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A closed half-space is a set of the form $\{x:a^{T}x\le b\}$ for a nonzero vector $a$; it is a convex set and is closed.

This is the non-strict version of a half-space; replacing the inequality by a strict one gives the open version.

A rooted tree whose leaves describe primitive regions and whose internal nodes describe Boolean operations. A half-space leaf can hold an implicit equation and an inside-sign convention. Evaluate membership from the leaves toward the root; interval arithmetic can propagate conservative inside/outside bounds over a box instead of a single point.

Every allocation $(x,y)$ in the convex feasible region for a resource budget is a convex combination of its three extreme points:The condition $x+2y\le 6$ is exactly ${\lambda }_1\ge 0$; nonnegative $x,y$ give the other two nonnegative coefficients. Thus the half-space description and the convex hull description encode the same triangle. At $(2,1)$ all three coefficients are $1/3$, and at $(3,9/8)$ they are $1/8$, $1/2$, $3/8$. The green and orange points in the budget diagram therefore have two complementary interpretations: direct resource constraints and mixtures of extreme choices.

The three extreme points alone are not a convex set, because their midpoints are absent. Their convex hull fills the whole triangle. More generally, Caratheodory theorem bounds how many generating points are needed for one convex combination in a finite-dimensional real vector space; its geometric content is that a complicated collection can describe each attainable average with a short representation.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

A polyhedron that is also a convex set. A bounded full-dimensional convex hull of finitely many points is a convex polyhedron; each face lies in a supporting plane, with the whole solid in its inward half-space.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

The evanescent field in total internal reflection oscillates and advances in wave phase along the boundary while its wave amplitude decays away from it. With the coordinates above and time dependence $e^{-i\omega t}$, a TE polarization has complex amplitudeThe physical electric field is $\mathrm{Re}[\mathbf{E}(x,z)e^{-i\omega t}]$. The decaying sign excludes a field that grows without bound at $z\to \infty$. Dielectric-interface boundary conditions determine $A$; for nongrazing incidence it is generally nonzero.^[ref]

Zero normal energy flux follows directly from Maxwell equations. The wavevector is $(q,0,i\kappa )$; the relation $\mathbf{H}=(\mathbf{k}\times \mathbf{E})/(\omega {\mu }_0)$ givesConsequently, the cycle-averaged Poynting vector isThe electric field and $H_x$ are a quarter-cycle apart, so their product has zero cycle average. Tangential energy flux survives. This is a local calculation for the single decaying evanescent wave in the infinite second half-space.

To check reflected power independently, put $a=n_1{k}_0\cos \theta >0$. The existing transverse-electric Fresnel reflection coefficient becomes $r=(a-i\kappa )/(a+i\kappa )$. Its numerator and denominator are complex conjugates, so $|r{|}^2=1$, while $1+r=2a/(a+i\kappa )\ne 0$ supplies the surface electric field relative to the incident wave amplitude. Thus a unit optical reflectance does not require a zero transmitted field. The same normal power balance holds for TM polarization; its reflected wave phase generally differs from the TE polarization phase. A mixture can therefore change its polarization of an electromagnetic wave on reflection even though all normal incident power returns.

Let $C\in {\mathbb{R}}^{m\times n}$ and $d\in {\mathbb{R}}^m$. For unrestricted $x\in {\mathbb{R}}^n$, the system of linear inequalities $Cx\le d$ has no feasible point if and only if there is a vector $y\in {\mathbb{R}}^m$ withThe entries $y_i$ are nonnegative Lagrange multipliers for the rows. Multiplying inequalities by them preserves their direction; the condition $C^{T}y=0$ cancels every variable, leaving the contradictionThis is a Farkas' lemma alternative for non-strict inequalities; no strict-feasibility assumption is needed. See Boyd and Vandenberghe, Convex Optimization (2004), §5.8.3, pp. 261–262.

Three individually possible requirements, jointly impossible. Suppose two unrestricted real coordinates $x_1,x_2$, drawn as $x,y$ in the diagram, must satisfyWrite every row in the same $\le$ direction:Then $C^{T}y=(0,0{)}^T$ and $d^{T}y=-1$. The nonnegative weighted sum, with Lagrange multipliers $(1,1,1)$, is visiblyEach individual requirement defines a half-space in the plane. Even every pair has a feasible point: $(1,1)$ satisfies the first two, $(1,0)$ the first and third, and $(0,1)$ the second and third. Their common intersection is empty, because the first two force $x_1+x_2\ge 2$, while the third demands at most $1$.

Why cancellation is necessary for unrestricted variables. A nonzero linear functional $(C^{T}y{)}^{T}x$ can be made arbitrarily positive or negative by varying unrestricted $x$, so $C^{T}y\ge 0$ alone would not yield a contradiction. Equality is essential here. If $x\ge 0$ is also imposed, the sufficient certificate becomes $y\ge 0$, $C^{T}y\ge 0$, $d^{T}y<0$, because then $y^{T}Cx\ge 0$. This is the sign convention used in robust infeasibility under uniform constraint relaxation. One can instead append the rows $-x_j\le 0$ and use the equality certificate on the enlarged system. Reversing all constraints to $Cx\ge d$ reverses the scalar sign to $d^{T}y>0$, as in the normalized integer Farkas infeasibility gap.

Checking a solver's infeasibility claim. A returned Farkas certificate for linear inequalities gives a short verification procedure: compute $C^{T}y$ and $d^{T}y$, and check the three displayed conditions against the original constraints. Any positive multiple is another certificate; division by $-d^{T}y$ normalizes $d^{T}y$ to $-1$. With rational numbers as data and certificate, the tests can be exact. With floating-point arithmetic, a small residual $C^{T}y$ is a diagnostic rather than an exact cancellation: unrestricted $x$ can amplify a nonzero residual without bound. A rigorous certificate therefore needs exact cancellation or justified error bounds and additional bounds on the variables. The strict negative scalar is the margin of contradiction.

The same weighted sums give dual bounds. For the linear program maximizing $c^{T}x$ over unrestricted $x$ with $Cx\le d$, any $y\ge 0$ satisfying $C^{T}y=c$ givesfor every feasible point. Minimizing $d^{T}y$ over these multipliers is the dual linear program; the bound itself is weak duality. For example, maximize $x_1+x_2$ subject to $-x_1\le 0$, $-x_2\le 0$, $x_1+x_2\le 1$. The multipliers $(0,0,1)$ give the upper bound $1$, attained at $(1,0)$. The infeasibility example above uses the same row directions, but lower bounds of $1$ on both coordinates. Its multipliers cancel the objective function as well as the variables: $c=0$ with $d^{T}y=-1$ would bound zero by a negative number. Thus impossibility and linear programming optimality certificates use the same arithmetic. See Boyd and Vandenberghe, Convex Optimization slides, §§5.32–5.35 for infeasibility as a dual alternative.

Why every inconsistent system has such a certificate. Introduce a slack variable vector $s\ge 0$ and write $Cx+s=d$. Split the unrestricted vector as $x=u-v$ with $u,v\ge 0$. Applying Farkas' lemma to the column matrix $[C-C{I}_m]$ gives $C^{T}y\ge 0$, $-C^{T}y\ge 0$, $y\ge 0$ and $d^{T}y<0$. The first two conditions together are exactly $C^{T}y=0$. This reduction explains both the equality and the nonnegative row multipliers.

To isolate finite size from quantum vortex penetration, keep the same weak-field London equations but replace the half-space by an infinite slab $-d/2\le z\le d/2$. Put equal tangential magnetic fields $B_0\hat{\mathbf{x}}$ at its two faces. With $z=0$ now at the slab center, the even solution and its signed current density areThe hyperbolic cosine profile has equal boundary values, while the electric currents on the two sides have opposite signs. In particular,For the stipulated thickness $d=2{\lambda }_L$, the center retains about $0.648B_0$ and the mean is about $0.762B_0$, even though this is a Meissner state. For $d\gg {\lambda }_L$, the center magnetic field is exponentially small and the mean is about $2{\lambda }_L{B}_0/d$, reflecting the two surface layers. For $d\ll {\lambda }_L$, the mean tends to $B_0$: little screening is possible across such a short distance. These are analytic model examples, not material measurements. Real finite objects also need their demagnetizing field and edge geometry accounted for; the slab calculation only isolates the effect of thickness.

Frustrated total internal reflection occurs when a nearby second high-index dielectric couples to the evanescent field in total internal reflection, allowing power to cross a finite low-index gap. The extra boundary changes the dielectric-interface boundary conditions; the gap field need not be a single decaying evanescent wave. Consequently, the zero normal Poynting vector calculation for two infinite half-spaces does not forbid transmission in this three-medium geometry.

Harvard’s actual demonstration uses two right-angle glass prisms, a helium-neon laser, a small clamp and two perpendicular screens as detectors. The contact faces are cleaned and placed opposite each other, with the clamp protecting the corners. Increasing pressure narrows the intervening air film; the observed beam routing changes gradually from total internal reflection toward onward transmission. The broad-gap reflecting configuration supplies the comparison for the narrow-gap transmitting configuration. This is a qualitative readout: the page explicitly says the gap is not measured.^[ref]

The inference is that the lower-index region can mediate electromagnetic energy transfer when another boundary is nearby. The demonstration does not determine a numerical evanescent penetration depth, resolve gap uniformity, or establish a three-layer transmission coefficient. For the constructed $50^{\circ }$ example, $\delta \approx 178\mathrm{nm}$ identifies the scale over which the isolated-interface field remains appreciable; substituting $e^{-2g/\delta }$ for a gap transmission fraction would omit the changed boundary fields and their interference. A quantitative gap calculation must solve the full Maxwell equations with both boundaries.

A half-plane is a half-space in a two-dimensional Euclidean space, consisting of points on one side of a line. For distinct centroids, each of the two Voronoi cells defined by their Euclidean distances is a closed half-plane bounded by their perpendicular bisector, with an explicit convention needed to assign boundary ties.

A closed half-space consists of a hyperplane and one of its two sides in a real affine space. Replacing the weak inequality by a strict inequality gives an open half-space. Half-spaces are convex sets, and finite intersections of closed half-spaces are linear polyhedra.

A linear inequality compares a linear functional of a vector of unknowns with a constant, for example $a^{T}x\le b$ or $a^{T}x<b$ over the real numbers. For $a\ne 0$, its solution set is a closed or open half-space, respectively. The inequality $x+y\le 1$ allows all points on one side of the line $x+y=1$, including the line. Multiplication by a negative real number reverses the inequality, so $x\ge 1$ is equivalently $-x\le -1$.

A linear functional cannot exceed its best original value by mixing. For a finite nonempty set $S$ and $x={\sum }_i{\lambda }_i{s}_i$ in its convex hull,Every original point is itself in the convex hull, so equality between the two maxima follows. Moreover, a mixture attains the maximum exactly when all its positive weights are assigned to maximizing original points. Multiple maximizers can therefore produce a whole line segment or a higher-dimensional region of equally good choices.

For the eight samples above, use the linear functional $f(x,y)=x+y$. Its largest sample value is $f(C)=6$, and $C$ is the only maximizing sample. Thus the unique maximum over the whole convex polygon is also at $C$. The line $x+y=6$ is a supporting line: every point of the convex hull lies in the half-space $x+y\le 6$, and $C$ touches its boundary.

The point $q=(3.8,2.5)$ passes the coordinate-wise bounds $0\le x\le 4$, $0\le y\le 3$, but $f(q)=6.3>6$. This single inequality certifies that $q$ is outside the convex hull. A weighted mixture could not have a larger linear functional value than all its ingredients.

This explains a basic feature of linear programming: searching all mixtures of finitely many candidate vectors cannot improve the objective beyond the best candidate, though mixing equally good candidates can give more optimal choices. The conclusion depends on linearity. For the concave function $g(x,y)=-(x-2{)}^2-(y-1{)}^2$, the new point $p=(2,1)$ instead attains zero, strictly above its value at every original sample. A nonlinear objective can favor a mixture.

Two subsets of a Euclidean space are weakly linearly separable when a hyperplane with nonzero normal vector places them in opposite closed half-spaces. Points on the separating hyperplane may belong to either subset. Strict separation instead requires strictly opposite signs, excluding all boundary ties.

For two distinct centroids in a Euclidean space, equal squared Euclidean distances expand to the displayed affine hyperplane. It is their perpendicular bisector; each closed half-space consists of points at least as close to the centroid on that side. This proves weak linear separability after a two-centroid nearest-assignment step, allowing ties on the boundary. Coincident centroids are an essential exception: every point ties and the distance comparison supplies no nonzero normal vector.

Solutions to all three questions of the paper dated 10 June 2015. The constructive solid geometry calculations use closed half-spaces with nonpositive defining functions inside; the B-spline calculations use unit-spaced simple spline knots and explicitly oriented parameters.^[ref]

Put $a_0=2-\sqrt{3}$ and define six closed half-spaces $H_i=\{(x,y):f_i(x,y)\le 0\}$. Their implicit equations and outward normal vectors are:

The six vertices of a polygon traverse the boundary counterclockwise. Each displayed normal vector therefore points to the right of the corresponding directed boundary line segment; it need not be a unit vector. The zero level sets are full straight lines, while the actual boundary retains only the appropriate line segments.

The inward corner at $C$ prevents using the set intersection of all six half-spaces: that would incorrectly discard the upper-left part of the seat. Instead takeThe last factor permits either $y\le 1$ or $x\le 1$, removing only the upper-right region. Another way to understand the same constructive solid geometry model is as the set union of two convex polygons, obtained by distributing the last set intersection over this set union. Its constructive solid geometry tree has an intersection at the root and a union child for $H_2,H_3$.

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.

Insert the union subtree from part (f) as another child of the original intersection: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.