= Solution
Choose a circular <fillet> that removes the sharp convex tip at $B$. The unit directions from $B$ toward $A$ and $C$ are $(-\sqrt3/2,-1/2)$ and $(-1,0)$. Therefore the prescribed tangent <points> are
$$
M=\left(2-\frac{\sqrt3}{4},\frac34\right),\qquad
N=\left(\frac32,1\right).
$$
The <circle> centre $K$ lies on the <normal line> to $BC$ through $N$, so $K=(3/2,k)$. Its radius to $M$ must be perpendicular to $AB$, whose <direction vector> is $(\sqrt3,1)$. Thus
$$
(M-K)\cdot(\sqrt3,1)=0
\ \Longrightarrow\
\sqrt3\left(\frac12-\frac{\sqrt3}{4}\right)+\frac34-k=0
\ \Longrightarrow\ k=\frac{\sqrt3}{2}.
$$
Consequently
$$
\boxed{K=\left(\frac32,\frac{\sqrt3}{2}\right),\qquad
r=1-\frac{\sqrt3}{2}=\frac{2-\sqrt3}{2}.}
$$
Indeed $|M-K|=|N-K|=r$. The interior angle $ABC$ is $30^\circ$, and the same radius follows from $r=\tfrac12\tan15^\circ$. The selected <circular arc> runs from $M$ to $N$ through angles $-60^\circ\le\theta\le90^\circ$ around $K$:
$$
\gamma(\theta)=K+r(\cos\theta,\sin\theta).
$$
The $150^\circ$ <circular arc> facing $B$ is the rounding; the complementary arc is not part of the boundary.
Define the disk and a half-plane through the chord $MN$ by
$$
D_r=\left\{f_c\le0\right\},\quad
f_c=\left(x-\frac32\right)^2+\left(y-\frac{\sqrt3}{2}\right)^2-r^2,
$$
$$
H_g=\{g\le0\},\qquad
g=x-\frac32+(2-\sqrt3)(y-1).
$$
Both $M,N$ satisfy $g=0$, and $g(B)=1/2>0$. The clipping <half-space> $H_g$ retains the part away from $B$, while the disk restores the rounded cap. The required local <constructive solid geometry tree> is
$$
\boxed{T_{\mathrm{blend}}=H_g\cup D_r,\qquad
P_{\mathrm{rounded}}=P\cap T_{\mathrm{blend}}.}
$$
Only the portion of $P$ on the $B$ side of $MN$ can be removed. To see the locality, $g$ is negative at every original <vertex of a polygon> except $B$, and vanishes on the two adjacent sides exactly at $M,N$. In the resulting <triangle> $BMN$, the disk keeps the curved cap next to $MN$ and discards the tip at $B$. A global <set intersection> with $D_r$ alone would destroy most of the seat. The outward <normal vector> on the new boundary is $\nabla f_c=2((x,y)-K)$; at $M,N$ it has the same direction as the outward <normal vectors> of $AB,BC$.
\Image[/past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/geometric-modelling-2015-fillet.png]
{title=Circular fillet at the sharp corner, tangent radii, chord clipping line, and local union subtree}
The tangent constraints alone do not insist on a <circle>; other <conic sections> can give different bulges. For example, a positive-weight <rational quadratic Bézier curve> with handles $M,B,N$ has the required tangent directions:
$$
R_w(s)=\frac{(1-s)^2M+2ws(1-s)B+s^2N}{(1-s)^2+2ws(1-s)+s^2},\qquad 0\le s\le1,\quad w>0.
$$
Varying $w$ changes the bulge while fixing the tangent <points>; the circular choice has $w=\cos75^\circ$. Once a circular <fillet> and these two tangent <points> are chosen, its centre and radius have no remaining freedom. The half-plane sign and arc selection are also essential: reversing them chooses a different retained region. This construction gives matching tangent directions, or first-order <geometric continuity>, at $M,N$; the <curvature> jumps from zero on each straight side to $1/r$ on the <circle>.
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