Inverting the tilted maps

We first obtain inverse labels anywhere in the wedge; the next subsection tests their lower-boundary inequalities. For fixed \(d\), both \(W_d\) and \(DW_d\) are bounded on the entire wedge. For \(|x|\le\tfrac12\), consider \[F_x(Y,T)=(Y,T)+\epsilon xW_d(Y,T).\] We claim that \(F_x\) maps \(\mathcal W\) onto itself for all sufficiently small \(\epsilon\), uniformly in \(x\). The same statement holds with \(-\epsilon\).

The wall functions \(r/2\pm Y\) change by \(\epsilon x(-b/2\pm a)\). Each latter expression vanishes on the corresponding wall and has bounded \(Y\)-derivative. Moving horizontally from that wall gives \[|-b/2\pm a|\le C_d(r/2\pm Y)\quad\text{on }\mathcal W.\] For small \(\epsilon\), therefore, each wall function retains its sign and vanishes precisely on the same wall. The interior maps into the interior and the boundary into the boundary. Also \[\det DF_x=1+\epsilon^2x^2\det DW_d>0\] uniformly on the wedge, so the image of its interior is open by the inverse function theorem. The image of the closed wedge is closed: a convergent sequence of images has bounded preimages because the displacement is bounded, and hence has a convergent subsequence of preimages in the wedge. Boundary points cannot map to the interior. Consequently the interior image is also closed relative to the connected wedge interior. It is nonempty, so it is the whole interior; closedness supplies the boundary. This proves the claim.

Why the two families leave no gap

For \((x,y,ht)\in K\), both \((y,t)\) and \((x,-t)\) lie in \(\mathcal W\). The preceding surjectivity supplies wedge labels satisfying \[\begin{align*} y&=Y+\epsilon xa(Y,T),&t&=T+\epsilon xb(Y,T),\\ x&=X-\epsilon ya(X,R),&-t&=R-\epsilon yb(X,R). \end{align*}\] If either label belongs to \(D_{\epsilon,q}\), the point is covered. Suppose both labels fail. Since \(P\) and \(q\) are nonnegative, this forces \(T,R<0\). Boundedness of \(W_d\) first gives \(|t|,|T|,|R|=O(\epsilon)\) and \(Y=y+O(\epsilon)\), \(X=x+O(\epsilon)\). All remainders below are uniform over \(K\). Since \(b(Y,0)=\alpha Y\), \[T=t-\epsilon\alpha xy+O(\epsilon^2),\qquad R=-t+\epsilon\alpha xy+O(\epsilon^2).\] Their sum is \(O(\epsilon^2)\) and both are negative, so separately \(T,R=O(\epsilon^2)\). The smooth formulas now give \[\begin{align*} Y&=y+\epsilon xc(y)+O(\epsilon^2),& X&=x-\epsilon yc(x)+O(\epsilon^2),\\ b(Y,T)&=\alpha y+\epsilon\alpha xc(y)+O(\epsilon^2),& b(X,R)&=\alpha x-\epsilon\alpha yc(x)+O(\epsilon^2). \end{align*}\] Thus \[ T+R=-\epsilon^2\alpha\bigl(x^2c(y)+y^2c(x)\bigr)+O(\epsilon^3). \tag{12}\] The choice of \(P\) is explained by the exact identity \[ P(x)+P(y)-\alpha\bigl(x^2c(y)+y^2c(x)\bigr) =\alpha c_2(x^2-y^2)^2\ge0. \tag{13}\] Replacing \(x,y\) by \(X,Y\) in \(P\) changes it by \(O(\epsilon)\). Equation (12) therefore gives \[T+R\ge-\epsilon^2(P(X)+P(Y))-C\epsilon^3.\] The cutoff at \(-1/4\) cannot fail when \(T,R=O(\epsilon^2)\). Failure of both remaining domain inequalities would give instead \[T+R<-\epsilon^2(P(X)+P(Y))-2q.\] This contradicts \(\epsilon^3=o(q)\). The two families therefore cover all of \(K\), including the boundary points handled by the inverse maps.

The area comparison and finite choices

We now prove the cost assertions in Proposition 3. The unperturbed label domain \(D_0\) has area \(1/8\). For \(|Y|\le1/4\), the added interval below \(T=0\) has length exactly \(\epsilon^2P(Y)+q\) for small \(\epsilon\). Outside that interval of \(Y\)-values, both the width and depth of the added portion are \(O(\epsilon^2)\), because \(q=O(\epsilon^2)\) and \(P\) is bounded on \(|Y|\le3/8\). Hence \[|D_{\epsilon,q}|=\frac18+\epsilon^2J_d+\frac q2+O(\epsilon^4).\] The field is bounded on these domains. Taylor expansion of the integrand in (9) has uniform remainder \(O(\epsilon^4)\), and replacing the domain of its quadratic term by \(D_0\) changes that term by \(O(\epsilon^4)\). This proves (10) with its stated remainder.

For the coefficient comparison we may now let \(d\downarrow0\). On \(T=0\) one has \(P(Y)\to(1/16+Y^2)Y^2\), giving \[J_d\longrightarrow\int_{-1/4}^{1/4}(Y^2/16+Y^4)\,dY =\frac1{960}.\] On \(D_0\), write \(z=Y/r\) away from the apex. Then \(|z|\le1/2\), \(a\to-(1/16+z^2/4)\) and \(b\to z/2\). The fields are uniformly bounded there for \(d>0\), so bounded convergence and \(dY\,dT=r\,dz\,dr\) yield \[I_d\longrightarrow\frac12\int_0^{1/2}r\,dr \int_{-1/2}^{1/2} \left[\left(\frac1{16}+\frac{z^2}4\right)^2 +\frac12\left(\frac z2\right)^2\right]dz =\frac{17}{15360}.\] This completes the proof of Proposition 3.

There are two useful parameter schedules. First choose \(d>0\) with \(\Delta_d=I_d-J_d>0\). For vanishing padding, take \(q=\epsilon^{5/2}\); then \[C_d(\epsilon,\epsilon^{5/2})/h =\frac18-\Delta_d\epsilon^2+o(\epsilon^2)<\frac18\] for all sufficiently small positive \(\epsilon\). For fixed quadratic padding, choose \(\eta>0\) with \(\eta/2<\Delta_d\), keep it fixed, and take \(q=\eta\epsilon^2\). In this case the sharper expression is \[C_d(\epsilon,\eta\epsilon^2)/h =\frac18-(\Delta_d-\eta/2)\epsilon^2+O(\epsilon^4)<\frac18.\] The parameter order is respectively \(d,\epsilon\) or \(d,\eta,\epsilon\).

For either schedule, fix such a positive \(\epsilon\). The label domain is compact, and the field is smooth on a neighborhood of it. The eigenvalues of \(D(\pm\epsilon W_d)\) are \[\pm\frac{\epsilon d}{4(r+d)^2}.\] Since \(d\) was fixed first, their magnitudes times the axial half-length \(1/2\) are strictly below one when \(\epsilon\) is small. Apply Theorem 1.2 of (OpenAI 2026) with \(V=+\epsilon W_d\) for \(S_+\), and with \(V=-\epsilon W_d\) followed by the isometry \((x,y,z)\mapsto(y,x,-z)\) for \(S_-\). Each application gives finitely many cylinders with cost at most \(C_d(\epsilon,q)+e\). Choose the sum of the two errors less than \(h/4-2C_d(\epsilon,q)>0\). Together the finite covers cover \(K\) and have total cost less than \(h/4=A_{\min}(K)/2\). The approximation theorem gives compact parallelogram bases. Enlarging them slightly in their own perpendicular planes also gives bounded open parallelogram bases, with the total added area chosen smaller than the remaining strict slack. No mesh uniform in \(\epsilon\) is needed. Similarity scales every base area and the minimum projection area by the same factor, proving Theorem 1 for every positive edge length.

Logarithmic-cell realization of the fixed-padding cover

For the field (17), direct differentiation gives \[ \mathop{\mathrm{tr}}Dv=0,\qquad \det Dv=-c_*^2\left(\frac{(1+d_0)d_0}{(s+d_0)^2}\right)^2<0. \tag{23}\] The determinant simplification uses \(gg''/2=(g')^2\). For fixed \(d_0\), the field is \(C^2\) on an open rectangle containing \(U_k\) for every sufficiently small \(k\). The boundary of \(U_k\) has area zero, since it lies in finitely many lines and a polynomial graph. The strict spectral condition \(2c_* k\sqrt{-\det Dv}<1\) also holds there for small \(k\). These are exactly the hypotheses of Corollary 4.4 of (OpenAI 2026). It supplies finite covers of each set in (19) with bounded open measurable bases, at cost arbitrarily close from above to \[ J_k=\int_{U_k}(1+k^2|v(u,s)|^2)^{-1/2}\,du\,ds. \tag{24}\] The second application uses \(-v\) and an isometry.

The exact cost comparison can also be checked in these coordinates. Put \(U_0=\{0\le s\le1, |u|\le c_*s\}\), of area \(c_*\). The same strip calculation as in Section 2.3 gives \[ J_k=c_*+k^2\left\{ \int_{-c_*}^{c_*}(a_0(u)+\eta_0)\,du -\tfrac12\int_{U_0}|v|^2\right\}+O(k^4). \tag{25}\] For fixed \(d_0,\eta_0\), the added horizontal and vertical extensions are \(O(k^2)\); their product is the fourth-order boundary remainder. Letting \(d_0\downarrow0\) only in the coefficient integrals, and setting \(u=sb\), gives \[\begin{align*} \tfrac12\int_{U_0}|v|^2&\longrightarrow \frac14\int_{-c_*}^{c_*} \left[\frac{(c_*^2+b^2)^2}{4}+b^2\right]db =\frac{c_*^3}{6}+\frac{7c_*^5}{30},\\ \int_{-c_*}^{c_*}a_0(u)\,du&\longrightarrow\frac{8c_*^5}{15}. \end{align*}\] Bounded convergence is valid because \(|u|\le c_*s\) bounds the fields uniformly for \(0<d_0<1\). The first limit exceeds the second by \(c_*^3/60>0\). Choose \(d_0>0\) such that the actual difference \[\Delta(d_0)=\tfrac12\int_{U_0}|v|^2 -\int_{-c_*}^{c_*}a_0(u)\,du\] is positive. Then choose \(0<\eta_0<\Delta(d_0)/(2c_*)\), and finally \(k\) small. Formula (25) gives \(J_k<c_*\). Only then choose the two finite errors with sum less than \(2(c_*-J_k)\). The resulting finite open-base cover has cost below \(2c_*=A_{\min}(K_0)/2\).

The \(C^2\) rectangle and strict spectral inequality give the open-base conclusion directly in the orthonormal \((u,s)\) coordinates. The scaling identity gives \(C_d(\epsilon,q)=J_k/8\), in agreement with Proposition 4.

Product-cell realization of the centered cover

For \(q=\epsilon^{5/2}\), use the original centered label set \(D_{\epsilon,q}\) and the two slopes \(V=\pm\epsilon W_d\). The set is compact with area-zero boundary, \(V\) is \(C^1\) on an open neighborhood, and \(DV\) has eigenvalues \(\pm\epsilon d/[4(r+d)^2]\). With \(M=1/2\) and \(h=1/\sqrt2\), the strict condition \(0<M\lambda<1\) follows after fixing \(d>0\) and taking \(\epsilon\) sufficiently small. Therefore Theorem 4.1 of (OpenAI 2026) gives each family a finite cover with bounded open measurable bases and cost at most \(C_d(\epsilon,\epsilon^{5/2})+e\).

The product-cell proof chooses its padding, product threshold, and local scale in that order, exhausts the interior by disjoint cells, and covers the remaining null set before taking a finite open subcover. These approximation choices occur after the present geometric parameters are fixed. By (10), choose \(d\) with \(I_d>J_d\), then \(\epsilon\) so that \(C_d<h/8\), and finally the two errors with sum below \(h/4-2C_d\). This gives the asserted open-base cover without altering either the fractional padding or the weighted physical metric. The compact parallelogram construction in Section 2.3 remains an independent finite realization of the same swept sets.

The field and the cutoff

Fix parameters in the order \[ 1<p<\frac{\sqrt{51}-5}{2},\qquad 1<\gamma<\frac1{2-p},\qquad d=c^\gamma\quad(c>0). \tag{26}\] The upper bound on \(p\) makes the quadratic area coefficient negative. The bounds on \(\gamma\) make the derivative of the tilted field uniformly small while keeping the tip area \(o(c^2)\). Only \(c\) will tend to zero. To obtain a field tangent to the cone walls \(b=\pm r\), take a stream function that vanishes on both walls. Multiplying \(b^2-r^2\) by \(r^{p-2}\) makes the stream function homogeneous of degree \(p\), so the resulting field has degree \(p-1\). The normalization below also gives \(Q(b,a)=b\), which will cancel the first-order overlap terms. For \(r>0\), define \[\begin{align*} \psi(b,r)&=\frac12(r/a)^{p-2}(b^2-r^2), &V=(P,Q)&=(-\partial_r\psi,\partial_b\psi),\\ P(b,r)&=\frac a2(r/a)^{p-1} \left[p+(2-p)(b/r)^2\right], &Q(b,r)&=(r/a)^{p-2}b. \tag{27}\end{align*}\] At the dividing level let \[ P_a(b)=P(b,a)=\frac{ap}{2}+\frac{2-p}{2a}b^2, \qquad \delta(b)=b^2P_a(b). \tag{28}\] Thus \(\delta\ge0\), \(Q(b,a)=b\), and \(Q=\pm P\) on \(b=\pm r\). Our compact label domain is \[ D_c=\{(b,r):d\le r\le2,\ |b|\le r, \ r\le a+c^2\delta(b)+c^{5/2}\}. \tag{29}\]

The field is smooth on \(r>0\), and direct differentiation gives \[ \operatorname{tr}DV=0,\qquad \det DV=(r/a)^{2p-4}\frac{p-1}{2} \left[(2-p)(b/r)^2-p\right]<0 \quad (|b|\le r). \tag{30}\] Indeed the determinant equals \(\det D^2\psi\); the bracket is at most \(2-2p<0\). On this cone, \(\|DV(b,r)\|\le C r^{p-2}\), with \(C\) depending only on the fixed parameters. Consequently \[ c\sup_{D_c}\|DV\|\le Cc d^{p-2} =C c^{1-\gamma(2-p)}\longrightarrow0, \qquad d^2=o(c^2). \tag{31}\] Thus, for sufficiently small \(c\), both \(cDV\) and \(-cDV\) have eigenvalues \(\lambda,-\lambda\) with \(0<\lambda<1\) at every label. Although no derivative bound is asserted at \(r=0\), each fixed \(D_c\) has a neighborhood on which the field is smooth.

Write \(s=1-t\). Consider the two sets of segments \[\begin{align*} E_c^-&=\{(x,y,\sqrt{2}\,t): |x|\le1, \ (y,t)=(b,r)+cxV(b,r),\ (b,r)\in D_c\},\\ E_c^+&=\{(x,y,\sqrt{2}\,t): |y|\le1, \ (x,s)=(b,r)-cyV(b,r),\ (b,r)\in D_c\}. \tag{32}\end{align*}\] We apply Theorem 1.2 of (OpenAI 2026) with \(h=\sqrt{2}\) and \(L=1\) to the first family with field \(cV\). Interchanging \(x,y\) and reflecting and translating the physical height gives the second family with field \(-cV\). Thus, for each fixed sufficiently small \(c\) and any \(e>0\), each family has a finite cylinder cover with compact parallelogram bases and total base area at most \[ J(c)+e, \qquad J(c)=\sqrt{2}\int_{D_c} \frac{db\,dr}{\sqrt{1+c^2(P^2+2Q^2)}}. \tag{33}\] The factor \(2\) multiplying \(Q^2\) records the physical height scale.

These two applications use the companion’s single square-tile construction. For the fixed compact cutoff domain \(D_c\), the fields \(\pm cV\) are \(C^1\) on a neighborhood. Equations (30) and (31) give \(\operatorname{tr}D(\pm cV)=0\) and \(-1<\det D(\pm cV)<0\) for sufficiently small \(c\), which is the theorem’s condition for \(L=1\). The tile orientations may vary, but their weights use the velocities in the original \((b,r)\) coordinates, giving (33) with the physical factor \(\sqrt2\). No condition on the area of \(\partial D_c\) or smallness of the operator norm is required by the theorem. The separate cylinders introduced next cover the omitted tips.

We add an \(x\)-parallel cylinder with triangular base \(\{(0,y,\sqrt{2}\,t):0\le t\le2d,\ |y|\le t\}\), and the analogous \(y\)-parallel cylinder at the upper end, obtained by projecting \(\{0\le s\le2d,\ |x|\le s\}\) to the plane \(y=0\). Each base has area \[ \sqrt{2}\int_0^{2d}2r\,dr =\sqrt{2}(2d)^2. \tag{34}\]

Inversion, including the cone boundary

All estimates below are uniform over the points under consideration, with \(p\) and \(\gamma\) fixed. Suppose \[2d\le T\le1,\qquad |B|\le T,\qquad |k_0|\le c.\] We first solve \[ (B,T)=(b,r)+k_0V(b,r) \tag{35}\] without imposing the last inequality of (29). In the closed Euclidean ball of radius \(T/4\) about \((B,T)\), one has \(3T/4\le r\le5T/4\) and \(|b|/r\le5/3\). Formula (27) gives bounds \(|V|\le C T^{p-1}\) and \(\|DV\|\le C T^{p-2}\) throughout the ball. The map \[(b,r)\longmapsto(B,T)-k_0V(b,r)\] maps this ball into itself and is a contraction for all sufficiently small \(c\), since \(cT^{p-2}\le c(2d)^{p-2}=o(1)\). Its fixed point satisfies \(d\le r\le2\).

To prove that this fixed point lies in the cone, use the exact identities \[ Q(b,r)\pm P(b,r) =\frac{r\pm b}{2}(r/a)^{p-2} \left[(2-p)b/r\pm p\right]. \tag{36}\] Equation (35) consequently gives \[T\pm B=(r\pm b)(1+\theta_\pm),\qquad |\theta_\pm|\le Cc r^{p-2}=o(1).\] The factors \(1+\theta_\pm\) are positive. Since \(T\pm B\ge0\), we obtain \(r\pm b\ge0\). This also handles equality on either wall: no boundary point is lost in passing to inverse labels.

The two families overlap sufficiently

Take a point of \(K_2\) outside the two tip regions, so that \(t,s\ge2d\). Apply (35) to \((B,T)=(y,t)\) with \(k_0=cx\), and to \((B,T)=(x,s)\) with \(k_0=-cy\). Let the resulting labels be \((b_-,r_-)\) and \((b_+,r_+)\). They already satisfy \(d\le r_\pm\le2\) and \(|b_\pm|\le r_\pm\). It remains to show that one label satisfies its ceiling in (29).

Suppose both ceilings fail. Since \(\delta\ge0\), this implies \(r_-,r_+>a\). The inverse equations give \[\begin{align*} b_-&=y-cxP(b_-,r_-),& b_+&=x+cyP(b_+,r_+),\\ r_-+r_+&=1+c\left[yQ(b_+,r_+)-xQ(b_-,r_-)\right]. \tag{37}\end{align*}\] All labels now lie in a fixed compact region with \(r\ge a\), on which the field and its derivatives are bounded. The last equation first gives \(r_-+r_+=1+O(c)\). Because both summands exceed \(a=1/2\), each is \(a+O(c)\). The first two equations also give \(b_-=y+O(c)\) and \(b_+=x+O(c)\). Using \(Q(b,a)=b\), the bracket in the last equation is therefore \(yx-xy+O(c)=O(c)\). This improves the sum to \(1+O(c^2)\) and each height to \(a+O(c^2)\).

Refining the first two equations and again using \(Q(b,a)=b\), we obtain \[b_-=y-cxP_a(y)+O(c^2),\qquad b_+=x+cyP_a(x)+O(c^2),\qquad Q(b_\pm,r_\pm)=b_\pm+O(c^2).\] Substitution into (37) yields \[ r_-+r_+ =1+c^2\left[x^2P_a(y)+y^2P_a(x)\right]+O(c^3). \tag{38}\] On the other hand, failure of both ceilings gives \[r_-+r_+> 1+c^2\left[\delta(b_-)+\delta(b_+)\right]+2c^{5/2}.\] Here \(\delta(b_-)+\delta(b_+)=\delta(y)+\delta(x)+O(c)\), and \[\begin{align*} \delta(y)+\delta(x)-x^2P_a(y)-y^2P_a(x) &=(x^2-y^2)\left[P_a(x)-P_a(y)\right]\\ &=\frac{2-p}{2a}(x^2-y^2)^2\ge0. \end{align*}\] Comparison with (38) would force \(2c^{5/2}\le Cc^3\), which is impossible for sufficiently small \(c\). Thus every middle-region point belongs to \(E_c^-\cup E_c^+\). Together with the two closed tip cylinders, the two families cover all of \(K_2\).

The exact quadratic saving

Let \[H=\{(b,r):0<r\le a,\ |b|\le r\},\qquad |H|=a^2.\] The deletion of \(r<d\) removes area \(d^2=o(c^2)\). The area added above \(r=a\) is \[c^2\int_{-a}^{a}\delta(b)\,db+o(c^2).\] Indeed, for \(|b|\le a\) its thickness is exactly \(c^2\delta(b)+c^{5/2}\). Since \(|b|\le r\le2\) and \(\delta\) is bounded there, the remaining portion has \(a<|b|\le a+O(c^2)\) and thickness \(O(c^2)\), hence area \(O(c^4)\). The field is uniformly bounded on \(|b|\le r\le2\) because \(p>1\). Taylor expansion of the integrand in (33) therefore has uniform remainder \(O(c^4)\). Also \(|D_c\mathbin{\triangle}H|\to0\), so the bounded quadratic coefficient may be integrated over \(H\). Including one tip cylinder, we conclude that \[ \frac{J(c)}{\sqrt{2}}+4d^2 =a^2+c^2\kappa_p+o(c^2),\qquad \kappa_p=\int_{-a}^a\delta(b)\,db -\frac12\int_H(P^2+2Q^2)\,db\,dr. \tag{39}\]

Both terms admit elementary evaluations. The first is \[\int_{-a}^a\delta(b)\,db =a^4\left(\frac p3+\frac{2-p}{5}\right) =\frac{2a^4(p+3)}{15}.\] For the second, put \(\eta=b/r\). Then \[P=\frac a2(r/a)^{p-1}[p+(2-p)\eta^2],\qquad Q=a(r/a)^{p-1}\eta,\qquad db\,dr=r\,d\eta\,dr.\] It follows that \[\begin{align*} \frac12\int_H(P^2+2Q^2)\,db\,dr &=\frac{a^4}{4p}\int_{-1}^{1} \left\{\frac14[p+(2-p)\eta^2]^2+2\eta^2\right\}\,d\eta\\ &=\frac{a^4(2p^2+2p+13)}{30p}. \end{align*}\] Thus the exact coefficient is \[ \boxed{\displaystyle \kappa_p=\frac{a^4(2p^2+10p-13)}{30p}<0.} \tag{40}\] The range of \(p\) in (26) is precisely the range above \(1\) on which this expression is negative. At \(p=1\) its continuous extension is \(-a^4/30\); the construction itself uses \(p>1\) to retain strict hyperbolicity.

Edge-one normalization and alternative cutoffs

The edge-one form allows a direct comparison with the rational field. First we transport the whole construction by a similarity; then we explain which cutoff and padding may be changed without changing its quadratic saving. The similarity halves physical lengths while leaving the height label unchanged: \[(x,y,\sqrt{2}\,t)\longmapsto(y/2,x/2,t/\sqrt{2}), \qquad (w,l)=(b/2,r),\qquad k=4c.\] Put \(h=1/\sqrt{2}\) and define \[\widehat V_1(w,l)=\tfrac14P(2w,l),\qquad \widehat V_2(w,l)=\tfrac12Q(2w,l),\qquad f(w)=w^2\widehat V_1(w,1/2).\] Equivalently, \(\widehat V=(-\partial_lH_*,\partial_wH_*)\) for \(H_*(w,l)=2^{p-3}(w^2l^{p-2}-l^{p}/4)\). The two segment families transform exactly, and \[k^2(\widehat V_1^2+h^2\widehat V_2^2) =c^2(P^2+2Q^2),\qquad c^2\delta(2w)=k^2 f(w).\] With \(S=\operatorname{diag}(1/2,1)\), the derivative identity is \[\widehat V=\tfrac12 S V\circ S^{-1},\qquad \tfrac12 kD\widehat V=cS(DV)S^{-1}.\] Thus the scaled eigenvalues for axial half-length \(L=1/2\) agree with those of \(cDV\). Under the fixed physical rescaling, the derivative is likewise similar to \(\pm kD\widehat V\). This is a spectral statement; the nonorthogonal matrix \(S\) does not preserve the ordinary operator norm. No operator-norm transfer is needed by the square-tile theorem. The already constructed physical cylinder cover also transports by the Euclidean similarity above.

Physical areas scale by \(1/4\). Therefore the per-family quadratic coefficient after dividing physical area by \(h\) is \(\kappa_p/32\); the positive saving in that normalization is \[-\frac{\kappa_p}{32} =\frac{13-10p-2p^2}{15360p}.\] For this exact similarity the cutoff becomes \(4^{-\gamma}k^\gamma\) and the ceiling padding becomes \(k^{5/2}/32\).

The estimates also permit independent choices of these two auxiliaries. Indeed, the preceding inverse, overlap, and area proofs hold with a positive cutoff \(\rho(c)\) and padding \(q(c)>0\) whenever \[ \rho(c)=o(c),\qquad c\rho(c)^{p-2}\longrightarrow0, \qquad q(c)=o(c^2),\qquad c^3=o(q(c)). \tag{41}\] The contraction and wall-sign estimates use only the second condition; the deleted label area and two tip areas use the first. The overlap contradiction compares \(2q(c)\) with \(O(c^3)\), and the coefficient is unchanged because \(q(c)=o(c^2)\). On the retained set \(r\le2\), the ceiling forces \(r\le1/2+O(c^2)\). Replacing its upper bound \(2\) by \(1\) therefore leaves the label set unchanged for sufficiently small \(c\).

We can now state the edge-one variant with its own cutoff. Choose \(1<\beta<1/(2-p)\) and put \(\sigma=k^\beta\). The field and padding polynomial are explicitly \[\begin{align*} \widehat V_1(w,l)&=\tfrac14(2l)^{p-1} [(2-p)(w/l)^2+p/4],\\ \widehat V_2(w,l)&=\tfrac12(2l)^{p-1}(w/l),\qquad f(w)=w^2[(2-p)w^2+p/16]. \tag{42}\end{align*}\] The image tetrahedron is \[\widetilde K_1=\{(x,y,hs):0\le s\le1,\ |x|\le s/2, \ |y|\le(1-s)/2\}.\] It is the translate \(K_1+(0,0,h/2)\) of the centered tetrahedron in (2). Its two segment families are \[(x,s)=(w,l)+ky\widehat V(w,l),\quad |y|\le1/2, \qquad (y,1-s)=(w,l)-kx\widehat V(w,l),\quad |x|\le1/2,\] with labels \[D_k=\{\sigma\le l\le1,\ |w|\le l/2, \ l\le1/2+k^2f(w)+k^{5/2}\}.\] Under \(c=k/4\), these new choices pull back to \(\rho(c)=(4c)^\beta\) and \(q(c)=(4c)^{5/2}\), which satisfy (41). Thus coverage follows from the full preceding proof, including its closed inversions and two tip cylinders. The scaled eigenvalue bound is now \(O(k\sigma^{p-2})=o(1)\), so Theorem 1.2 of (OpenAI 2026) applies with \(L=1/2\). For every positive error \(e\), each finite family costs at most \(I(k)+e\), where \[ I(k)=h\int_{D_k} (1+k^2(\widehat V_1^2+h^2\widehat V_2^2))^{-1/2}\,dw\,dl. \tag{43}\] There are exactly two additional tip cylinders, each of area \(2h\sigma^2\).

The coefficient has the following direct edge-one evaluation. On \(D_0=\{0\le l\le1/2,\ |w|\le l/2\}\), put \[A(p)=\int_{-1/4}^{1/4}f(w)\,dw =\frac{2-p}{2560}+\frac{p}{1536}, \qquad G(p)=\frac12\int_{D_0}(\widehat V_1^2+h^2\widehat V_2^2)\,dw\,dl.\] Substituting \(R=w/l\) gives \[\begin{align*} G(p)&=\frac1{16p} \left\{\frac{(2-p)^2/80+p(2-p)/24+p^2/16}{16} +\frac1{96}\right\},\\ A(p)-G(p)&=\frac{2p^2+10p-13}{15360p}. \tag{44}\end{align*}\] Here \(\int_0^{1/2}l(2l)^{2p-2}\,dl=1/(8p)\) and the second and fourth moments on \([-1/2,1/2]\) are \(1/12\) and \(1/80\). In particular \(A(1)=16/15360\) and \(G(1)=17/15360\) are continuous coefficient limits. They coincide with the rational limits of \(J_d\) and \(I_d\) in (11). We still fix \(p>1\), then \(\beta\), then a positive \(k\), and finally the two finite errors. This gives \(2I(k)+4h\sigma^2<h/4\) before those errors. The minimum projection area of \(\widetilde K_1\) is \(h/2\) by the similarity above.

At the endpoint \(p=1\), writing \(\theta=b/r\), the original edge-two field in (27) becomes \[P(b,r)=\frac{1+\theta^2}{4},\qquad Q(b,r)=\frac{\theta}{2}.\] These formulas depend only on the angle of the label. The cutoff requirements (41) would now demand both \(\rho(c)=o(c)\) and \(c/\rho(c)\to0\), which are incompatible. The next section fixes a truncated label set, applies the square-zero case of the same finite-approximation theorem, and covers the apex segments by a finite velocity net. The field agrees with the endpoint above; the padding and the treatment of the apex are chosen anew.

Choosing the angular field

Write a nonzero label in the upper half-plane as \(b=(pt,t)\), where \(t>0\). For a field of the form \(G(b)=(v(p),w(p))\), differentiation gives \[DG(b)=\frac1t \begin{pmatrix}v'(p)\\w'(p)\end{pmatrix} \begin{pmatrix}1&-p\end{pmatrix}.\] Thus \(v'(p)=p w'(p)\) makes \((DG)^2=0\): the row in this product annihilates its column. We also want the velocity tangent to the boundary rays \(p=\pm1\), which requires \(v(p)=p w(p)\) there. Fix \(c=1/2\) and choose \(w(p)=cp\). Integrating the differential relation with these endpoint conditions determines \(v\). We also define \(a\) by \[ v(p)=\frac c2(1+p^2),\qquad w(p)=cp, \qquad a(p)=c^2p^2v(p). \tag{47}\] The last function specifies the quadratic overlap we will allow. For \(0<k<1\), put \[H(p)=c+k^2a(p)+k^3,\qquad -1\le p\le1.\] The extra cubic term provides a strict margin in the coverage proof. In normalized coordinates define two sets swept out by segments: \[ \begin{aligned} \mathcal E_+&=\{(s,pt+ksv(p),t+ksw(p)): |s|\le1,\ |p|\le1,\ 0\le t\le H(p)\},\\ \mathcal E_-&=\{(pt-ksv(p),s,1-t+ksw(p)): |s|\le1,\ |p|\le1,\ 0\le t\le H(p)\}. \end{aligned} \tag{48}\] At \(k=0\) the two sweeps cover the lower and upper halves of the tetrahedron by axis-parallel segments. For positive \(k\) we must check that their varying tilts do not leave a gap between them.

The cost predicted by the sweeps

We have proved complete coverage by two continuous families. We next compute their limiting cost, so that a definite positive tolerance is available when making the cover finite.

For a planar tile \(T\) and a constant velocity \(g=(g_1,g_2)\), write \[\mathcal C(T,g)=\{(s,\xi):s\in\mathbb R,\ \xi\in T+sg\}, \qquad J_h(g)=(1+g_1^2+h^2g_2^2)^{-1/2}.\] The two physical maps \[ F_+(s,\xi)=(s,\xi_1,h\xi_2),\qquad F_-(s,\xi)=(\xi_1,s,h(1-\xi_2)) \tag{53}\] send this auxiliary cylinder to cylinders with perpendicular base area \(h|T|J_h(g)\). Indeed, the transverse tile has physical area \(h|T|\), and projection to the plane perpendicular to \((1,g_1,hg_2)\) has area factor \(J_h(g)\). The second map differs from the first by a Euclidean isometry. A square \(T\), of any orientation, therefore gives a compact convex parallelogram base.

Under these maps the two sweeps have the common auxiliary form \[ (s,(pt,t)+s\sigma k(v(p),w(p))),\qquad |s|,|p|\le1,\quad 0\le t\le H(p), \qquad \sigma\in\{1,-1\}. \tag{54}\] The Jacobian of \((p,t)\mapsto(pt,t)\) is \(t\). The integral predicting the weighted tile cost of either sweep is consequently \[ I(k)=\int_{-1}^1\frac{H(p)^2}{2} \bigl(1+k^2(v(p)^2+h^2w(p)^2)\bigr)^{-1/2}\,dp. \tag{55}\]

A finite cover, including the apex

We now realize the saving by finitely many cylinders. On the open half-plane \(b_2>0\) define \[G_\sigma(b)=\sigma k \bigl(v(b_1/b_2),w(b_1/b_2)\bigr),\] using the same polynomial formulas for \(v,w\) at every real argument. This is a smooth field, and the earlier differential computation gives \[ DG_\sigma(pt,t)=\frac{\sigma kc}{t} \begin{pmatrix}p\\1\end{pmatrix} \begin{pmatrix}1&-p\end{pmatrix}, \qquad (DG_\sigma)^2=0. \tag{57}\] At \(t=0\) every label \((pt,t)\) is the origin, but its assigned velocity still depends on \(p\). Thus the field has no continuous extension to the apex. That part of the family must be covered separately.

The remaining finite approximation uses the same square-tile theorem, Theorem 1.2 of (OpenAI 2026), with \(L=1\). A square-zero planar matrix has trace and determinant zero, so it satisfies the theorem’s condition \(\operatorname{tr}DG=0\) and \(-1<\det DG\le0\). For a fixed compact set \(D\), a \(C^1\) field \(G\) on a neighborhood of \(D\), fixed \(h>0\), and positive tolerance, the theorem supplies finitely many compact nondegenerate square tiles, with possibly different orientations, and velocities covering \(\{(s,b+sG(b)):b\in D,\ |s|\le1\}\). Their weighted cost is at most \(\int_D J_h(G)\) plus that tolerance. The weights use the original velocity coordinates. Neither a small-shear condition nor a boundary regularity assumption is required.

Second moments of the face normals

Projection area from overlap volumes

The same comparison area can be obtained from an identity for every nondegenerate tetrahedron. The simplex chord–projection identity is attributed to Martini and Weissbach in (Heinrich 2014, 8268). The following proof differentiates the volume of a translated intersection.