 Solution

Choose a circular fillet that removes the sharp convex tip at $B$. The unit directions from $B$ toward $A$ and $C$ are $(-\sqrt{3}/2,-1/2)$ and $(-1,0)$. Therefore the prescribed tangent points areThe 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 $(\sqrt{3},1)$. ThusConsequentlyIndeed $|M-K|=|N-K|=r$. The interior angle $ABC$ is $30^{\circ }$, and the same radius follows from $r=\frac{1}{2}\tan 15^{\circ }$. The selected circular arc runs from $M$ to $N$ through angles $-60^{\circ }\le \theta \le 90^{\circ }$ around $K$: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 $MN$ byBoth $M,N$ 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 isOnly 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 $AB,BC$.

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:Varying $w$ changes the bulge while fixing the tangent points; the circular choice has $w=\cos 75^{\circ }$. 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.