SLOPE-FIELD PERTURBATIONS OF THE TWO-CYLINDER COVERING
A complete and expanded proof
PREFACE
This file gives a full, self-contained proof that a regular tetrahedron has a finite cylinder
cover whose total perpendicular base area, and whose directionwise normalized area, are both
strictly below one half of the natural reference value. It repairs and expands the earlier
draft. The reader needs only vectors, partial derivatives, and single-variable integration.
One external result is used, the finite-approximation theorem for ruled sets from the companion
paper "Finite cylinder approximation of ruled sets". Every time it is used we state its
hypotheses and check them, so no step is hidden.
We work in R^3 with coordinates (x,y,z). For a unit vector u and a convex body K, write
pi_{u^perp} K for the orthogonal projection of K onto the plane u^perp (perpendicular to u),
and |pi_{u^perp} K| for its two-dimensional area. A set of the form B + R u = {b + s u :
b in B, s real}, where B lies in u^perp, is a cylinder with perpendicular base B and base area
|B|. A finite cylinder cover is a finite list of cylinders whose union contains K. The symbols
det, tr, sqrt, log keep their usual meaning; "area" of a planar set is its 2-volume.
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1. THE PROBLEM, THE CONJECTURE, AND WHAT WE PROVE
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Let K be a three-dimensional convex body and let
A_min(K) = min over unit u of |pi_{u^perp} K|,
the smallest area of any orthogonal shadow of K.
Bang and Bezdek asked the half-area question: does every finite cylinder cover of K satisfy
sum_i |B_i| >= (1/2) A_min(K) ?
The regular tetrahedron with two edge-parallel cylinders is the equality example: its total base
area equals exactly half the minimum projection area. A stronger, directionwise form uses
relative cost R(C;K) = sum_i |B_i| / |pi_{u_i^perp} K|,
conjecture (Bezdek-Khan): R(C;K) >= 1/2 for every finite cylinder cover in dimension 3.
Because each denominator |pi_{u_i^perp} K| is at least A_min(K),
R(C;K) <= ( sum_i |B_i| ) / A_min(K),
so any cover with total base area below A_min(K)/2 automatically has relative cost below 1/2.
We prove such covers exist for the regular tetrahedron, hence disprove both the half-area
question and the directionwise 1/2 conjecture.
MAIN THEOREM. Every regular tetrahedron K in R^3 of positive edge length admits
(a) a finite cylinder cover whose perpendicular bases are compact parallelograms, and
(b) a finite cylinder cover whose perpendicular bases are bounded open measurable sets,
such that in both cases
sum_i |B_i| < (1/2) A_min(K) and sum_i |B_i| / |pi_{u_i^perp} K| < 1/2.
COROLLARY. Every nondegenerate tetrahedron T in R^3 admits a finite cylinder cover with
compact parallelogram bases, and another with bounded open measurable bases, each having
relative cost below 1/2.
The corollary costs nothing extra because relative cost is unchanged by invertible affine maps
(proved in Section 10). It therefore suffices to treat one regular tetrahedron.
Strategy, which is the heart of the whole argument. Start from the equality cover by two
perpendicular families of parallel segments. Replace the fixed axis direction of each family by
a slowly varying slope field W. On one hand, tilting a bundle of near-parallel segments shrinks
the perpendicular area of each small patch of base, by roughly the factor 1/sqrt(1+|W|^2),
which lowers the cost quadratically in the tilt. On the other hand, the tilted segments drift,
so the two families may stop meeting; to restore coverage we must enlarge the label domain, which
raises the cost, also quadratically in the tilt. The construction is tuned so the shrink wins.
The precise bookkeeping is the cost expansion of Section 7, and the decisive fact is that two
explicit integrals differ by a positive amount that survives at every scale.
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2. THE REFERENCE TETRAHEDRON AND ITS MINIMUM PROJECTION AREA
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Take the regular tetrahedron of edge one, placed as
K_1 = { (x, y, h t) : -1/2 <= t <= 1/2, |x| <= (1/2 + t)/2, |y| <= (1/2 - t)/2 },
with
h = 1/sqrt(2).
Its four vertices are
(0, -1/2, -h/2), (0, 1/2, -h/2), (-1/2, 0, h/2), (1/2, 0, h/2).
Any two are at distance 1: between the first two, 1 in y; between the last two, 1 in x; between a
lower and an upper vertex, (1/2)^2 + (1/2)^2 + h^2 = 1/4 + 1/4 + 1/2 = 1. Hence K_1 is the
regular tetrahedron of edge 1, and its diameter is exactly 1.
Volume. At level t the horizontal section is the rectangle with full widths (1/2 + t) in x and
(1/2 - t) in y, of area 1/4 - t^2. The physical height element is h dt, so
vol(K_1) = h * integral from -1/2 to 1/2 of (1/4 - t^2) dt = h * (1/6) = h/6.
We next compute A_min(K_1) = h/2.
LEMMA 2.1 (projection by entering faces). For a tetrahedron K with vertices v_1,...,v_4 and
affine barycentric coordinates lambda_1,...,lambda_4, and for a unit vector u, let
d_i = directional derivative of lambda_i in direction u (a constant). Then
|pi_{u^perp} K| = 3 vol(K) * sum_i max(d_i, 0).
Proof. Index faces by the opposite vertex. Face i has outward unit normal n_i and area F_i.
The projection equals the total projected area of the faces that "face" u:
|pi_{u^perp} K| = sum over i with u.n_i > 0 of (u.n_i) F_i.
(The front faces tile the shadow; overlaps have area zero.) Now lambda_i is 1 at vertex i and 0
on face i, so grad(lambda_i) points from face i toward vertex i, and n_i = -grad(lambda_i) /
|grad(lambda_i)|. Moreover vol = F_i * (distance from vertex i to face i) / 3, and that distance
is 1/|grad(lambda_i)|, so F_i = 3 vol |grad(lambda_i)|. Finally d_i = u.grad(lambda_i), so
u.n_i = -d_i / |grad(lambda_i)|. Thus on a front face (u.n_i > 0, i.e. d_i < 0),
(u.n_i) F_i = ( -d_i / |grad lambda_i| )( 3 vol |grad lambda_i| ) = -3 vol d_i.
Summing over d_i < 0 and using sum_i d_i = 0 gives
|pi_{u^perp} K| = 3 vol * sum_{d_i < 0} (-d_i) = 3 vol * sum_i max(d_i,0).
[]
Now finish the area computation. Let S = sum_i max(d_i, 0). Since sum_i d_i = 0, the numbers
d_i/S over the positive part sum to 1, so P = sum_{d_i>0} (d_i/S) v_i is a point of K (convex
combination of vertices), and similarly N = sum_{d_i<0} (-d_i/S) v_i is a point of K. The identity
u = sum_i d_i v_i
holds because sum_i lambda_i = 1 and sum_i lambda_i v_i = (position); taking directional
derivatives gives sum_i d_i v_i = u. It rewrites as
u = sum_{d_i>0} d_i v_i + sum_{d_i<0} d_i v_i = S P - S N = S (P - N).
So u/S = P - N is a difference of two points of K, of length at most the diameter 1. As |u| = 1,
1/S = |u/S| <= 1, hence S >= 1.
Therefore |pi_{u^perp} K_1| = 3 vol * S = 3 (h/6) S = (h/2) S >= h/2. An edge direction has
barycentric derivatives (1,-1,0,0), so S = 1 there and the bound is attained. Hence
A_min(K_1) = h/2.
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3. THE EQUALITY COVER (TWO PERPENDICULAR FAMILIES OF SEGMENTS)
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We record the classical equality example, which is the seed of everything.
In the label plane with coordinates (Y,T), let
D_0 = { (Y,T) : 0 <= T <= 1/2, |Y| <= (1/2 - T)/2 }.
This is an isosceles triangle: at level T its horizontal half-width is (1/2 - T)/2, so its total
width is (1/2 - T), shrinking to 0 at T = 1/2. Its planar area is
|D_0| = integral from 0 to 1/2 of (1/2 - T) dT = 1/8.
First cylinder. Let
C_+ = { (x, Y, h T) : (Y,T) in D_0, x any real }.
This is the triangle { (0,Y,hT) : (Y,T) in D_0 } in the plane x = 0, translated along the
x-axis; its perpendicular base is that triangle, perpendicular to the x-axis, of area
h * |D_0| = h/8.
It contains every point (x,y,ht) of K_1 with 0 <= t <= 1/2: taking T = t and Y = y we need
|y| <= (1/2 - T)/2, which is exactly the t >= 0 half of K_1's y-bound, and x is unrestricted here.
Second cylinder. Let
C_- = { (X, y, -h R) : (X,R) in D_0, y any real }.
This is the triangle { (X,0,-hR) : (X,R) in D_0 } in the plane y = 0, translated along the
y-axis; its perpendicular base is that triangle, perpendicular to the y-axis, also of area h/8.
It contains every point (x,y,ht) of K_1 with -1/2 <= t <= 0: write t = -R with R in [0,1/2],
take X = x, and note |x| <= (1/2 - R)/2 = (1/2 + t)/2 is exactly the x-bound of K_1.
Coverage. The half t >= 0 is covered by C_+, the half t <= 0 by C_-, and the common slice t = 0
by both. So K_1 is contained in C_+ union C_-.
Cost. The two bases are disjoint planes; their areas add to h/8 + h/8 = h/4 = (1/2) A_min(K_1).
This is the equality configuration: two edge-parallel cylinders meeting the half-area threshold
exactly. The whole problem is whether a small deformation can push the total strictly below h/4
while still covering K_1.
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4. THE PERTURBING SLOPE FIELD
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We deform each family by replacing its constant axis direction with a slowly varying direction,
encoded by a slope field. A point of K_1 at height h t has two complementary descriptions:
t (used by the x-family, whose labels are (Y,T) with T = t), and
-t = 1/2 - t - 1/2 ... measured instead by R = -t and the reflected height 1/2 ... .
It is cleaner to use the single distance-to-the-top coordinate
r = 1/2 - T, so that T = 1/2 - r.
At the top edge (t = 1/2) we have r = 0; going down, r grows. In terms of r the body is
W = { (Y,T) : |Y| <= r/2 }, i.e. T <= 1/2 - 2|Y|,
a wedge with apex at (Y,r) = (0,0). The two straight walls are r = 2Y and r = -2Y (that is,
Y = r/2 and Y = -r/2). Every point of K_1 projects into this wedge.
Fix a smoothing parameter d > 0. On r > -d define the stream function and the field
psi(Y,T) = (Y^2 - r^2/4) / (4 (r + d)), r = 1/2 - T,
W_d = (a, b) = ( -d(psi)/dT, d(psi)/dY ).
Carrying out the differentiations (using dr/dT = -1 and dr/dY = 0) gives explicitly
a(Y,T) = - Y^2 / (4 (r+d)^2) - r (r + 2d) / (16 (r+d)^2),
b(Y,T) = Y / (2 (r+d)).
So W_d is a genuine pair of smooth functions of (Y,T), nice on the whole wedge for fixed d.
Why this field. Two properties drive the construction.
(Tangency.) The polynomial Y^2 - r^2/4 vanishes exactly on the two walls r = 2|Y|. Hence
psi = 0 on both walls. The rotated-gradient field W_d = (-psi_T, psi_Y) is everywhere tangent to
the level lines of psi, so W_d is tangent to both walls of the wedge. This means the tilted
straight-line maps we build below will keep the wedge inside itself, so no point of the label set
is ever pushed off the body: coverage can only be lost by a genuine gap between the families, not
by a label sliding off the edge.
(Spectral shape.) Direct differentiation of a, b gives the two quantities we need:
trace: a_Y + b_T = 0,
determinant: a_Y b_T - a_T b_Y = - d^2 / (16 (r+d)^4) < 0.
For a 2x2 matrix with zero trace and negative determinant, the eigenvalues are real and opposite:
eigenvalues of D W_d = + d / (4 (r+d)^2) and - d / (4 (r+d)^2).
Let us verify the determinant and note the ingredients. From psi one computes
psi_YY = 1/(2(r+d)), psi_Yr = -Y/(2(r+d)^2), psi_rr = (4Y^2 - d^2)/(8(r+d)^3).
Since d/dT = -d/dr, the Jacobian of W_d = (psi_r, psi_Y) is
D W_d = [ [psi_rY, -psi_rr], [psi_YY, -psi_Yr] ].
Its trace is psi_rY - psi_Yr = 0, and its determinant is
-psi_rY^2 + psi_YY psi_rr = -Y^2/(4(r+d)^4) + (4Y^2 - d^2)/(16(r+d)^4)
= -d^2/(16(r+d)^4).
This is negative for every d > 0, so the eigenvalues are +-d/(4(r+d)^2), one positive and one
negative, with equal magnitude. Finally, for fixed d the field W_d and its derivative DW_d are
bounded on the whole wedge (because r + d >= d > 0), which is what makes the small-tilt expansion
below uniform.
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5. THE TILTED FAMILIES AND THE LABEL DOMAIN
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We now fix the small parameters. Choose a smoothing d > 0 (it will be taken small later). At the
dividing level T = 0 (that is r = 1/2) single out the following data. Put
alpha = 1/(1 + 2d),
c_0 = (1/2)(1/2 + 2d) / (16 (1/2 + d)^2),
c_2 = 1 / (4 (1/2 + d)^2),
c(Y) = c_0 + c_2 Y^2, (so c(Y) = - a(Y, 0)),
P(Y) = alpha c(Y) Y^2 = - alpha a(Y,0) Y^2.
Two elementary identities will be used repeatedly:
b(Y,0) = Y / (1 + 2d) = alpha Y, c(Y) = -a(Y,0). (5.1)
The label domain. For a tilt amount eps > 0 and a padding q = q(eps) >= 0, define the label set
D_{eps,q} = { (Y,T) in W : T >= -1/4 and T >= - eps^2 P(Y) - q }.
It is D_0 shrunk a little and pushed down: the second inequality is the new lower boundary, and the
first keeps us inside a fixed bounded band. Since P >= 0 and q >= 0, D_{eps,q} contains D_0 for
every eps. It is compact for each fixed eps, and the field W_d is smooth on a neighborhood of it.
The swept sets. Each label (Y,T) in D_{eps,q} carries a straight segment; the families are the
unions of these segments over the label set and over the axial coordinate.
S_+ = { ( x, Y + eps x a(Y,T), h ( T + eps x b(Y,T) ) ) : (Y,T) in D_{eps,q}, |x| <= 1/2 },
S_- = { ( X - eps y a(X,R), y, - h ( R - eps y b(X,R) ) ) : (X,R) in D_{eps,q}, |y| <= 1/2 }.
At eps = 0 these are exactly the two triangular prisms C_+, C_- of Section 3. For eps > 0 each
segment is tilted: its direction, as x varies, is the derivative of (x, Y + eps x a, h T + h eps x b),
namely (1, eps a, h eps b). So its slope in the (y,z)-directions, seen along x, is (eps a, h eps b).
Cost of one family. Suppose a small label patch has parameter area dY dT. Its physical transverse
image in the plane x = 0 is the patch (0, Y, hT) of physical area h dY dT. The cylinder over this
patch is perpendicular to the unit direction proportional to (1, eps a, h eps b); projecting the
transverse patch onto that perpendicular plane multiplies its area by the cosine of the angle with
e_0 = (1,0,0), namely 1/sqrt(1 + eps^2(a^2 + h^2 b^2)). Summing over the whole label set, the
perpendicular base area of one continuous family is
C_d(eps, q) = h * integral over D_{eps,q} of dY dT / sqrt(1 + eps^2 (a^2 + h^2 b^2)). (5.2)
This is the target that the finite-approximation theorem will match to any prescribed error.
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6. COVERAGE: THE TWO FAMILIES STILL FILL THE TETRAHEDRON
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The whole difficulty is to show that for small eps the two tilted families S_+ and S_- still cover
all of K_1. There are two equivalent proofs; we give the main one (a two-failed-labels comparison)
and, in Section 6.3, the shorter threshold version for fixed quadratic padding.
Fix a point P_0 = (x, y, h t) of K_1. Since |y| <= (1/2 - t)/2 and |x| <= (1/2 + t)/2, both
labels (y, t) and (x, -t) lie in the wedge W. (Check: for (y,t), |y| <= (1/2 - t)/2 is exactly
|Y| <= (1/2 - T)/2; for (x,-t), |x| <= (1/2 + t)/2 = (1/2 - (-t))/2.) If either inverse label
can be pushed into D_{eps,q}, the point is covered. The plan: (6.1) both inverse maps exist and
take W onto W; (6.2) if both resulting labels were outside D_{eps,q}, a positive surplus would give
a contradiction. Hence at least one label is inside, and the point is covered.
6.1 The tilted label maps are bijections of the wedge. ---------------------
Fix d > 0 and |x| <= 1/2. Consider
F_x(Y,T) = (Y + eps x b(Y,T), T + eps x a(Y,T)) (we write (Y', T') for the image).
(Recall W_d = (a,b) with a the T-velocity and b the Y-velocity; here a multiplies eps x in the
second coordinate, matching the definition of S_+.)
Step 1: walls go to walls, and the wedge stays inside itself. Let
f_-(Y,T) = r/2 - Y, f_+(Y,T) = r/2 + Y, r = 1/2 - T,
so f_- = 0 is the lower wall Y = r/2 and f_+ = 0 is the upper wall Y = -r/2. Directly,
f_-' = f_- - eps x (b + a/2), f_+' = f_+ + eps x (b - a/2).
The two combinations
g_-(Y,T) = b + a/2, vanishes on the wall f_- = 0;
g_+(Y,T) = b - a/2, vanishes on the wall f_+ = 0;
(these are exactly the tangency of W_d to the two walls of Section 4). Because a, b are smooth and
bounded on the wedge with bounded first derivatives, and g_- vanishes on the line f_- = 0 while
g_+ vanishes on f_+ = 0, there is a constant C_d with
|g_-| <= C_d f_-, |g_+| <= C_d f_+ on the wedge
(vanishing on a line plus a local Lipschitz bound gives this bound everywhere on the wedge). For
eps < 1/(2 C_d) each wall function keeps its sign and vanishes on the same wall. So F_x maps W
into W, sends each wall to the same wall, and the interior to the interior.
Step 2: F_x is onto W. Its Jacobian determinant is
det D F_x = det ( I + eps x D W_d ) = 1 + eps^2 x^2 det D W_d
= 1 - eps^2 x^2 d^2/(16(r+d)^4) > 0
(the last estimate holds for small eps, uniformly on W). So F_x is a local diffeomorphism; the
image of the open interior is open. The image is also closed in the interior: the displacement
eps x W_d is bounded, so a convergent sequence of images has bounded preimages, which have a
convergent subsequence in the closed wedge. A connected, nonempty subset of the interior that is
both open and closed there is the whole interior, and Step 1 gives the boundary. Hence F_x is a
bijection of W onto W. The same argument, with eps replaced by -eps, handles the inverse for S_-.
Consequently, for each point of K_1 there are unique wedge labels (Y,T), (X,R) with
y = Y + eps x a(Y,T), t = T + eps x b(Y,T), (6.1a)
x = X - eps y a(X,R), -t = R - eps y b(X,R). (6.1b)
The point is covered by S_+ if (Y,T) in D_{eps,q}, and by S_- if (X,R) in D_{eps,q}.
6.2 Two failed labels are impossible. ------------------------------------
From (6.1) we read off the two height equations
T = t - eps x b(Y,T), R = - t + eps y b(X,R). (6.2)
Suppose, toward a contradiction, that neither label is in D_{eps,q}. Since P >= 0 and q >= 0
the failing inequality in the definition of D_{eps,q} forces T < 0 and R < 0 (the T >= -1/4
constraint will be addressed at the end).
First-order size. Because W_d is bounded on W, (6.1) gives |T - t|, |R + t| = O(eps), hence
|T|, |R| = O(eps) and Y = y + O(eps), X = x + O(eps). Using b(Y,0) = alpha Y and |T| = O(eps),
T = t - eps x (alpha y + O(eps)) = t - eps alpha xy + O(eps^2),
R = - t + eps y (alpha x + O(eps)) = - t + eps alpha xy + O(eps^2).
Adding, T + R = O(eps^2); since both are negative, each of |T|, |R| is O(eps^2).
Second-order refinement. With |T| = O(eps^2) we now improve the label Y. From (6.1a),
Y = y - eps x a(Y,T), and a(Y,T) = a(y,0) + O(eps) = -c(y) + O(eps). Hence
Y = y + eps x c(y) + O(eps^2), and likewise X = x - eps y c(x) + O(eps^2).
Feeding these into (6.2) through b(Y,T) = alpha Y + O(eps^2) and b(X,R) = alpha X + O(eps^2):
b(Y,T) = alpha y + eps alpha x c(y) + O(eps^2),
b(X,R) = alpha x - eps alpha y c(x) + O(eps^2).
Substituting into T + R = eps[ y b(X,R) - x b(Y,T) ] gives the key asymptotic
T + R = - eps^2 alpha ( x^2 c(y) + y^2 c(x) ) + O(eps^3). (6.3)
The surplus identity. A direct expansion using c(Y) = c_0 + c_2 Y^2 gives the exact algebraic fact
P(x) + P(y) - alpha ( x^2 c(y) + y^2 c(x) ) = alpha c_2 (x^2 - y^2)^2 >= 0. (6.4)
Equivalently, -alpha x^2 c(y) - alpha y^2 c(x) >= -(P(x) + P(y)). Combining with (6.3),
T + R >= - eps^2 ( P(x) + P(y) ) - C eps^3.
Replacing x, y by X, Y in P changes each value by O(eps), so
T + R >= - eps^2 ( P(X) + P(Y) ) - C eps^3. (6.5)
The contradiction. If both labels are outside D_{eps,q}, then both lower P-inequalities fail:
T < - eps^2 P(Y) - q, R < - eps^2 P(X) - q,
so adding,
T + R < - eps^2 ( P(X) + P(Y) ) - 2 q. (6.6)
Compare (6.5) and (6.6): this forces 2 q < C eps^3, i.e. 2q = O(eps^3). As soon as the padding
is larger than the cubic error, that is eps^3 = o(q), this is impossible.
Finally, the constraint T, R >= -1/4 cannot fail: we have just shown |T|, |R| = O(eps^2), so for
small eps both are comfortably above -1/4. Therefore the failing inequality is only the P-bound,
and (6.5)-(6.6) apply. We conclude that, whenever 0 < eps is small and q = q(eps) satisfies
q(eps) = O(eps^2) and eps^3 = o(q(eps)), (6.7)
every point of K_1 has at least one label in D_{eps,q}. Hence
K_1 is contained in S_+ union S_-.
6.3 A threshold proof for fixed quadratic padding. ----------------------
For fixed quadratic padding q = eta eps^2 with eta > 0, there is a second, more explicit way to
locate the overlap: give each family a definite height below which it certainly covers. Fix
d, eta > 0 and take q = eta eps^2. For |x|, |y| <= 1/2 define the threshold
Theta_x(y) = eps alpha x y + eps^2 ( alpha x^2 c(y) - P(y) ). (6.8)
We claim the following sufficient condition for the first family:
(y,t) in W, -1/2 <= t <= 1/2, t >= Theta_x(y)
implies (y,t) = F_x(Y,T) for some (Y,T) in D_{eps,q}. (6.9)
Consider the rectangle R = [-1,1] x [-1, 1/2 + d/2] (this avoids the singular line r = -d since
r = 1/2 - T >= -d/2 there). On R the field W_d is smooth with a finite Lipschitz constant L_d.
For small eps the map (Y,T) -> (y,t) - eps x W_d(Y,T) is a contraction of R into itself whenever
|y| <= 1/2 and -1/2 <= t <= 1/2, so the inverse label (Y,T) exists and is unique in R.
It preserves the wedge directly. The wall combinations satisfy (from the tangency in Section 4)
-b/2 + a = -(r/2 + Y) * [ (2Y + 2d + r)/(8(r+d)^2) ],
-b/2 - a = (r/2 - Y) * [ (-2Y + 2d + r)/(8(r+d)^2) ],
and the bracketed factors are bounded on R. So the target wall functions (1/2 - t)/2 +- y are the
label wall functions r/2 +- Y times positive factors when eps is small; nonnegativity of the former
gives nonnegativity of the latter, including equality on either wall. In particular the label lies
in W.
For fixed (x,y), the inverse height T is strictly increasing in t. (If two target heights differ
by dt > 0 and p, p' are their labels, then |(p' - p) - (0,dt)| <= (eps L_d/2)|p' - p|, giving
|p' - p| <= (eps L_d/2)/(1 - eps L_d/2) dt with coefficient < 1.) At the threshold t = Theta_x(y)
the inverse equations first give Y = y + O(eps), T = O(eps); using b(y,0) = alpha y improves this to
T = O(eps^2). Then Y = y + eps x c(y) + O(eps^2) and b(Y,T) = alpha y + eps alpha x c(y) + O(eps^2),
and substituting into T = t - eps x b(Y,T) at t = Theta_x(y) gives
T = - eps^2 P(y) + O(eps^3) = - eps^2 P(Y) + O(eps^3). (6.10)
Throughout the whole height range [Theta_x(y), 1/2], boundedness of the field on R keeps
Y = y + O(eps), so P(Y) changes from P(y) by only O(eps). Combining with the monotonicity of T,
for every t >= Theta_x(y) under consideration,
T >= - eps^2 P(Y) - C eps^3 >= - eps^2 P(Y) - eta eps^2 = - eps^2 P(Y) - q,
and also T >= -1/4, when eps is small. So the label is in D_{eps,q}, proving (6.9).
Applying the same argument with x replaced by -y gives a sufficient condition -t >= Theta_{-y}(x)
for the second family. But, using the surplus identity (6.4),
Theta_x(y) + Theta_{-y}(x) = - eps^2 alpha c_2 (x^2 - y^2)^2 <= 0. (6.11)
Hence for every point of K_1 at least one of t >= Theta_x(y) and -t >= Theta_{-y}(x) holds, so one
family covers it. This proves coverage again, and explains why the same quartic polynomial P
appears in both the overlap and the threshold arguments.
------------------------------------------------------------------------------
7. THE COST: EXPANSION AND THE DECISIVE POSITIVE DIFFERENCE
------------------------------------------------------------------------------
We now compute the continuous cost C_d(eps,q) of one family, formula (5.2), to second order in eps.
Throughout, d > 0 is fixed, the padding obeys q = O(eps^2) and eps^3 = o(q), and all remainders are
uniform (their constants may depend on d and on a fixed bound for q/eps^2).
The integrand expands uniformly on D_{eps,q} (the field is bounded there) as
1/sqrt(1 + eps^2 (a^2 + h^2 b^2)) = 1 - (1/2) eps^2 (a^2 + h^2 b^2) + O(eps^4). (7.1)
The label area. The unperturbed triangle D_0 has area 1/8. For |Y| <= 1/4 the added interval
below T = 0 has length exactly eps^2 P(Y) + q (for small eps), and
integral over |Y| <= 1/4 of (eps^2 P(Y) + q) dY = eps^2 J_d + q/2,
where J_d = integral from -1/4 to 1/4 of P(Y) dY.
Outside |Y| <= 1/4 the added width and depth are both O(eps^2) (because P is bounded and
q = O(eps^2)), so they contribute O(eps^4). Hence
|D_{eps,q}| = 1/8 + eps^2 J_d + q/2 + O(eps^4). (7.2)
The quadratic correction. Replacing the domain of the O(eps^2) term of (7.1) by D_0 changes that
term by O(eps^4) (again because the added region has area O(eps^2), width and depth O(eps^2)).
So the quadratic term is -eps^2 I_d with
I_d = (1/2) integral over D_0 of (a^2 + h^2 b^2) dY dT.
Putting (7.1)-(7.2) together,
C_d(eps,q)/h = 1/8 + eps^2 ( J_d - I_d ) + q/2 + O(eps^4). (7.3)
Coefficient limits as d falls to 0. These are the numbers that decide everything.
On T = 0 one has c(Y) -> 1/16 + Y^2 and alpha -> 1, so P(Y) -> Y^2/16 + Y^4, giving
J_d -> integral from -1/4 to 1/4 of (Y^2/16 + Y^4) dY = 16/15360 = 1/960. (7.4)
(Indeed 2[ (1/4)^3/48 + (1/4)^5/5 ] = 2[1/3072 + 1/5120] = 16/15360.)
On D_0 write z = Y/r, where r = 1/2 - T, so |z| <= 1/2 and dY dT = r dz dr. As d -> 0,
a -> -(1/16 + z^2/4), b -> z/2, h^2 = 1/2,
and the fields are uniformly bounded for d > 0 on D_0, so bounded convergence gives
I_d -> (1/2) integral_0^{1/2} r dr integral_{-1/2}^{1/2}
[ (1/16 + z^2/4)^2 + (1/2)(z/2)^2 ] dz = 17/15360. (7.5)
(Compute the inner integral: (1/16+z^2/4)^2 + z^2/8 = 1/256 + 5z^2/32 + z^4/16, whose integral over
[-1/2,1/2] is 68/7680; times (1/2)(1/8) = 1/16 gives 136/122880 = 17/15360.)
Therefore
lim_{d -> 0} ( I_d - J_d ) = (17 - 16)/15360 = 1/15360 > 0. (7.6)
This one positive number is the entire source of the counterexample.
------------------------------------------------------------------------------
8. CHOOSING THE PARAMETERS AND MAKING THE COVER FINITE
------------------------------------------------------------------------------
We now fix the numbers, in order, and obtain an actual finite cover.
The parameter order.
(1) Choose d > 0 small enough that Delta_d := I_d - J_d is positive. This is possible by (7.6),
because the limit of I_d - J_d is 1/15360 > 0. Fix this d for the rest of the proof; the
field W_d and its bounds are now frozen.
(2) Choose eta > 0 with eta/2 < Delta_d. Keep eta fixed. We use the fixed quadratic padding
q = eta eps^2. (This satisfies the coverage requirement (6.7): q = O(eps^2), and
eps^3 = o(eta eps^2) because eps -> 0.)
(3) Choose eps > 0 small, to be specified. The cost of one family, by (7.3), is
C_d(eps, eta eps^2)/h = 1/8 + eps^2 (J_d - I_d) + eta eps^2 / 2 + O(eps^4)
= 1/8 - ( Delta_d - eta/2 ) eps^2 + O(eps^4) < 1/8 (8.1)
for all sufficiently small eps > 0, since Delta_d - eta/2 > 0. Fix such an eps. Also
shrink eps if necessary so that the coverage statement of Section 6 holds and so that the
companion theorem's spectral hypotheses hold (checked below). Now all of d, eta, eps are
fixed, and we have the strict margin
h/4 - 2 C_d(eps, eta eps^2) > 0. (8.2)
The companion finite-approximation theorem. We use the following result (it is Theorem 1.2 of the
companion paper "Finite cylinder approximation of ruled sets"); we state it in the precise form we
need and then check every hypothesis.
THEOREM 8.1 (finite approximation, compact parallelogram bases). Let h > 0 and L > 0, let
D be a compact subset of a plane with boundary of area zero, and let V be a C^1 field on an open
neighborhood of D such that
(H1) tr DV = 0 everywhere, and
(H2) -L^{-2} < det DV <= 0 everywhere.
Then for every choice of the base class "compact parallelogram" and every error e > 0, the ruled
set of segments { (s, b + s V(b)) : b in D, |s| <= L } (placed in R^3 with the physical height
factor h) admits a finite cover by cylinders with compact parallelogram perpendicular bases whose
total base area is at most
e + h * integral over D of ( 1 + V_1(b)^2 + h^2 V_2(b)^2 )^{-1/2} db.
In particular the total base area can be made less than that integral plus e.
Here "b" is a planar label, V(b) = (V_1(b), V_2(b)) its assigned slope, and the integral is the
continuous cost of the ideal family (identical in form to (5.2)).
Applying Theorem 8.1 to each family. Take D = D_{eps, eta eps^2} (writing eps for the fixed value
and dropping it from the label). This set is compact; its boundary lies in finitely many straight
lines (the wedge walls and the level T = -1/4) and one polynomial graph T = -eps^2 P(Y) - q, hence
has area zero. Take V = eps W_d. Because d > 0 is fixed, V is C^1 on an open neighborhood of D
(the field is smooth on the whole half-plane r > -d, and D lies in r >= 0 there). Check:
(H1) tr D(eps W_d) = eps tr D W_d = 0, by Section 4.
(H2) det D(eps W_d) = eps^2 det D W_d = - eps^2 d^2 / (16 (r+d)^4).
On D we have r >= 0, so r + d >= d, hence (r+d)^4 >= d^4 and
0 >= det D(eps W_d) >= - eps^2 d^2 / (16 d^4) = - eps^2 / (16 d^2).
With the axial half-length L = 1/2 (the |x| <= 1/2 in S_+), L^{-2} = 4. For small eps,
eps^2/(16 d^2) < 4, so -L^{-2} < det D(eps W_d) <= 0. (H2) holds.
So Theorem 8.1 applies to the first family S_+, with the axial coordinate x playing the role of s
(|x| <= 1/2 = L) and the label (Y,T) playing the role of b. Its continuous cost is exactly
C_d(eps, eta eps^2) of (5.2). The theorem gives, for any e_+ > 0, a finite cover of S_+ by
cylinders with compact parallelogram bases and total base area at most C_d + e_+.
The second family S_- has the same label set and field -eps W_d (recall the isometry
(x,y,z) -> (y,x,-z) carries the first construction to the second, with the label R playing the role
of T). The same three checks hold verbatim (trace 0, and the same determinant bound since the
determinant is invariant under V -> -V). So Theorem 8.1 gives, for any e_- > 0, a finite cover of
S_- with compact parallelogram bases and total base area at most C_d + e_-.
Finishing the compact-base cover. Choose e_+ and e_- positive with e_+ + e_- < h/4 - 2 C_d (possible
by (8.2)). The two finite lists together have total base area at most
2 C_d + e_+ + e_- < h/4 = (1/2) A_min(K_1). (8.3)
By Section 6, K_1 is contained in S_+ union S_-, hence in the union of the two finite covers. So we
have a finite cylinder cover of K_1 with compact parallelogram bases and total base area < h/4.
Every denominator |pi_{u_i^perp} K_1| in the relative cost is at least A_min(K_1) = h/2, so
sum_i |B_i| / |pi_{u_i^perp} K_1| <= ( sum_i |B_i| ) / (h/2) < (h/4)/(h/2) = 1/2. (8.4)
This proves part (a) of the Main Theorem for K_1, with both inequalities.
No mesh uniform in eps is needed: eps is fixed before e_+, e_- are chosen, and Theorem 8.1 is applied
to that single fixed compact label set. Enlarging each base by a tiny amount in its own
perpendicular plane also produces bounded open parallelogram bases with the total added area smaller
than the remaining strict slack h/4 - 2 C_d - e_+ - e_-; this is a harmless alternative base class,
and we take it up properly through the product-cell theorem next.
------------------------------------------------------------------------------
9. BOUNDED OPEN MEASURABLE BASES
------------------------------------------------------------------------------
Part (b) of the Main Theorem asks for bounded open measurable bases rather than compact
parallelograms. The cleanest route is the product-cell theorem from the companion paper, which
applies to the very same continuous families, so we do not redo the geometry.
THEOREM 9.1 (finite approximation, bounded open bases). Let h > 0, let D be a compact subset of a
plane with boundary of area zero, and let V be C^1 on an open neighborhood of D such that DV has
two real eigenvalues +lambda and -lambda with
0 < M lambda < 1 (H3)
for some fixed M > 0. Then for every error e > 0 the ruled set of segments over D (placed with
physical height factor h) admits a finite cover by cylinders whose perpendicular bases are bounded
open measurable sets, with total base area at most
e + h * integral over D of ( 1 + V_1(b)^2 + h^2 V_2(b)^2 )^{-1/2} db.
Check the hypothesis (H3) for V = eps W_d on D = D_{eps, eta eps^2}. The eigenvalues of
D(eps W_d) are the numbers +- eps d / (4 (r+d)^2) (Section 4, scaled by eps). Take M = 1/2. Since
r >= 0 gives r + d >= d,
0 < M lambda = (1/2) eps d / (4 (r+d)^2) = eps d / (8 (r+d)^2) <= eps / (8 d) < 1
for small eps. So (H3) holds. The remaining hypotheses (D compact, area-zero boundary, V of class
C^1 on a neighborhood) are the same ones already verified in Section 8. Hence Theorem 9.1 applies to
each of S_+, S_- with field +-eps W_d, giving, for any e_+', e_-' > 0, finite covers with bounded
open measurable bases and total base area at most C_d + e_+' and C_d + e_-'.
Choosing e_+', e_-' with e_+' + e_-' < h/4 - 2 C_d (possible by (8.2)), the union is a finite cover
of K_1 by cylinders with bounded open measurable bases and total base area < h/4 = (1/2) A_min(K_1);
and the same relative-cost estimate (8.4) gives total relative cost < 1/2. This proves part (b) of
the Main Theorem for K_1.
(For completeness, the companion paper also offers a logarithmic-cell realization after passing to
the orthonormal coordinates (u,s) = (the label rescaled and rotated); it yields the same continuous
cost up to the exact factor 1/8 from the similarity that takes the edge-two tetrahedron to K_1, and
hence the same strict conclusion. We do not need it, since Theorem 9.1 already supplies bounded
open bases directly.)
------------------------------------------------------------------------------
10. FROM ONE REGULAR TETRAHEDRON TO ALL TETRAHEDRA
------------------------------------------------------------------------------
We prove the Corollary. The key point, due to Bezdek and Litvak, is that relative cylinder cost is
invariant under invertible affine maps. Because the absolute half-area statement is only asserted
for regular tetrahedra (where A_min is known explicitly), the corollary is about relative cost only,
which is exactly what affine invariance preserves.
LEMMA 10.1 (affine invariance of relative cost). Let A be an invertible affine self-map of R^3 and
let C = ( B_i + R u_i )_i be a finite cylinder cover of a body K. Then A C = ( A B_i + R (A u_i) )_i
is a finite cylinder cover of A K, its bases A B_i are perpendicular to A u_i, the base classes
(compact parallelogram, bounded open measurable) are preserved, and
R( A C ; A K ) = R( C ; K ).
Proof. Affine maps send cylinders to cylinders and preserve covers. Translation does not affect any
area, so suppose A is linear. For a unit direction u put v = A u / |A u| (unit) and define the
linear map between perpendicular planes
Q_u = pi_{v^perp} A restricted to u^perp : u^perp -> v^perp.
It is invertible: if Q_u p = 0 then A p is parallel to A u, hence p is parallel to u, and since
p is in u^perp this forces p = 0. The image of B_i + R u_i is (A B_i) + R v_i, with perpendicular
base Q_u B_i, and pi_{v^perp}(A K) = Q_u ( pi_{u^perp} K ). The same planar Jacobian |det Q_u| > 0
multiplies numerator and denominator of |B_i| / |pi_{u_i^perp} K|, so the ratio is unchanged term by
term. Summing gives R(A C; A K) = R(C; K). Both base classes are preserved by an invertible linear
map. []
Proof of the Corollary. Any nondegenerate tetrahedron T is an invertible affine image of a regular
one, say T = A K_1. By part (a) of the Main Theorem there is a finite cover C of K_1 with compact
parallelogram bases and R(C; K_1) < 1/2. By Lemma 10.1, A C is a finite cover of T with compact
parallelogram bases and R(A C; T) = R(C; K_1) < 1/2. The same argument with the bounded-open cover
of part (b) gives the bounded-open conclusion. This is the Corollary. []
==============================================================================
PART II. TWO FURTHER CONSTRUCTIONS OF THE SAME PHENOMENON
==============================================================================
Sections 8-10 already prove the Main Theorem and Corollary in full. The two constructions that
follow are independent proofs, developed because they illuminate the mechanism from different
angles and because each is short and self-contained. Both produce finite covers of a regular
tetrahedron with total base area strictly below half the minimum projection area. A reader
satisfied with the Main Theorem may skip to Section 13; nothing below is used by Part I.
Both use the same general device, so we state it once. A "ruled set over a label set D with slope
field V" is
{ (s, b + s V(b)) : b in D, |s| <= L },
placed in R^3 with a fixed physical height factor h; each label b carries a segment in direction
(1, V_1(b), h V_2(b)), and small label patches have perpendicular base area
h / sqrt(1 + V_1(b)^2 + h^2 V_2(b)^2) db.
Theorem 8.1 (and its open-base twin, Theorem 9.1) turn such a set into a finite cylinder cover whose
total base area is arbitrary-close to the integral of that density, provided the field satisfies the
stated spectral hypotheses on a neighborhood of the compact label set D. So the whole task is to
exhibit a slope field whose ideal cost is strictly below the threshold and whose segments cover the
tetrahedron.
------------------------------------------------------------------------------
11. A SINGULAR POWER-LAW FIELD
------------------------------------------------------------------------------
Here we keep a genuine power-law singularity in the derivative at the apex and simply remove a small
tip region, covering it by two cheap triangular-base cylinders. The exponent is tuned so the tip
cost is o(c^2) while the quadratic area saving is strictly negative.
11.1 The edge-two tetrahedron and its splitting cylinders. -----------------
Use the regular tetrahedron of edge two,
K_2 = { (x, y, sqrt2 t) : 0 <= t <= 1, |x| <= 1 - t, |y| <= t },
whose physical metric is dx^2 + dy^2 + 2 dt^2 (because z = sqrt2 t has dz = sqrt2 dt). At height
parameter t the section is the rectangle of half-widths (1-t) in x and t in y, of area 4t(1-t). Thus
vol(K_2) = sqrt2 * integral_0^1 4t(1-t) dt = 2 sqrt2 / 3.
Its four vertices (+-1,0,0), (0,+-1,sqrt2) are pairwise at distance exactly 2, so its diameter is 2.
By Lemma 13.1 and Lemma 13.2 below, A_min(K_2) = sqrt2 (this is the value we aim below).
Split K_2 at t = a := 1/2. The x-parallel cylinder with base
{ (0, y, sqrt2 t) : 0 <= t <= a, |y| <= t } and the analogous y-parallel cylinder at the top split
it exactly. Each base is a triangle of area sqrt2 * a^2 (in the (y,t) plane the triangle has area
a^2, scaled by sqrt2), so the two together have base area 2 sqrt2 a^2 = sqrt2/2 = (1/2) A_min(K_2):
the equality configuration, now at edge two. We perturb the two axis directions.
11.2 The field and the cutoff. ---------------------------------------------
Fix parameters in this order:
1 < p < (sqrt(51) - 5)/2 (approx 1.07), 1 < gamma < 1/(2-p), d = c^gamma (c>0). (11.1)
Only c tends to zero. The upper bound on p is exactly the range making the quadratic coefficient
below negative; the bounds on gamma make the tilted derivative vanish while keeping the removed tip
area o(c^2). To get a field tangent to the cone walls b = +-r, take a stream function vanishing on
both walls: multiply b^2 - r^2 by r^{p-2}. For r > 0 set
psi(b,r) = (1/2) (r/a)^{p-2} (b^2 - r^2),
V = (P, Q) = ( -d(psi)/dr, d(psi)/db ),
and a short computation gives
P(b,r) = (a/2) (r/a)^{p-1} [ p + (2-p)(b/r)^2 ],
Q(b,r) = (r/a)^{p-2} b. (11.2)
These are chosen so that Q(b,a) = b (the "first-order cancellation" condition) and Q = +- P on the
cone walls b = +- r. At the dividing level put
P_a(b) = P(b,a) = a p/2 + (2-p) b^2 /(2a), delta(b) = b^2 P_a(b) >= 0. (11.3)
The compact label domain, with the removed tip r >= d and the enlarged ceiling, is
D_c = { (b,r) : d <= r <= 2, |b| <= r, r <= a + c^2 delta(b) + c^{5/2} }. (11.4)
11.3 Hyperbolicity and the small-tilt estimates. --------------------------
Direct differentiation gives
tr DV = 0, det DV = (r/a)^{2p-4} * (p-1)/2 * [ (2-p)(b/r)^2 - p ] < 0 on |b| <= r,
because the bracket is at most 2 - 2p < 0. On the cone |b| <= r one has
||DV(b,r)|| <= C r^{p-2}
with C depending only on p. Hence
c * sup_{D_c} ||DV|| <= C c d^{p-2} = C c^{1 - gamma(2-p)} -> 0,
d^2 = c^{2 gamma} = o(c^2), (11.5)
where the exponent 1 - gamma(2-p) > 0 by (11.1). So for small c, the matrices c DV and -c DV have
eigenvalues +lambda and -lambda with 0 < lambda < 1 at every label; the strict spectral hypothesis
(-1 < det < 0 with L = 1, plus tr = 0) of Theorem 8.1 holds on D_c. The field is smooth on the
neighborhood r > 0 of each fixed D_c (which lies in r >= d > 0).
11.4 The two families and their cost. -------------------------------------
Write s = 1 - t. The two families of tilted segments are
E_c^- = { (x, y, sqrt2 t) : |x| <= 1, (y,t) = (b,r) + c x V(b,r), (b,r) in D_c },
E_c^+ = { (x, y, sqrt2 t) : |y| <= 1, (x,s) = (b,r) - c y V(b,r), (b,r) in D_c }. (11.6)
Apply Theorem 8.1 with h = sqrt2 and L = 1 to E_c^- (field cV). Interchanging x and y and
reflecting and translating the height carries E_c^- to E_c^+, which is thus the same statement with
field -cV. So for each fixed small c and every e > 0, each family has a finite cover with compact
parallelogram bases and total base area at most
J(c) + e, J(c) = sqrt2 * integral over D_c of db dr / sqrt(1 + c^2(P^2 + 2Q^2)). (11.7)
The factor 2 in front of Q^2 records the physical height scale (h^2 = 2).
11.5 Inversion, including the cone boundary. ------------------------------
All estimates are uniform with p, gamma fixed. Suppose 2d <= T <= 1, |B| <= T, |k_0| <= c. We solve
(B,T) = (b,r) + k_0 V(b,r) (11.8)
for the inverse label (b,r). In the closed ball of radius T/4 about (B,T) one has 3T/4 <= r <= 5T/4
and |b|/r <= 5/3, and (11.2) bounds |V| <= C T^{p-1}, ||DV|| <= C T^{p-2} there. The map
(b,r) -> (B,T) - k_0 V(b,r) sends the ball into itself and is a contraction for small c, since
c T^{p-2} <= c (2d)^{p-2} = o(1). Its fixed point has d <= r <= 2.
That fixed point lies in the cone |b| <= r. Indeed the exact factorization
Q(b,r) +- P(b,r) = ((r +- b)/2) (r/a)^{p-2} [ (2-p) b/r +- p ] (11.9)
together with (11.8) gives
T +- B = (r +- b)(1 + theta_+-), |theta_+-| <= C c r^{p-2} = o(1).
The factors 1 + theta_+- are positive; since T +- B >= 0 we get r +- b >= 0. This also covers
equality on either wall, so no boundary point is lost.
11.6 The families overlap in the middle. ----------------------------------
Take a point of K_2 outside the two tip regions, so t, s >= 2d. Apply (11.8) to (B,T) = (y,t) with
k_0 = c x, and to (B,T) = (x,s) with k_0 = -c y. Let the labels be (b_-, r_-) and (b_+, r_+); both
satisfy d <= r_+- <= 2 and |b_+-| <= r_+-. It remains to show that one of them satisfies its
ceiling in (11.4).
Suppose both ceilings fail. Since delta >= 0 this forces r_-, r_+ > a. The inverse equations give
b_- = y - c x P(b_-,r_-), b_+ = x + c y P(b_+,r_+),
r_- + r_+ = 1 + c [ y Q(b_+,r_+) - x Q(b_-,r_-) ]. (11.10)
Now all labels lie in the fixed compact region r >= a where the field and its derivatives are
bounded. The last equation first gives r_- + r_+ = 1 + O(c). Since both summands exceed a = 1/2,
each is a + O(c). The first two equations give b_- = y + O(c), b_+ = x + O(c). Using Q(b,a) = b,
the bracket in (11.10) is y x - x y + 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,
b_- = y - c x P_a(y) + O(c^2), b_+ = x + c y P_a(x) + O(c^2), Q(b_+-,r_+-) = b_+- + O(c^2).
Substitution into (11.10) yields
r_- + r_+ = 1 + c^2 [ x^2 P_a(y) + y^2 P_a(x) ] + O(c^3). (11.11)
On the other hand, failure of both ceilings gives
r_- + r_+ > 1 + c^2 [ delta(b_-) + delta(b_+) ] + 2 c^{5/2},
and delta(b_-) + delta(b_+) = delta(y) + delta(x) + O(c). The exact identity
delta(y) + delta(x) - x^2 P_a(y) - y^2 P_a(x) = (x^2 - y^2)(P_a(x) - P_a(y))
= ((2-p)/(2a)) (x^2 - y^2)^2 >= 0 (11.12)
then shows that comparing with (11.11) would force 2 c^{5/2} <= C c^3, impossible for small c
(because c^{5/2}/c^3 = c^{-1/2} -> infinity). Thus every middle point lies in E_c^- union E_c^+,
and together with the two closed tip cylinders the two families cover all of K_2.
11.7 The exact quadratic saving. ------------------------------------------
Let H = { (b,r) : 0 < r <= a, |b| <= r }, of area a^2. The deletion r < d removes area d^2 = o(c^2).
The area added above r = a is c^2 * integral_{-a}^a delta(b) db + o(c^2): for |b| <= a its thickness
is exactly c^2 delta(b) + c^{5/2}, and the remaining part has a < |b| <= a + O(c^2) and thickness
O(c^2), hence area O(c^4). The field is uniformly bounded on |b| <= r <= 2 (p > 1), so Taylor
expansion of the integrand in (11.7) has uniform remainder O(c^4). As |D_c triangle H| -> 0, the
bounded quadratic coefficient may be integrated over H. Including one tip cylinder,
J(c)/sqrt2 + 4 d^2 = a^2 + c^2 kappa_p + o(c^2),
kappa_p = integral_{-a}^a delta(b) db - (1/2) integral_H (P^2 + 2Q^2) db dr. (11.13)
Evaluate the two integrals. The first is
integral_{-a}^a delta(b) db = a^4 ( p/3 + (2-p)/5 ) = 2 a^4 (p+3)/15.
For the second put eta = b/r, so P = (a/2)(r/a)^{p-1}[p + (2-p)eta^2], Q = a(r/a)^{p-1} eta, and
db dr = r deta dr; then
(1/2) integral_H (P^2 + 2Q^2) db dr
= (a^4/(4p)) integral_{-1}^1 { (1/4)[p + (2-p)eta^2]^2 + 2 eta^2 } deta
= a^4 (2p^2 + 2p + 13) / (30p).
Subtracting gives the boxed coefficient
kappa_p = a^4 (2p^2 + 10p - 13) / (30p). (11.14)
The range (11.1) for p is exactly the range on which 2p^2 + 10p - 13 < 0, so kappa_p < 0. At p -> 1
its continuous value is -a^4/30; the construction uses p > 1 only to keep strict hyperbolicity.
11.8 Completion. ----------------------------------------------------------
Fix p in the range (11.1) and then gamma. Choose c > 0 small enough for the derivative bounds, for
all coverage statements, and for
2 J(c) + 8 sqrt2 d^2 = sqrt2/2 + 2 sqrt2 kappa_p c^2 + o(c^2) < sqrt2/2. (11.15)
(Use 2J(c) + 8 sqrt2 d^2 = 2 sqrt2 ( J(c)/sqrt2 + 4 d^2 ) and (11.13).) The strict difference is
some Delta > 0. Only now apply Theorem 8.1 to each family, choosing the two additive errors with
sum below Delta/2. Those two finite covers and the two triangular tip cylinders give a finite
cylinder cover of K_2 with compact parallelogram bases and total base area strictly below
A_min(K_2)/2 = sqrt2/2. This is an independent proof of the Main Theorem (for K_2), with a
different local mechanism: a power-law derivative blow-up tamed by a cutoff and two cheap tips.
------------------------------------------------------------------------------
12. AN ANGULAR FIELD WITH SQUARE-ZERO DIFFERENTIAL
------------------------------------------------------------------------------
The singular field of Section 11 needs p > 1 to stay strictly hyperbolic away from the apex. At the
limiting exponent p = 1 the field becomes purely angular: it depends only on direction, and its
differential is nilpotent (square-zero). The differential then satisfies tr = 0 and det = 0, which
is still within the companion theorem's range (-1 < det <= 0 with L = 1). What fails is smoothness
at the apex: the field cannot be extended continuously there, because a single point carries
segments of every slope. We handle this by cutting out a small tip, applying the theorem to the
truncated label set, and covering the omitted tip segments by a finite net of fixed-slope cylinders.
12.1 The angular field. --------------------------------------------------
Return to K_2 = { (x, y, h z) : 0 <= z <= 1, |x| <= 1 - z, |y| <= z } with h = sqrt2, so that
A_min(K_2) = h and, in the "similarity" language, (x,y,z) are normalized coordinates with Euclidean
metric dx^2 + dy^2 + 2 dz^2 (physical height h z). Write a nonzero label of the upper half-plane as
b = (p t, t) with t > 0. For a field of the form G(b) = (v(p), w(p)) (a function of the slope p
only), differentiation gives
D G(p t, t) = (1/t) * (column (v'(p), w'(p))) * (row (1, -p)). (12.1)
The product of this rank-one matrix with itself is (1/t^2) (column)(row)(column)(row); the middle
scalar (row)(column) = v'(p) - p w'(p). So (DG)^2 = 0 exactly when v'(p) = p w'(p). We also want
the velocity tangent to the boundary rays p = +-1, that is v(p) = p w(p) there. Take
c = 1/2, w(p) = c p, and integrate v'(p) = p w'(p) = c p with v(+-1) = +-1 * c:
v(p) = c/2 + c p^2 / 2 = (c/2)(1 + p^2). (12.2)
Then v(p) = p w(p) at p = +-1 indeed ( v(1) = c, w(1)=c; v(-1)=c, p w(p) = (-1)(-c)=c). Define the
"allowed quadratic overlap"
a(p) = c^2 p^2 v(p) = (c^3/2) p^2 (1 + p^2). (12.3)
For 0 < k < 1 put
H(p) = c + k^2 a(p) + k^3, -1 <= p <= 1. (12.4)
The cubic term k^3 supplies a strict margin in the coverage estimate.
12.2 The two sweeps and their coverage. ----------------------------------
In normalized (x,y,z) coordinates define two sets swept by segments:
E_+ = { (s, p t + k s v(p), t + k s w(p)) : |s| <= 1, |p| <= 1, 0 <= t <= H(p) },
E_- = { (p t - k s v(p), s, 1 - t + k s w(p)) : |s| <= 1, |p| <= 1, 0 <= t <= H(p) }. (12.5)
At k = 0 these are the lower and upper halves of K_2 split by axis-parallel segments. The claim is
that they still cover K_2 for 0 < k < 1.
LEMMA 12.1. For every 0 < k < 1, every normalized point of K_2 lies in E_+ union E_-.
Proof. First an elementary inversion fact. If z >= 0, d is real, and |y| <= z, the equation
y = r z + d (1 - r^2)
has a solution r in [-1,1] with z - 2 d r >= 0. (For d > 0 the nondecreasing branch of this
concave quadratic runs from -z at r = -1 to at least z; for d < 0 the nondecreasing branch of the
convex quadratic runs from at most -z at r = -1 to z at r = 1; for d = 0 it is linear, and if z = 0
then y = 0 and any r works.) Apply it twice to a normalized point (x,y,z) of K_2 to get r, q in
[-1,1] with
y = r z + (k c x / 2)(1 - r^2), t = z - k c x r >= 0,
x = q(1 - z) - (k c y / 2)(1 - q^2), t' = 1 - z + k c y q >= 0. (12.6)
Equivalently,
y = r t + k x v(r), x = q t' - k y v(q). (12.7)
If t <= H(r), the first sweep contains the point with axial parameter s = x; if t' <= H(q), the
second contains it with s = y. In both cases |s| <= 1.
Suppose both inequalities fail, so t, t' > c. From (12.6), t + t' = 1 + k c (y q - x r) <= 1 + 2 k c.
Since 2c = 1, both t - c and t' - c lie in (0, 2kc]. The bounds |x|,|y| <= 1 and 0 < v <= c together
with (12.7) give the center error estimate
|x - c q| <= 3 k c, |y - c r| <= 3 k c. (12.8)
Substituting (12.7) into t + t' - 1 and simplifying gives the exact identity
(1 - k c r q)(t - c) + (1 + k c r q)(t' - c) = k^2 c ( x q v(r) + y r v(q) ). (12.9)
Both coefficients on the left are positive. Because both cutoff inequalities failed, the left side
is strictly greater than
k^2 (a(r) + a(q)) + (2 + c r q (a(q) - a(r))) k^3 >= k^2 (a(r) + a(q)) + (2 - c^4) k^3,
using 0 <= a <= c^3. On the right, (12.8) together with 0 < v <= c bounds it by
k^2 c^2 ( q^2 v(r) + r^2 v(q) ) + 6 c^3 k^3 <= k^2 (a(r) + a(q)) + 6 c^3 k^3,
where the last inequality uses (r^2 - q^2)(v(r) - v(q)) >= 0 (both factors have the sign of
|r| - |q|, since v is increasing in |p|). The two bounds are contradictory because they would force
6 c^3 >= 2 - c^4; but at c = 1/2, 6 c^3 = 3/4 and 2 - c^4 = 31/16, so 6 c^3 < 2 - c^4.
All inversions used closed intervals, so the argument includes faces, edges, and vertices. []
12.3 The cost predicted by the sweeps. -----------------------------------
For a planar tile T and a constant velocity g = (g_1, g_2), write
C(T,g) = { (s, xi) : s real, xi in T + s g }, J_h(g) = (1 + g_1^2 + h^2 g_2^2)^{-1/2}.
The two physical maps
F_+(s,xi) = (s, xi_1, h xi_2), F_-(s,xi) = (xi_1, s, h(1 - xi_2)) (12.10)
send this auxiliary cylinder to a physical cylinder with perpendicular base area h |T| J_h(g):
the transverse tile has physical area h |T|, and projection perpendicular to (1, g_1, h g_2) scales
it by J_h(g). The second map differs from the first by a Euclidean isometry, so a square T of any
orientation gives a compact convex parallelogram base. Under these maps both sweeps (12.5) have the
common auxiliary form
(s, (p t, t) + s sigma k (v(p), w(p))), |s| <= 1, |p| <= 1, 0 <= t <= H(p), sigma in {+1,-1}.
The Jacobian of (p,t) -> (p t, t) is t, so the integral predicting the weighted tile cost of either
sweep is
I(k) = integral_{-1}^1 ( H(p)^2 / 2 ) ( 1 + k^2 ( v(p)^2 + h^2 w(p)^2 ) )^{-1/2} dp. (12.11)
LEMMA 12.2. As k -> 0, I(k) = 1/4 - k^2/480 + O(k^3); in particular I(k) < 1/4 for all small k.
Proof. Expand (1 + k^2(v^2 + h^2 w^2))^{-1/2} = 1 - (k^2/2)(v^2 + h^2 w^2) + O(k^4) and
H(p)^2 = c^2 + 2 c k^2 a(p) + O(k^3), uniformly on [-1,1]. With c = 1/2, c^2 = 1/4, and h^2 = 2,
I(k) = c^2 + k^2 ( c * integral a dp - (c^2/4) * integral (v^2 + 2 w^2) dp ) + O(k^3).
Direct polynomial integration gives
integral_{-1}^1 a dp = 8 c^3/15, integral_{-1}^1 (v^2 + 2 w^2) dp = 34 c^2/15.
The coefficient of k^2 is therefore c * 8c^3/15 - (c^2/4) * 34c^2/15 = (16 - 17)/30 * c^4 = -c^4/30.
At c = 1/2 this is -(1/16)/30 = -1/480. []
12.4 A finite cover, including the apex. ---------------------------------
On the open half-plane b_2 > 0 define, for either sign sigma,
G_sigma(b) = sigma k ( v(b_1/b_2), w(b_1/b_2) ),
using the same polynomial formulas for v, w at every real argument. This is a smooth field, and the
differential computation above gives, for its label (p t, t),
D G_sigma(p t, t) = (sigma k c / t) * (column (p, 1)) * (row (1, -p)),
(D G_sigma)^2 = 0. (12.12)
At t = 0 every label (p t, t) is the origin, but the assigned velocity still depends on p; so G_sigma
has no continuous extension to the apex, and the tip must be covered separately.
LEMMA 12.3 (a finite apex net). Fix 0 < k < 1 and 0 < rho <= 1. For either sign sigma, the part of
the signed sweep with 0 <= t <= rho is covered by at most ceil(2/rho) + 1 auxiliary cylinders whose
tiles are squares of side 2(1 + k c) rho; their total weighted tile area is at most
16 (1 + k c)^2 rho.
Proof. Put m = ceil(2/rho) and p_j = -1 + 2j/m for 0 <= j <= m. Every p in [-1,1] is within rho of
some p_j. Use the common tile T_j = [-(1+kc)rho, (1+kc)rho]^2 and velocity g_j = sigma k(v(p_j),
w(p_j)). Since |v'| <= c and |w'| = c on [-1,1], each coordinate of
(p t, t) + s sigma k ( (v(p),w(p)) - (v(p_j),w(p_j)) )
has absolute value at most (1 + kc) rho for 0 <= t <= rho and |s| <= 1, so the cylinders cover these
segments, including t = 0. Finally J_h <= 1 and m + 1 <= 2/rho + 2 <= 4/rho, so the weighted cost is
at most (4/rho) * 4 (1 + kc)^2 rho^2 = 16 (1 + kc)^2 rho. []
The remaining finite approximation uses Theorem 8.1 with L = 1. A square-zero planar matrix has
trace and determinant zero, so the field G_sigma satisfies tr = 0 and -1 < det = 0 <= 0 on the open
half-plane. Theorem 8.1 therefore supplies, for a fixed compact set D in that half-plane, a fixed
h = sqrt2, and any positive tolerance, finitely many cylinders with compact square (hence
parallelogram) bases covering the truncated sweep { (s, b + s G_sigma(b)) : b in D, |s| <= 1 }, with
total weighted cost at most integral_D J_h(G_sigma) plus that tolerance. No small-shear or boundary
regularity hypothesis is needed.
PROPOSITION 12.4. K_2 admits a finite cylinder cover with compact convex parallelogram perpendicular
bases whose total area is less than (1/2) A_min(K_2) = h/2.
Proof. Choose and fix 0 < k < 1 with I(k) < c^2 = 1/4 (Lemma 12.2), then choose
0 < eps < c^2 - I(k), and then 0 < rho < min{ c, eps / (32 (1 + kc)^2 ) }.
By Lemma 12.3 the apex net costs less than eps/2 in weighted tile area, for each sign. Now form the
truncated label set
D_rho = { (p t, t) : -1 <= p <= 1, rho <= t <= H(p) }.
It is compact, lies in the open half-plane b_2 > 0, and G_sigma is C^1 on a neighborhood of it with
square-zero differential there (12.12). Apply Theorem 8.1 with this fixed D_rho, h = sqrt2, unit
axial interval. Although the derivative can grow like 1/t as rho shrinks, no hypothesis bounds it by
a fixed constant; the finite approximation is chosen after rho is fixed. Since the weight is
independent of t and of the sign,
integral_{D_rho} J_h(G_sigma) = integral_{-1}^1 ((H(p)^2 - rho^2)/2) (1 + k^2(v^2 + 2w^2))^{-1/2} dp
<= I(k).
Combining the finite approximation with the apex net gives a finite physical cylinder cover of each
entire sweep with cost at most h (I(k) + eps). All its perpendicular bases are compact convex
parallelograms. By Lemma 12.1 the two lists together cover all of K_2, and their total area is at
most
2 h (I(k) + eps) < 2 h c^2 = h/2 = (1/2) A_min(K_2).
The explicit apex count controls the omitted endpoint; finiteness of the other lists is supplied at
the fixed tolerance, with no uniform bound needed as k -> 0. []
This completes the third and last construction. It is the most delicate at the apex, but it needs
only the square-zero (nilpotent) endpoint of the companion theorem on the truncated set plus a cheap
finite net, so it needs neither strict hyperbolicity nor a smooth extension at the point where the
slope field is genuinely undefined.
==============================================================================
PART III. APPENDIX
==============================================================================
------------------------------------------------------------------------------
13. THE MINIMUM PROJECTION AREA OF THE EDGE-TWO TETRAHEDRON
------------------------------------------------------------------------------
Section 2 computed A_min(K_1) = h/2 for the edge-one tetrahedron by the entering-face argument
(Lemma 2.1). For the edge-two tetrahedron K_2 = { (x,y,sqrt2 t) : 0 <= t <= 1, |x| <= 1-t,
|y| <= t } used in Sections 11 and 12 we need A_min(K_2) = sqrt2. By similarity
K_2 = 2 K_1 scaled, so A_min(K_2) = 4 A_min(K_1) = 4 (h/2) = 2h = sqrt2 with h = 1/sqrt2; but we
give two independent direct proofs, both used above.
LEMMA 13.1 (second moments of the face normals). A_min(K_2) = sqrt2.
Proof. The four outward unit face normals of K_2 are
n_1, n_2 = (0, +-sqrt2, -1)/sqrt3, n_3, n_4 = (+-sqrt2, 0, 1)/sqrt3,
and each face has area sqrt3 (the face of an edge-two regular tetrahedron). For a unit vector u the
front faces project onto pi_{u^perp} K_2 with disjoint interiors (overlaps have area zero), and the
back faces give the same area, so
|pi_{u^perp} K_2| = (sqrt3 / 2) * sum_{i=1}^4 | u . n_i |. (13.1)
Next, because sum_i n_i = 0 and sum_i n_i n_i^T = (4/3) I_3, the numbers a_i = u . n_i satisfy
sum_i a_i = 0, sum_i a_i^2 = u^T (sum_i n_i n_i^T) u = (4/3) |u|^2 = 4/3.
If S = sum_i max(a_i, 0), then the sum of the absolute values of the negative parts is also S (their
signed sum is 0), so sum_i |a_i| = 2S, and sum_i a_i^2 <= (sum_i |a_i|)^2 / 2? More directly,
sum_i a_i^2 <= 2 S^2 because the positive and negative parts each contribute at most S^2. Hence
4/3 <= 2 S^2, i.e. S >= sqrt(2/3), and (13.1) gives
|pi_{u^perp} K_2| = (sqrt3/2)(2S) = sqrt3 * S >= sqrt3 * sqrt(2/3) = sqrt2.
Equality holds for u = (1,0,0) (then two of the u.n_i vanish and the other two are equal in size).
Hence A_min(K_2) = sqrt2. []
LEMMA 13.2 (projection area from a longest chord). Let T be a nondegenerate tetrahedron, u a unit
vector, and L_u the length of a longest chord of T parallel to u. Then
|pi_{u^perp} T| = 3 vol(T) / L_u. (13.2)
Consequently min_{|u|=1} |pi_{u^perp} K_2| = 3 vol(K_2) / (diameter of K_2) = 2 sqrt2 / 2 = sqrt2.
Proof. Let lambda_1,...,lambda_4 be the affine barycentric coordinates of T, and set
Delta_i = lambda_i(x+u) - lambda_i(x). These differences are independent of x and sum to 0, and they
are not all zero (barycentric coordinates distinguish points). Hence
M = sum_i max(0, -Delta_i) > 0.
For e >= 0, the conditions x in T and x + e u in T are equivalent to lambda_i(x) >= e max(0,-Delta_i)
for every i; summing these lower bounds over i uses total barycentric mass 1, giving the further
requirement e M <= 1. So T intersect (T - e u) is nonempty exactly when e <= M^{-1}, and for
e < M^{-1} it is a translate of the homothetic copy of T of ratio 1 - eM. This proves L_u = M^{-1}
and, for 0 <= e < M^{-1},
vol(T intersect (T - e u)) = (1 - e M)^3 vol(T). (13.3)
Now let ell(q) be the length of the fiber of T above q in pi_{u^perp} T. Fubini gives
vol(T intersect (T - e u)) = integral over pi_{u^perp} T of (ell(q) - e)_+ dq.
Every interior point of the projection has a fiber of positive length (homothetic copies of T about an
interior point, with ratio increasing to 1, lie inside T and their projections fill the projection up
to its area-zero boundary). The difference quotient of (ell - e)_+ at e = 0 lies between -1 and 0,
and tends to -1 where ell > 0, so dominated convergence makes the right-derivative of the integral at
e = 0 equal to -|pi_{u^perp} T|. Differentiating (13.3) gives -3 M vol(T) = -|pi_{u^perp} T|, i.e.
(13.2) since M = 1/L_u.
For K_2: its volume is 2 sqrt2/3 (Section 11.1), its four vertices are pairwise at distance 2 so its
diameter is 2, and no chord parallel to any unit direction can exceed the diameter; an edge direction
attains 2. Hence max_u L_u = 2 and (13.2) gives min_u |pi_{u^perp} K_2| = 3 (2 sqrt2/3) / 2 = sqrt2.
[]
These two independent proofs agree, and together with the similarity K_2 = 2 K_1 they confirm
A_min(K_1) = h/2 of Section 2 as well.
------------------------------------------------------------------------------
14. WHAT HAS BEEN PROVED
------------------------------------------------------------------------------
Part I (Sections 1-10) gave a complete proof of the Main Theorem:
* For the regular edge-one tetrahedron K_1, the reference minimum projection area is h/2
(Section 2);
* the two perpendicular-segment families with the rational slope field W_d still cover K_1 once the
label domain is enlarged by the specific quartic padding (Section 6), and their continuous cost
expands as 1/8 + eps^2 (J_d - I_d) + q/2 + O(eps^4) with I_d - J_d -> 1/15360 > 0 (Section 7);
* choosing d, then eta < 2 (I_d - J_d), then eps, makes each family cost strictly less than h/8, and
the companion finite-approximation theorem realizes each family by finitely many cylinders with
either compact parallelogram bases (Section 8) or bounded open measurable bases (Section 9),
keeping the strict inequality;
* therefore sum |B_i| < h/4 = (1/2) A_min(K_1) and the relative cost is below 1/2 (Section 8);
* invertible affine maps preserve relative cost, so every nondegenerate tetrahedron admits such a
cover with relative cost below 1/2 (Section 10).
Part II (Sections 11-12) gave two further, independent constructions of the same phenomenon, both at
edge two: a singular power-law field cut off near the apex with two cheap triangular tip cylinders
(coefficient kappa_p = a^4 (2p^2 + 10p - 13)/(30p) < 0), and an angular square-zero field whose only
delicate point, the apex, is covered by a finite velocity net. Each independently produces a finite
cover of K_2 of total base area below A_min(K_2)/2.
Note on sharpness of the method. The guaranteed strict improvement per family is of order eps^2 with
coefficient (I_d - J_d) - eta/2, so it is real for every sufficiently small eps but has no asserted
quantitative lower bound independent of eps. What is certain is the sign: the tilt decreases base
area faster than the required padding increases it. Combined with the lower bound of Bezdek and
Litvak, this places the infimum relative cost of a finite cylinder cover of any nondegenerate
tetrahedron in [1/3, 1/2): the directionwise half-bound conjecture fails.
END OF FILE