{O_2-O}{|O_2-O|}
=\frac{64}7\sqrt2+16\i
,\\
T&=O_1+r_1\cdot\frac{O_2-O_1}{|O_2-O_1}
=-\frac{144}{119}\sqrt2+\frac{144}{17}\i
.
\end{align}
Given these known points, the corresponding areas are
\begin{align}
s_1=S_{TFD}&=\frac{10368}{187}\sqrt2
=\frac{288}{187}\sqrt2\cdot 36
,\\
s_2=S_{TDE}&=\frac{864}{17}\sqrt2
=\frac{288}{187}\sqrt2\cdot 33
,\\
s_3=S_{TEG}&=\frac{1152}{17}\sqrt2
=\frac{288}{187}\sqrt2\cdot 44
,\\
s_4=S_{TGF}&=\frac{11520}{187}\sqrt2
=\frac{288}{187}\sqrt2\cdot 40
,
\end{align}
so the solution was obvious.
$\endgroup$
A: 
Let $EM=a$ be tangential to both circles at the point $E$ and the angle $\theta$. Then, the area $[ABE] = 33= a^2\sin\theta$ and
\begin{align}
[ABCD]&=153 =[AOD]+[BOC]=[COD] \\
& = \frac12Rr_1\cos\alpha + \frac12Rr_2\cos\beta + \frac12R^2\sin(\alpha+\beta)
\end{align}
Substitute $\sin\alpha = \frac{r_1}{R-r_1}$, $\sin\beta= \frac{r_2}{R-r_2}$ into above equation and, after simplifing
\begin{align}
153=aR\left( \frac1{(1-\frac{r_1}{R})(1-\frac{r_2}{R})}-1\right)\tag1
\end{align}
Note that $r_1 = a \tan\frac{\theta}2$, $r_2 = a \cot\frac{\theta}2$ and $AB=AO +OB$, or
\begin{align}
2a & = \sqrt{(R-r_1)^2-r_1^2} + \sqrt{(R-r_2)^2-r_2^2} 
= \sqrt{R^2-2aR\tan\frac{\theta}2}+ \sqrt{R^2-2aR\cot\frac{\theta}2}
\end{align}
which reduces to
$$R = \frac a{\sqrt2-\csc\theta}$$
Then, along with $a^2=33\csc\theta$, substitute $r_1$, $r_2$ and $R$ derived above into (1) to obtain the equation for $\theta$
$$\frac{153}{33} = \frac{\csc\theta}{\sqrt2-\csc\theta}\left(\frac1{3-4\sqrt2\csc\theta+3\csc^2\theta} -1\right)
$$
Solve to get the valid solution $\csc\theta = \frac{17}{12\sqrt2}$ and, in turn, the radius
$$R = \frac{\sqrt{33\csc\theta}}{\sqrt2-\csc\theta} = \frac67\sqrt{187\sqrt2}$$
