The reference area

For a polytope with face areas \(S_j\), outward unit normals \(n_j\), and a unit vector \(u\), Cauchy’s projection formula takes the form \[ \operatorname{area}(\pi_{u^\perp}K)=\frac12\sum_j S_j|n_j\cdot u|. \tag{10}\] See Martini (Martini 1991, 83, equation (1)) for the projection-body formulation. The elementary polytope proof is to count the two endpoints of almost every line parallel to \(u\): projecting all faces counts its shadow twice, whereas edges and faces parallel to \(u\) contribute zero area.

Each face of (9) has area \(4\sqrt3\), and its normal is one of the normalizations of \((\pm1,0,-1/h)\) and \((0,\pm1,1/h)\). For a unit vector \(u=(u_1,u_2,u_3)\), formula (10) and \(|a+b|+|a-b|=2\max(|a|,|b|)\) give \[\operatorname{area}(\pi_{u^\perp}K) =4h\left(\max\{|u_1|,|u_3|/h\} +\max\{|u_2|,|u_3|/h\}\right).\] If the two maxima are \(X,Y\), then \(X^2+Y^2\ge u_1^2+u_2^2\) and \(XY\ge u_3^2/h^2\). Since \(h^2=2\), it follows that \((X+Y)^2\ge1\). Equality is attained at \(u=(1,0,0)\), so \[ A_{\min}(K)=4h. \tag{11}\] The upper half of \(K\) is covered by an \(x\)-parallel cylinder of base area \(h\), and the lower half by a \(y\)-parallel cylinder of the same area. Their total cost is \(2h=A_{\min}(K)/2\).

The perturbed segments cover the whole tetrahedron

For \(q\in[-1,1]\), define \[G(q)=\frac{1+q^2}{2},\qquad H(q)=\frac{q^2+q^4}{2},\qquad R_\varepsilon(q)=1+\varepsilon^2\bigl(H(q)+5\varepsilon\bigr),\quad \varepsilon>0.\] The upper sweep is \[ U_\varepsilon=\{(x,\rho q-\varepsilon xG(q), h(1-\rho+\varepsilon xq)): |x|\le2,\ |q|\le1,\ 0\le\rho\le R_\varepsilon(q)\}. \tag{12}\] Its intercept ray starts at \((0,0,h)\) in direction \((0,q,-h)\). The identity \(G'(q)=q\) makes the derivative of its segment direction \((1,-\varepsilon G(q),h\varepsilon q)\) equal to \(-\varepsilon(0,q,-h)\), exactly the radial alignment required by Theorem 1. The constant term in \(G\) gives \(G(q)-q^2=(1-q^2)/2\), which vanishes at both angular endpoints. This will keep the endpoint levels unchanged when we select a line through a point of \(K\).

The sweep is compact. Let \(\mathcal R(x,y,z)=(y,-x,-z)\), an orthogonal map preserving \(K\). Its image \(\mathcal R U_\varepsilon\) is the lower sweep, which can equivalently be written as \[ \{(\sigma p+\varepsilon yG(p),y,h(-1+\sigma+\varepsilon yp)): |y|\le2,\ |p|\le1,\ 0\le\sigma\le R_\varepsilon(p)\}. \tag{13}\] Indeed, apply \(\mathcal R\) in (12) and replace its free coordinate \(x\) by \(-y\). We will show that the two sweeps overlap enough to cover \(K\).

Finite bases and the weighted cost

We now hold \(\varepsilon\) fixed. In the intercept plane \(P=\{x=0\}\), the upper sweep has the data \[p_0=(0,0,h),\quad e=(1,0,0),\quad a=(0,0,-h),\quad b=(0,1,0),\quad L=2,\] and \[w(q)=(0,-\varepsilon G(q),h\varepsilon q),\qquad w'(q)=-\varepsilon(a+qb).\] Thus the hypothesis of Theorem 1 holds with \(\lambda=-\varepsilon\), \(\kappa=h\), and radial cap \(R_\varepsilon\). Also \[\lVert e+w(q)\rVert^2=1+\varepsilon^2D_0(q),\qquad D_0(q)=G(q)^2+2q^2.\] For every partition into \(N\) intervals, the theorem and the orthogonal map \(\mathcal R\) cover \(K\) by \(2N\) cylinders with nondegenerate compact triangular bases. As the mesh tends to zero, their total areas converge to \[ 2hI(\varepsilon),\qquad I(\varepsilon)=\int_{-1}^1 \frac{R_\varepsilon(q)^2}{2\sqrt{1+\varepsilon^2D_0(q)}}\,dq. \tag{18}\] Taking actual swept projections instead gives compact bases with the same limiting total area, for any choice of freezing tags. Their finite areas may differ from those of the triangles.

The saving and the final finite choice

It remains to put the limiting cost strictly below \(2h\). The expansion is uniform for \(q\in[-1,1]\): \[\frac{R_\varepsilon(q)^2}{2\sqrt{1+\varepsilon^2D_0(q)}} =\frac12+\varepsilon^2\left(H(q)-\frac{D_0(q)}4\right) +5\varepsilon^3+O(\varepsilon^4).\] Direct integration gives \[\int_{-1}^1H(q)\,dq=\frac8{15},\qquad \int_{-1}^1\frac{D_0(q)}4\,dq=\frac{17}{30}.\] Therefore \[ I(\varepsilon)=1-\frac{\varepsilon^2}{30}+10\varepsilon^3+O(\varepsilon^4)<1 \tag{19}\] for all sufficiently small positive \(\varepsilon\). The cubic cap padding that guaranteed overlap is smaller than the quadratic projection saving.

Choose one such \(\varepsilon\) and put \(\Delta=2h(1-I(\varepsilon))>0\). For the triangular covers, the limit (18) gives a sufficiently fine finite partition, with \(N\) intervals, whose cost is less than \(2hI(\varepsilon)+\Delta/2<2h\). Proposition 4 and Theorem 1 give coverage of every point of \(K\) for this same \(N\). The actual swept projections also give finite covers within this gap by their convergence to the same limit.

By (11), the triangular cover has cost strictly below \(A_{\min}(K)/2\). A Euclidean similarity transports it to every regular tetrahedron of positive edge length, preserving triangularity and multiplying both areas by the same factor. Finally each individual projection area is at least \(A_{\min}(K)\). For the axes and bases of any one of these finite covers, \[\sum_i\frac{\operatorname{area}(B_i)}{\operatorname{area}(\pi_{u_i^\perp}K)} \le\frac{\sum_i\operatorname{area}(B_i)}{A_{\min}(K)}<\frac12.\] This proves the two negative resolutions stated in the introduction.

Explicit polygonal alternatives with the same limiting cost are given in Appendix 7.

Euclidean coordinates and the cost normalization

All notation in this section is local. We construct the cover for a regular tetrahedron of edge length \(2\sqrt{2}\); similarity then gives the cover at every positive edge length, since both the covering cost and \(A_{\min}\) scale by the square of the similarity ratio. Put \(\beta=1/\sqrt{2}\) and write \[P(x,y,t)=(\beta x,\beta y,t),\qquad K=\{P(x,y,t):-1\le t\le1,\ |x|\le1-t,\ |y|\le1+t\}.\] The map \(P\) specifies physical Euclidean coordinates; \((x,y,t)\) are parameters, so Euclidean lengths and areas must be computed after applying \(P\). The four vertices are \(P(\pm2,0,-1)\) and \(P(0,\pm2,1)\). This is the image of the edge-four tetrahedron in Section 3 under the Euclidean similarity \((X,Y,Z)\mapsto(\beta X,\beta Y,-\beta Z)\). Thus all six edges have length \(2\sqrt2\), and (11) gives \(A_{\min}(K)=\beta^2(4\sqrt2)=2\sqrt2\). In the computations below we measure areas in units of \(\beta^2/\sqrt2\); this is a fixed cost normalization, distinct from dividing each cylinder area by the projection in its own direction: \[ A_{\min}(K)=2\sqrt{2},\qquad \text{normalized cost}=\frac{\text{perpendicular base area}}{\beta^2/\sqrt{2}}. \tag{20}\] The desired threshold in normalized cost is \(4\).

We will repeatedly use the following exact conversion. For a compact planar set \(S\subset\mathbb R^2\) and \(v\in\mathbb R^2\), the cylinder through \(P(S,0)\) in direction \((\beta v,1)\) has perpendicular base \(\pi_{(\beta v,1)^\perp}P(S,0)\). Its normalized cost is \[ |S|\,g(v),\qquad g(v)=\frac{\sqrt{2}}{\sqrt{1+\beta^2|v|^2}}. \tag{21}\] Indeed, the horizontal area is \(\beta^2|S|\), and orthogonal projection multiplies it by the absolute inner product of the two unit plane normals, namely \((1+\beta^2|v|^2)^{-1/2}\). Projection changes neither the cylinder nor its direction. Normalizing that direction gives a unit axis as required.

Compact beams and their finite coverings

For \(0<a<1/2\), define \[R_a=2-a,\qquad r(p)=(p,1-p),\qquad v(p)=(1,-1)+aR_a(-p^2,(1-p)^2),\quad 0\le p\le1.\] With \(c=(-1,-1)\), set \[\begin{align*} H^-&=\{P(c+h r(p)+t v(p),t):0\le p\le1,\ 0\le h\le R_a,\ |t|\le1\},\\ H^+&=\{P(-x,-y,t):P(x,y,t)\in H^-\}. \end{align*}\] The radial length \(R_a=2-a\) shortens each central triangle by \(a\). Both beam regions are compact images of compact parameter sets. At \(t=0\) they are triangles on opposite corners of the square section \([-1,1]^2\); Figure 2 displays their central gap. The regions \(H^-\) and \(H^+\) are the sets to be covered, whereas each approximating cylinder will have one fixed direction.

The points left between the beams

At \(t=0\), the missed set is the diagonal band \(|x+y|<a\) shown in Figure 2. At other heights we enclose the missed set in a tilted band with a uniform cubic error, using the transverse coordinate \(s\) and the coordinate \(l\) along the diagonal: \[s=x+y,\qquad l=(x-y)/2,\qquad f(t)=2at-2a^2t^3.\] For a constant \(M>0\), to be chosen uniformly in the next lemma, define \[\begin{align*} W(t)&=\bigl(a(1-2t^2) +a^2t^2\{1+(1-|t|)^2+t^2\}+Ma^3\bigr)_+, \tag{24}\\ G&=\{P(x,y,t)\in K:|s+f(t)l|\le W(t)\}, \end{align*}\] where \(q_+=\max(q,0)\).

Finite horizontal cylinders for the strip

The two beams now cover every point outside \(G\). It remains to give \(G\) a finite covering whose cost can be combined with the beam cost.

The strict saving survives finite approximation

A different reason that an edge direction minimizes the shadow

The chord identity is not needed to locate the minimum for a regular tetrahedron. There is an independent argument based on the piecewise linear expression in Cauchy’s face formula (10). Indeed, some minimizing direction for any tetrahedron is parallel to an edge; for a regular tetrahedron all edge directions give the same area. To see this directly, argue as follows. The continuous positive projection-area function \(A\) attains its minimum on the unit sphere. Suppose that a minimizing \(u\) is perpendicular to at most one facet normal. Choose a unit \(v\perp u\), also perpendicular to that normal if it exists. For sufficiently small \(\theta>0\), all nonzero signs in the face formula stay unchanged at \[u_\pm=(\cos\theta)u\pm(\sin\theta)v,\] and a zero scalar product remains zero. Consequently \[A(u_+)+A(u_-)=2\cos\theta\,A(u)<2A(u),\] contradicting minimality. Equivalently, along this great-circle arc the area is a fixed linear form and satisfies \(A''=-A<0\), so it cannot have a local minimum. A minimizing direction is therefore perpendicular to at least two facet normals and hence parallel to the common edge. For a regular tetrahedron, isometries permute the edges. In the edge-\(2\) model \[\{(x,y,\sqrt2 t):0\le t\le1,\ |x|\le t,\ |y|\le1-t\},\] projection in the \(x\)-direction has area \(\sqrt2\int_0^1 2(1-t)\,dt=\sqrt2\). Scaling gives \(\ell^2/(2\sqrt2)\) for every edge direction.

A common criterion and the exact cost

For reference, the following rational fixed margins all give strict saving. Several appear in the explicit feet and coordinate realizations of Appendices 7–8. Their exact quadratic cost coefficients are

The square-root choices \(\beta=1/120,1/125\) have the same displayed quadratic coefficients as their linear counterparts, but their caps and finite covers are different. Cubic choices \(c=5,12,20,64\) contribute, respectively, \(10,24,40,128\) at order \(\varepsilon^3\). Fractional choices \(c=1,4\sqrt2,8\sqrt2\) contribute \(2,8\sqrt2,16\sqrt2\) at order \(\varepsilon^{5/2}\). Each constant here specifies an actual cap rather than an identification based only on a leading coefficient.

Alternative identities for the overlap

The coverage and saving for all four caps have now been proved. The following arguments are alternatives to the common criterion, not additional requirements. The first uses the monotonicity of \(G\) in the squared angular variable; the second keeps positive weights on the radial parameters; the third avoids expanding a square-root cap.

A quantitative complementary calculation at half height

An exact weighted calculation gives another explicit cubic certificate. Use coordinates \((X,Y,Z)\) with \(|X|\le1-Z\), \(|Y|\le Z\), and their physical realization \((X/2,Y/2,Z/\sqrt2)\). For \(r\in[-1,1]\) define \[\alpha(r)=(1+r^2)/4,\quad \beta(r)=r/2,\quad F(r)=r^2(1+r^2)/16.\] The two line families are \[\begin{align*} Y&=ur+\delta X\alpha(r),& Z&=u+\delta X\beta(r),\\ X&=vs-\delta Y\alpha(s),&1-Z&=v-\delta Y\beta(s). \end{align*}\] The crossing lemma supplies \(u,v\ge0\). Both families use the cap \(R_\delta(r)=1/2+\delta^2F(r)+4\delta^3\). Their exact complementary identity is \[ (u-1/2)(1-\delta Xr)+(v-1/2)(1+\delta Ys) =\delta^2\{X^2\alpha(r)+Y^2\alpha(s)\}. \tag{45}\] To verify it, expand \(Z-\delta XY=1/2+(u-1/2)(1-\delta Xr)-\delta^2X^2\alpha(r)\) and its complementary equation, and add.

Suppose both caps fail and \(0<\delta\le1/2\). The weights in (45) are at least \(1/2\), since \(|X|,|Y|\le1\). Its right side is at most \(\delta^2\), so each positive radial excess is at most \(2\delta^2\). It follows that \(|Y-r/2|,|X-s/2|\le2\delta^2+\delta/2\le3\delta/2\). Therefore \[X^2\alpha(r)+Y^2\alpha(s) \le\frac{s^2\alpha(r)+r^2\alpha(s)}4+\frac94\delta \le F(r)+F(s)+\frac94\delta.\] Here \(|X+s/2|,|Y+r/2|\le3/2\), \(\alpha\le1/2\), and the last inequality is monotonicity in \(r^2,s^2\). Failed caps, however, make (45) divided by \(\delta^2\) exceed \[(1-\delta)(F(r)+F(s)+8\delta) \ge F(r)+F(s)+\frac{15}{4}\delta,\] because \(F\le1/8\). The incompatible constants \(9/4\) and \(15/4\) prove coverage throughout \(0<\delta\le1/2\).

Replacing \(4\delta^3\) by \(\delta^{5/2}\) gives a second useful version for all sufficiently small positive \(\delta\). To see the fractional cancellation directly, set \(a=1/2\), \(P(\xi)=a/2+\xi^2/(2a)\) and \(g(\xi)=\xi^2P(\xi)\). The identity before (45) becomes \[Z=a+\delta XY-\delta^2P(ar)X^2 +(u-a)(1-\delta rX).\] With both caps failed, \(|Y-ar|,|X-as|=O(\delta)\) uniformly and the positive factors are bounded away from zero. Bounded derivatives of \(P,g\) consequently give \[\begin{align*} Z&>a+\delta XY+\delta^2(g(Y)-P(Y)X^2) +\delta^{5/2}-C\delta^3,\\ Z&<a+\delta XY-\delta^2(g(X)-P(X)Y^2) -\delta^{5/2}+C\delta^3. \end{align*}\] Their incompatibility follows from the exact identity \[g(X)+g(Y)-P(X)Y^2-P(Y)X^2 =\frac{(X^2-Y^2)^2}{2a}\ge0.\] Thus the complementary and the positive-weight arguments both preserve fractional padding with its proper, non-fourth-order leading remainder. The affine change \(x=2Y\), \(y=2X\), \(t=2Z-1\) takes these two half-height families to the earlier canonical families with \(\varepsilon=\delta/2\). The first family becomes the lower family with radial parameter \(2u\), angular parameter \(r\), and free coordinate \(y=2X\); the second becomes the upper family with radial parameter \(2v\), angular parameter \(s\), and free coordinate \(x=2Y\). The cubic and fractional caps become, respectively, \(1+\varepsilon^2H+64\varepsilon^3\) and \(1+\varepsilon^2H+8\sqrt2\varepsilon^{5/2}\).

One intercept identity, several enclosures

Use a cap \(R_\varepsilon\) and fix a positive tilt for which Appendix 6 gives coverage. Retain \(h=\sqrt2\). For a closed interval \(I\subset[-1,1]\) of positive length \(d\), a tag \(c\in I\), and \(M=\max_I R_\varepsilon\), the upper-sweep point with parameters \(q,\rho,x\) has frozen intercept \[ u=\rho-\varepsilon x(q-c),\qquad Y=qu+\frac{\varepsilon x}{2}(q-c)^2, \qquad 0\le\rho\le M,\quad |x|\le2. \tag{46}\] The physical intercept point is \((0,Y,h(1-u))\). Its area element is \(h\,dY\,du\), and projection perpendicular to the frozen direction multiplies area by \((1+\varepsilon^2D_0(c))^{-1/2}\). The lower family is obtained by the same orthogonal map as before.

Let \(m\) be the midpoint of \(I\). For positive constants \(A,B\), an absolute-value footprint is \[ -Ad\le u\le M+Ad,\qquad |Y-mu|\le \frac d2|u|+Bd^2. \tag{47}\] For any tag, it contains (46) when \(A\ge2\varepsilon\) and \(B\ge\varepsilon\). For midpoint tags, the sharper sufficient conditions are \(A\ge\varepsilon\) and \(B\ge\varepsilon/4\). Indeed \(|q-c|\le d\) for any tag and \(|q-c|\le d/2\) at the midpoint; (46) then bounds the radial displacement by \(2\varepsilon|q-c|\) and the extra transverse displacement by \(\varepsilon|q-c|^2\). The angular sector itself has width \(d|u|\). The exact coordinate area is \[ \frac{dM^2}{2}+(A+2B)Md^2+(A^2+4AB)d^3. \tag{48}\] This follows by integrating \(d|u|+2Bd^2\) from \(-Ad\) to \(M+Ad\). The footprint is a compact polygonal set. With a negative collar it is generally nonconvex: the upper-boundary points at \(u=\pm Ad\) have a midpoint above the upper boundary at \(u=0\).

Two exact formulas at unit edge length

It is useful to write two of the finite enclosures in coordinates in which the tetrahedron has edge one. Let \(h_1=1/\sqrt2\) and fix a sufficiently small \(\eta>0\). For the square-root cap \[T_\eta(s)^2=\frac14+\eta^2\left(s^2/4+s^4+1/960\right), \qquad |s|\le\frac12,\] use the tetrahedron \(|x|\le(1-t)/2\), \(|y|\le t/2\) in physical coordinates \((x,y,h_1t)\). The first line family is \[y=sp+\eta x(1/4+s^2),\qquad t=p+2\eta xs, \qquad 0\le p\le T_\eta(s),\] and the second exchanges \(x,y\), \(t,1-t\), and changes the sign of the tilt. Under \(X=4y\), \(Y=4x\), \(Z=2t-1\) this is cap (S) with \(\varepsilon=\eta/2\) and \(\beta=1/120\), so the coverage proof is already complete: the first family becomes the canonical lower family with angle \(2s\) and radius \(2p\); the second becomes the upper family. For midpoint intervals of length \(\Delta=1/N\), midpoint \(s_j\), and maximum \(T_j\), the intercept triangles are \[ -\eta\Delta\le c\le T_j+\eta\Delta, \qquad |b-s_jc|\le\frac\Delta2(c+\eta\Delta). \tag{54}\] Their physical area is exactly \[ \frac{h_1\Delta}{2}(T_j+2\eta\Delta)^2. \tag{55}\] For completeness, subtracting the frozen direction from a represented point gives \(c=p+2\eta x(s-s_j)\) and \(b-s_jc=(s-s_j)p+\eta x(s-s_j)^2\). Since \(|x|\le1/2\), \(|c-p|\le\eta\Delta/2\) and \(|b-s_jc|\le\Delta(p+\eta\Delta/4)/2 \le\Delta(c+\eta\Delta)/2\). Projecting these intercept triangles gives \(2N\) triangular bases whose cylinders cover for every \(N\ge1\), not only after mesh refinement. Projection divides their physical intercept area by \(\sqrt{1+\eta^2((1/4+s_j^2)^2+2s_j^2)}\). Summing over both families gives the exact limit \[h_1\int_{-1/2}^{1/2} \frac{1/4+\eta^2(s^2/4+s^4+1/960)} {\sqrt{1+\eta^2((1/4+s^2)^2+2s^2)}}\,ds =h_1\left(\frac14-\frac{\eta^2}{960}+O(\eta^4)\right).\] This is a particular triangle formula in addition to the arbitrary-tag triangles furnished by Theorem 1.

For the half-height construction of Appendix 6, take intervals of half-length \(\rho=1/N\), midpoint \(p_j\), and maximum cap \(M_j\). Put \(e=\delta\rho/2\). In its normalized intercept coordinates define \[ D_j=\{(v,w):-e\le w\le M_j+e,\quad |v-p_jw|\le\rho(w+2e)\}. \tag{56}\] Its exact coordinate area is \[ |D_j|=\rho(M_j^2+6eM_j+8e^2). \tag{57}\] An actual first-family intercept has \((v,w)=(ur+\sigma(r+p_j)/2,u+\sigma)\) with \(\sigma=\delta X(r-p_j)/2\), so \(|\sigma|\le e\) and \(|v-p_jw|\le\rho(u+e/2)\le\rho(w+2e)\). The second-family formula has \(\sigma=-\delta Y(s-p_j)/2\) and gives the same bounds. The perpendicular projections of these trapezoids give \(2N\) compact convex bases whose cylinders cover the tetrahedron. The physical foot area multiplier is \(h_1/2\), and its projection factor is \((1+\delta^2(\alpha(p_j)^2+2\beta(p_j)^2))^{-1/2}\). Consequently the cubic cap gives total area limit \[\frac{h_1}{2}\int_{-1}^1 \frac{R_\delta(r)^2}{\sqrt{1+\delta^2(\alpha(r)^2+2\beta(r)^2)}}\,dr =\frac{h_1}{2}\left(\frac12-\frac{\delta^2}{240}+O(\delta^3)\right).\] The exact cubic coefficient can also be retained by (36). Coverage holds for \(0<\delta\le1/2\); strict saving uses a sufficiently small such \(\delta\) before selecting \(N\). The count is exactly two cylinders per interval and is not asserted to remain bounded as the tilt tends to zero.

Beam bases from neighborhoods of two triangles

We finish with a different geometry for the beam bases of Section 4. Fix \(a\in(0,1/2)\) and recall \[R_a=2-a,\quad c=(-1,-1),\quad r(p)=(p,1-p),\quad v(p)=(1,-1)+aR_a(-p^2,(1-p)^2).\] The physical map is \(P(x,y,t)=(\beta x,\beta y,t)\) with \(\beta=1/\sqrt2\), and the first beam \(H^-\) consists of \(P(c+h r(p)+t v(p),t)\) for \(0\le p\le1\), \(0\le h\le R_a\), and \(|t|\le1\). Its reflection in the vertical axis is the second beam. Using the projection factor \(g\) from (21), the combined limiting normalized cost is \(C_1=R_a^2\int_0^1g(v(p))\,dp\). We construct bases from neighborhoods of two triangular sectors; their finite areas differ from the triangular bases in the main proof.

Partition \([0,1]\) into \(k\) closed intervals \(I\) of length \(\delta=1/k\), and freeze the direction at each midpoint \(p_I\). Sliding a point of \(H^-\) to height zero along \((\beta v(p_I),1)\) leaves the planar intercept \(c+h r(p)+t(v(p)-v(p_I))\). For \(e=p-p_I\), the quadratic formula for \(v\) gives the exact identity \[ v(p)-v(p_I)=-2aR_a e\,r(p)+aR_a e^2(1,-1). \tag{58}\] The first term changes only the radial parameter \(h\). Indeed, \(|2aR_a et|\le\delta\) and \(|aR_a e^2t(1,-1)|\le\sqrt2\delta^2/4<\delta^2\) because \(aR_a<1\) and \(|e|\le\delta/2\). Thus every intercept lies in \(c+S_I\), where \(S_I\) is the closed \(\delta^2\)-neighborhood of \[Q_I=\{d r(p):p\in I,\ -\delta\le d\le R_a+\delta\}.\] The set \(Q_I\) consists of two triangles, with total area \[|Q_I|=\frac{(R_a+\delta)^2+\delta^2}{2}\,\delta.\] This follows either from their vertex determinants or from the Jacobian \(|d|\) of \((d,p)\mapsto d r(p)\). Their perimeters are uniformly bounded. For a triangle, a radius-\(\rho\) neighborhood adds at most its perimeter times \(\rho\) plus \(\pi\rho^2\) in area: decompose the added region into edge rectangles and vertex sectors, whose angles sum to \(2\pi\). Applying this to both triangles, with \(\rho=\delta^2\), gives \[|S_I|\le \frac{R_a^2}{2}\delta+O(\delta^2).\] The constant is uniform over the intervals. Projecting \(P(c+S_I,0)\) onto the perpendicular plane produces a compact base whose cylinder covers the corresponding part of \(H^-\). By (21), its normalized cost is \(|S_I|g(v(p_I))\). The unpadded sector \(\{d r(p):p\in I,\ 0\le d\le R_a\}\) lies in \(Q_I\subset S_I\) and has area \(R_a^2\delta/2\). Combining this lower bound with the upper estimate gives the first beam’s total normalized cost as \[\frac{R_a^2}{2}\sum_I\delta\,g(v(p_I))+O_a(\delta) \longrightarrow \frac{C_1}{2}.\] Since \(g\circ v\) is smooth on \([0,1]\), the midpoint sum differs from its integral by \(O_a(\delta)\). The orthogonal map \((X,Y,T)\mapsto(-X,-Y,T)\) supplies the reflected covering at the same cost. Thus, for every positive integer \(k\), these \(2k\) cylinders cover the two beams, and their total normalized cost is \(C_1+O_a(k^{-1})\). The quadratic transverse remainder in (58) makes the summed padding cost vanish. For the small tilts used in Proposition 8, combining sufficiently fine beam covers of this form with the strip cover of Lemma 7 gives another finite tetrahedron cover of total area below \(A_{\min}/2\).

A scaled coordinate model

Let \(b>0\) and \(H_b=\sqrt2 b\). In physical coordinates \((x,y,H_bt)\), define the regular edge-\(2b\) tetrahedron \[K_b=\{(x,y,H_bt):0\le t\le1,\quad |x|\le bt,\quad |y|\le b(1-t)\}.\] Its paired line families, with angular variable \(s\in[-b,b]\), are \[\begin{align*} &(x,\rho s-\eta x(s^2+b^2)/2, H_b(1-\rho+\eta xs)),\\ &(\rho s+\eta y(s^2+b^2)/2,y, H_b(\rho+\eta ys)). \end{align*}\] Their free coordinate satisfies \(|x|\le b\) or \(|y|\le b\), and \(0\le\rho\le T_\eta(s)\). The transformation to the earlier physical model is \[ X=2x/b,\quad Y=2y/b,\quad Z=2t-1, \qquad r=s/b,\quad R=2\rho,\quad \varepsilon=\eta b^2. \tag{59}\] The canonical physical point is \((X,Y,\sqrt2 Z)\). This map is a translation followed by a Euclidean similarity of ratio \(2/b\). Substituting in the first displayed line gives \((X,Rr-\varepsilon XG(r),1-R+\varepsilon Xr)\) in the \((X,Y,Z)\) coordinates; substitution in the second gives the lower line. This checks the two families and their radial bounds, not only the tetrahedron. The common upper bound is \(R_\varepsilon(r)=2T_\eta(br)\).

For example, cap (L) and cap (S), respectively, become \[\begin{align*} T_\eta(s)&=\frac12+\eta^2 A_{b,\beta}(s),\tag{60}\\ T_\eta(s)&=\sqrt{\frac14+\eta^2 A_{b,\beta}(s)},\tag{61}\\ A_{b,\beta}(s)&=\frac{s^4+b^2s^2}{4}+\frac{b^4\beta}{2}. \end{align*}\] The exact two-family area integral in these coordinates is \[ H_b\int_{-b}^b \frac{T_\eta(s)^2}{\sqrt{1+\eta^2((s^2+b^2)^2/4+H_b^2s^2)}}\,ds =\frac{bH_b}{2}Q_R(\eta b^2). \tag{62}\] Indeed \(ds=b\,dr\), \(T=R/2\), and the denominator becomes \(\sqrt{1+\varepsilon^2D_0(r)}\). Both sides express the same weighted cost in physical area units. The multiplier \(bH_b/2\) is half the minimum projection area of \(K_b\). Thus the coefficient in these coordinates, relative to this reference area, is exactly \((2\beta-1/30)\eta^2b^4\).

The following examples use tetrahedra of edge lengths one, \(\sqrt2\), and two. Keeping the physical scale and angular scale together lets each row determine both the finite bases and their cost coefficients. The parameter \(\tau\) denotes the tilt used in the indicated row.

For instance, the \(b=1/\sqrt2\) row has additive polynomial \((s^4+s^2/2)/4+1/1000\) and relative quadratic coefficient \(-13\tau^2/3000\). The \(b=1/2\), \(\eta=\tau\) row has additive polynomial \(s^4/4+s^2/16+1/5000\), and its limiting total area is \[2H_b\left(\frac18-\frac{77}{480000}\tau^2+O(\tau^4)\right).\] For \(b=1/2\), \(\eta=\tau/2\), write \(c=1/2\) and \(a(s)=c(1/8+s^2/2)\); the cap can equivalently be written \(c+\tau^2(c^2s^2a(s)+c^3/1920)\). Its limiting total area is \(H_b/4-(H_bc^4/960)\tau^2+O(\tau^4)\). The monotone-polynomial calculation in Appendix 6 applies with precisely these constants.

Under (59), the generously padded absolute-value footprint \(A=B=\sqrt2+2\varepsilon\) and the midpoint affine footprint \(A=B=4\varepsilon\) remain explicit finite enclosures. Equations (48) and (52) compute their finite areas; the sector lower bound in Theorem 1 shows that actual swept projections have the same limiting total area. A similarity preserves that limit, the possible differences between the finite areas, convexity, and the number \(2N\) of cylinders.

A barycentric realization and its boundary points

In this subsection \(k\) denotes a new positive tilt parameter, independent of the fixed conversion constant used above. It will be chosen sufficiently small.

Consider the nonnegative simplex \(\Delta\) in the affine Euclidean space \[\{(a,b,c,d)\in\mathbb R^4:a+b+c+d=1\}.\] Its edge length is \(\sqrt2\). Set \(s=a+b\), \(q=a-b\), and \(p=c-d\). The induced squared length element is \(ds^2+(dq^2+dp^2)/2\). Hence \[ (X,Y,Z)=(2q,2p,2s-1) \tag{63}\] gives a Euclidean similarity of ratio \(2\sqrt2\) to the canonical physical model, where the physical point is \((X,Y,\sqrt2 Z)\). The minimum projection area of \(\Delta\) is therefore \(1/\sqrt2\).

Define, for \(0\le t\le1\), \[V(t)=(t^2,-(1-t)^2),\quad w(t)=2t-1,\quad F(v)=v^2+4v^4+1/1000,\quad B_k(t)=1/2+k^2F(w(t)/2).\] For the first sweep set \[ (a,b)=S(t,1-t)+kpV(t),\quad 0\le S\le B_k(t),\quad |p|\le1, \tag{64}\] with \(c,d\) determined by their sum and difference. For the second set \[ (c,d)=T(r,1-r)-kqV(r),\quad 0\le T\le B_k(r),\quad |q|\le1. \tag{65}\] Under (63), the first family is the canonical lower family and the second is the upper family, with angular parameters \(2t-1\) and \(2r-1\), radial parameters \(2S,2T\), and free coordinates \(2p,2q\). Their cap is precisely \(1+k^2(H+1/500)\). This already gives coverage for small \(k\). The following direct proof explains the positive radial parameters and the closure at boundary points in the original affine simplex.

For an interior point of \(\Delta\), the function \(a/t-b/(1-t)\) decreases continuously from \(+\infty\) to \(-\infty\). Choose its root at level \(kp\) and put \[S=a/t-kpt=b/(1-t)+kp(1-t).\] Multiplying the two equal expressions by \(1-t\) and \(t\), respectively, and adding proves \(S>0\). Apply the analogous construction to \(c,d\) and level \(-kq\) to obtain \(r\) and \(T>0\). Writing \(w=2t-1\), \(z=2r-1\), the sum and difference equations are \[1=S+T+kpw-kqz,\qquad q=Sw+kpG(w),\qquad p=Tz-kqG(z).\] If both caps fail, \(S,T>1/2\) and boundedness first gives \(S,T=1/2+O(k)\), \(w=2q+O(k)\), \(z=2p+O(k)\). Substitution in the sum improves \(S+T\) to \(1+O(k^2)\) and therefore each radial parameter to \(1/2+O(k^2)\). The remaining equations yield \[w=2q-kp(1+4q^2)+O(k^2),\qquad z=2p+kq(1+4p^2)+O(k^2).\] All remainders are uniform on the bounded parameter set. It follows that \(S+T=1+k^2(p^2+q^2+8p^2q^2)+O(k^3)\). The two failed caps require \(S+T>1+k^2(F(p)+F(q))+O(k^3)\), whereas \[F(p)+F(q)-(p^2+q^2+8p^2q^2) =4(p^2-q^2)^2+2/1000>0.\] This proves coverage of the interior for sufficiently small \(k\). Each sweep is a continuous image of a compact parameter domain and is therefore closed. Their union contains the closure of the interior, which is all of \(\Delta\). This final compactness step includes the faces and vertices without extending the fractions \(a/t,b/(1-t)\) to endpoints.

The barycentric finite enclosure

Fix such a \(k\) and divide \([0,1]\) into \(N\) equal closed intervals of length \(\ell\). For a midpoint \(t_0\), freeze the first-family direction and take the perpendicular projection of that interval’s compact swept set as the base. Moving along the frozen direction to the plane \(p=0\) gives the \((a,b)\)-coordinate intercept \[S(t,1-t)+kp(V(t)-V(t_0)).\] With \(d=t-t_0\) one has the exact identity \[ V(t)-V(t_0)=2d(t,1-t)-d^2(1,-1). \tag{66}\] The intercept is therefore within coordinate distance \(\sqrt2 k\ell^2\) of the sector \[\{\rho(t,1-t):t\in I,\ -2k\ell\le\rho\le M_I+2k\ell\}, \qquad M_I=\max_I B_k.\] This sector is the union of two triangles meeting at the origin. The absolute Jacobian of \((\rho,t)\mapsto\rho(t,1-t)\) is \(|\rho|\); its area is consequently \(\ell M_I^2/2+O_k(\ell^2)\). Its two triangle perimeters are bounded independently of \(I\) and \(N\). Thickening a triangle by a distance \(r\) adds at most its perimeter times \(r\) plus \(\pi r^2\): cover the added region by the rectangles along its edges and the circular sectors at its vertices. Apply this separately to the two triangles with \(r=\sqrt2 k\ell^2\). The resulting bound for the actual compact intercept image is \(\ell M_I^2/2+O_k(\ell^2)\). This enclosure retains the geometric thickening method; it need not equal an absolute-value polygonal foot.

The plane \(p=0\) is parametrized by \((a,b,(1-a-b)/2,(1-a-b)/2)\). Its Gram determinant is two, so physical area is \(\sqrt2\) times its \((a,b)\) coordinate area. Along a frozen line parametrized by \(p\), the rates are \(ds/dp=kw(t_0)\) and \(dq/dp=k(t_0^2+(1-t_0)^2)\). Using the metric above, its projection cosine is \[[1+k^2 E(t_0)]^{-1/2},\qquad E(t)=2w(t)^2+(t^2+(1-t)^2)^2.\] There is also a matching lower estimate. Taking \(p=0\) in the first sweep includes the sector \(\{S(t,1-t):t\in I,\ 0\le S\le B_k(t)\}\) in its intercept image. Its coordinate area is \(\tfrac12\int_I B_k(t)^2\,dt\). Therefore the actual perpendicular base has area at least \[\frac{\sqrt2}{2\sqrt{1+k^2E(t_0)}} \int_I B_k(t)^2\,dt.\] This lower bound and the preceding geometric upper bound have the same sum limit. Exchanging the coordinate pairs and changing the sign of the tilt gives the identical argument for the second family. Thus \(2N\) compact-base cylinders cover \(\Delta\), and their total area converges to \(2J\), where \[J=\sqrt2\int_0^1 \frac{B_k(t)^2/2}{\sqrt{1+k^2E(t)}}\,dt, \qquad J/\sqrt2=\frac18-\frac{11}{3000}k^2+O(k^4).\] For the coefficient, integrate \(F(w(t)/2)/2-E(t)/16\); the two relevant averages are \(\int_0^1 F(w/2)=2/15+1/1000\) and \(\int_0^1E=17/15\). Equivalently the exact change of variable \(r=2t-1\) gives \(2J=(\sqrt2/4)Q_R(k)\) with \(\beta=1/500\). Choose \(k\) small enough for coverage and \(2J<\sqrt2/4\), then take a finite sufficiently fine partition. The thickened sectors and actual swept projections need not have equal finite areas; the matching sector lower bound proves equality of their limiting costs.