= Solution
Insert the union subtree from part (f) as another child of the original intersection:
$$
\boxed{P_{\mathrm{rounded}}=
H_5\cap H_6\cap H_4\cap H_1\cap(H_2\cup H_3)\cap(H_g\cup D_r).}
$$
Thus the <constructive solid geometry tree> has one intersection root, four original <half-space> leaves and two union children. The first union handles the inward corner at $C$; the second embeds the circular <fillet> at $B$. The <line segments> $AB$ and $BC$ now end at $M$ and start at $N$, with the chosen <circular arc> between them. All other boundary <line segments> are retained.
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