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A fillet rounds a corner or edge by inserting a tangent transition. In a planar circular construction, intersect normal lines through the prescribed tangent points to find a candidate circle centre. The two resulting radii must agree; otherwise those prescribed points are incompatible with one circular fillet. Choosing the retained side and local clipping region is part of the design: intersecting the entire original solid with the fillet disk generally removes unrelated geometry. Tangent joining need not make curvature continuous.

A normal section of a regular embedded surface at a point $p$ is its local intersection with a plane containing the normal line at $p$ and a chosen tangent line. Its curvature at $p$ measures the normal curvature in that direction. The plane need not contain the surface normal vector at other points of the section.

A stronger condition does imply a geodesic: if one fixed plane contains the surface normal vector at every point of a regular intersection curve, then its arc-length parametrization has acceleration in that plane and perpendicular to the curve tangent. It is therefore normal to the surface, by the ambient acceleration criterion for a surface geodesic.

A normal section need not have that stronger property. For $z=x^2/2+y^2$, the plane $y=x$ contains the surface normal at the origin and gives $\gamma (t)=(t,t,3t^2/2)$. Away from $t=0$, the determinant of ${\gamma }^{\prime }$, ${\gamma }^{\prime \prime }$ and the surface normal $(-t,-2t,1)$ is $3t\ne 0$. No change of parameter can then make the acceleration surface-normal, so this section is not a geodesic near the origin.

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.

A critical point satisfies $df=0$; it is nondegenerate when the Hessian on its two-dimensional tangent space is nonsingular. For a smooth function with only nondegenerate critical points on a compact surface, the Morse theory formula iswhere $m_i$ counts the critical points whose Hessian has $i$ negative eigenvalues.

For $f(x)=|x{|}^2$, differentiation in a tangent direction $v$ gives $d{f}_x(v)=2x\cdot v$. This vanishes for all $v\in T_{x}X$ exactly when the position vector is perpendicular to that tangent plane, equivalently the normal line at $x$ contains the origin. This includes $x=0$, if it is on the surface. On a torus $\chi =0$, while a minimum and a maximum give $m_0,m_2\ge 1$. Thus $m_1=m_0+m_2\ge 2$ and the total is at least four. Hence at least four surface normals pass through the origin under the stated nondegeneracy assumption.

Write the generating circle as center $(a,R)$ and radius $r$, with $R>r>0$. A parametrization and unit normal areTake $P=(a,R/2,0)$. If $P-X=tn$, its first coordinate gives $(r+t)\cos u=0$. The alternative $t=-r$ would force the radial distance of $P$ from the axis to be $R$, contrary to $R/2$. Thus $\cos u=0$. The remaining equations and the nonzero radial distance of $P$ force $v=0$ or $\pi$. Each of the two choices of $u$ and two of $v$ works, giving exactly four distinct points. Take $Q=(a,0,0)$ instead. All points on each of the circles $u=\pi /2$ and $u=3\pi /2$ have radial normals through $Q$, giving infinitely many.

Using the normal vector from i, the normal line isFor $0<\theta <\pi$, the $y$- and $z$-coordinates vanish together exactly when $\lambda =-1$. At that point, the $x$-coordinate isThusAt $\theta =0$ or $\theta =\pi$, both the position vector and normal vector lie along the $x$-axis, so $L$ is the whole $x$-axis in either case.

The required lines are the $x$-axis and every line through the origin in the plane $x=0$. Explicitly,The $x$-axis meets the ellipsoid at its two poles and is a normal line at each. A line of the second kind meets it at the antipodal points $(0,\pm \cos \varphi ,\pm \sin \varphi )$, whose normal vectors are parallel to that line.

To prove completeness, suppose a common normal line passes through distinct points $P=\mathbf{r}(\theta ,\varphi )$ and $Q=\mathbf{r}({\theta }^{\prime },{\varphi }^{\prime })$. If either is a pole, b forces the line to be the $x$-axis. Otherwise b says that the line meets the $x$-axis at $(-\cos \theta /\sqrt{2},0,0)$ and at $(-\cos {\theta }^{\prime }/\sqrt{2},0,0)$. A line other than the $x$-axis has at most one such intersection, so $\cos \theta =\cos {\theta }^{\prime }$.

The points $P,Q$ consequently have equal $x$-coordinates. Since they are distinct, their joining line has zero $x$-component in its direction. Its normal vector at $P$ therefore has $\sqrt{2}\cos \theta =0$. Hence $P,Q$ lie in $x=0$, and the normal line there passes through the origin. This gives exactly the listed family.

A $C^1$ embedded hypersurface $M\subseteq {\mathbb{R}}^n$ is locally carried by a $C^1$ coordinate diffeomorphism to an open subset of ${\mathbb{R}}^{n-1}\times \{0\}$. Equivalently, near each of its points it has the form $F^{-1}(0)$ for a $C^1$ function with nonzero gradient there.

The Lagrange multiplier theorem states that if a differentiable function $g$ has a local extremum on such a hypersurface at $x_0$, then $dg(x_0)$ annihilates the tangent space. In a regular local level-set description this isfor some real $\lambda$. The regularity condition $\nabla F(x_0)\ne 0$ is essential; it is what identifies the normal line and permits the local constraint chart.