<s>[INST] 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)

 [/INST] 
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
$$R = \frac{\sqrt{33\csc\theta}}{\sqrt2-\csc\theta} = \frac67\sqrt{187\sqrt2}$$
</s>