J. geom. 74 (2002) 78 – 85
0047–2468/02/020078 – 08
© Birkhäuser Verlag, Basel, 2002
Illuminating a convex lateral surface by horizontal directions
Boris V. Dekster∗
Abstract. This paper deals with a modification of the following well-known Illumination Problem.
Let C be a convex body in E d , d ≥ 2. Let x be a point in ∂C and u be a direction. Consider the axis passing
through x and having the direction u. One says that u illuminates x if contains a point y ∈ int C which succeeds
x. If each point of a part of ∂C is illuminated by at least one direction from a set U , this part is said to be illuminated
by U . When the whole ∂C is illuminated by U , we say that C is illuminated by U .
The Hadwiger Conjecture, still unproved, states that each convex body in E d can be illuminated by 2d directions.
There are numerous partial results in this area.
If a restriction is applied to the directions, one naturally might need for illumination more than 2d such directions.
def
Fix a pair P , Q of parallel (“horizontal”) supporting hyperplanes and call the set L(C) = ∂C\[(C ∩P )∪(C ∩Q)]
the lateral surface of C. If P and Q are chosen orthogonal to a diameter of C, then C ∩ P and C ∩ Q are points,
and two directions will illuminate them. Try now to illuminate the lateral surface L(C) by horizontal directions
only. It turns out that, for d ≥ 3, no finite number of horizontal directions might be sufficient for this. Moreover,
a body C is constructed such that, if a set of horizontal directions illuminates L(C), it must be dense in the sphere
S d−2 of all horizontal directions. This is a part of a more general theorem here.
Mathematics Subject Classification (2000): 52A20, 52A37, 52C17.
Key words: Illumination, Hadwiger conjecture.
This paper deals with a modification of the following well-known Illumination Problem.
Let C be a convex body (compact convex set with a non-empty interior) in E d, d ≥ 2. Let
x be a point in ∂C and u be a direction (unit vector). Consider the axis passing through
x and having the direction u. We say that u illuminates x if contains a point y ∈ int C
which succeeds x. If each point of a part of ∂C is illuminated by at least one direction from
a set U , we say that this part is illuminated by U . When the whole ∂C is illuminated by U ,
we say that C is illuminated by U .
The Hadwiger Conjecture, still unproved, states that each convex body in E d can be
illuminated by 2d directions. There are numerous partial results in this area. Levi [4]
proved the Conjecture for d = 2. Boltyanski reported his proof of the Conjecture for
d = 3 at two conferences in spring 2000 in Oberwolfach and Balatonföldvár, Hungary.
(A complete proof in writing is not available yet.) Some upper bounds for the minimum
number of illuminating directions have been published. Other results involve restrictions
∗ Supported by a Canadian NSERC Grant
78
Vol. 74, 2002
Illuminating a convex lateral surface by horizontal directions
79
on the bodies. Levi [4] and Boltyanski [1] proved the Conjecture for smooth bodies (in
which case d + 1 directions suffice), Lassak [3] - for centrally symmetric bodies. A detailed
description of the history of the problem and results in this area can be found in [2, §34].
If a restriction is applied to the directions, one naturally might need for illumination more
than 2d such directions. For each convex body under consideration, we will fix a pair P , Q
of parallel supporting hyperplanes and call them horizontal. The set L(C),
def
L(C) = ∂C\[(C ∩ P ) ∪ (C ∩ Q)],
(1)
will be called the lateral surface of C. Suppose the parts C ∩ P and C ∩ Q of ∂C are
illuminated by a few directions. (If P and Q are chosen orthogonal to a diameter of C, then
C ∩ P and C ∩ Q are points and two directions will illuminate them.) Try now to illuminate
the lateral surface L(C) by horizontal directions only. It turns out that, for d ≥ 3, no finite
number of horizontal directions might be sufficient for this. Moreover, we construct a body
C such that, if a set of horizontal directions illuminates L(C), it should be dense in the
sphere S d−2 of all horizontal directions. This is a part of a little more detailed theorem
below.
Let Kα with α in an index set A be a family of convex bodies in E m , m ≥ 2, and let Lm
be the set of all nondegenerate linear transformations : E m → E m . For fixed α and ,
consider 2 closed concentric balls, one lying in (Kα ) and the other one containing (Kα ).
The ratio (≤ 1) of their radii obviously attains its maximum ρ(α, ) ≤ 1 over all such pairs
of balls. Put
σ () = inf ρ(α, );
α∈A
τ = sup σ ()
∈Lm
(≤ 1).
(2)
(3)
REMARK 1. Obviously τ is an affine invariant of the family Kα . If the family consists of
only one centrally symmetric set K, definition (3) will yield just the Banach-Mazur distance
between K and a ball.
THEOREM. With C, P , Q, L(C) as above and the dimension d ≥ 3, let h be the distance
between P and Q and let Pα be the horizontal hyperplane between P and Q at the distance
αh from P , α ∈ (0, 1). Put Kα = C ∩ Pα . Identifying Kα with its vertical projection onto
P = E d−1 , define τ = τ (Kα ) as in (3), with m = d − 1.
A. Suppose that τ > 0 and consider the standard unit sphere S d−2 in E d−1 . Let N (τ )
be the minimum number of open (d − 1)-dimensional balls of radius τ with their
centers on S d−2 sufficient to cover the sphere S d−2 . Then L(C) can be illuminated
by N (τ ) horizontal directions.
80
Boris V. Dekster
J. Geom.
B. There exists a convex body C0 with C0 ∩ P being a point such that, if a set U of
horizontal directions illuminates L(C0 ), then U is dense in the sphere S d−2 of all
horizontal directions.
C. For any k ∈ {0, 1, . . . , d − 3}, there exists a body Ck with dim Ck ∩ P = k such
that no finite number of horizontal directions can illuminate L(Ck ).
REMARK 2. Some upper bounds for N (τ ) will be given at the end of the paper.
REMARK 3. When both C ∩ P and C ∩ Q are (d − 1)-dimensional, this is obviously case
A of the Theorem. We do not know if infinitely many horizontal directions can be required
when one of the two has dimension d − 2 and the other d − 2 or d − 1. Neither do we know
if a dense set of horizontal directions can be required when none of the two is a point and
one has a dimension ≤ d − 3.
Due to A, the equality τ = 0 holds for the body C0 in B. In general, this equality does not
imply that infinitely many horizontal directions are required: consider a tetrahedron with
two horizontal edges.
Proof. A. Take an ∈ (0, τ ) and, by (3), choose ∈ Ld−1 such that
σ () > τ − > 0.
(4)
Treating E d as the metric product E d−1 × E 1 with E d−1 = P , consider the nondegenerate
linear transformation ∗ : E d → E d defined by the formula
∗ ((x, z)) = ((x), z).
We illuminate now the lateral surface L(∗ (C)).
Take an arbitrary point p = (y, z) ⊂ L(∗ (C)) and put α = z/h. Obviously
p ∈ ∂(∗ (C) ∩ Pα ) = ∂(∗ (C ∩ Pα )) = ∂((Kα )).
(5)
By (2), there exist 2 (d − 1)-dimensional balls b and B of radii r and R, r ≤ R, centered
at the same point c such that
b ⊂ (Kα ) ⊂ B
and
r
= ρ(α, ) ≥ σ () > τ − ,
R
(6)
see (4). Then ∂B, for sufficiently small , can be covered by N = N (τ ) copies bi0 of int b
whose centers ci ∈ ∂B, i = 1, 2, . . . , N. Let cq be the radius of B passing through p and
suppose q ∈ bi0 . The distance between q and the radius cci is < r. Then the distance from
p to cci is also < r. Therefore the line through p having the direction ui of the segment
Vol. 74, 2002
Illuminating a convex lateral surface by horizontal directions
81
ci c crosses int b and hence int ((Kα )) beyond the point p. Thus p is illuminated by ui as
a point of ∂((Kα )) in the horizontal plane Pα . Hence p is illuminated by the horizontal
direction ui as a point of L(∗ (C)).
This means that the horizontal directions u1 , u2 , . . . , uN illuminate L(∗ (C)). Then the
directions −1
∗ (ui ), i = 1, 2, . . . , N, illuminate
−1
−1
∗ (L(∗ (C))) = L(∗ (∗ (C))) = L(C).
(7)
B. We construct the body C0 as follows. Let x1 , . . . , xd be the standard coordinates in E d
and denote by H0 the (horizontal) hyperplane xd = 0. Consider the standard unit sphere
S d−2 in H0 . Let points ui ∈ S d−2 , i = 1, 2, . . . , form a dense set in S d−2 . We construct
first an increasing sequence of convex bodies Ci . Take an arbitrary convex body F0 in H0
with the origin in int F . Consider a cone A0 with the base F0 and a vertex on the xd -axis
above F0 . Cut off a top part of A0 by a horizontal plane H1 . In the cross-section H1 ∩ A0 ,
construct the convex hull F1 of a closed (d − 1)-dimensional ball centered at the point z1 on
the xd -axis and a point v1 outside the ball. Choose them so that the segment v1 z1 have the
direction u1 and that the angle ϕ1 between the segment v1 z1 and the rulings of the conical
part of ∂F1 be < 1. Finally, put
C1 = conv (F0 ∩ F1 ).
Thus, for i = 1,
a) the convex body Ci has a top horizontal face Fi with an interior point zi on the
xd -axis. A part of ∂Fi is a circular cone with a vertex vi and the segment vi zi being
a part of its axis. Its rulings form with vi zi an angle ϕi < 1/ i. The segment vi zi has
the direction ui .
Another property of Ci , i = 1, is easy to establish.
b) Let p ∈ ∂Fi and S ⊂ Hi be a supporting (d − 2)-dimensional plane of Fi at p.
Consider the steepest hyperplane S̃ that passes through S and supports Ci . Let β(p, S)
be the angle between S̃ and a horizontal plane. Then
β(p, S) ≥ βi > 0
(8)
where βi depends only on the body Ci .
Suppose now that one has constructed a body Ci , i ≥ 1, satisfying conditions a) and b).
Then Ci+1 fitting a) and b) can be constructed as follows. Consider a cone Ai with the base
Fi and a vertex on the xd -axis above Fi . Due to b), this vertex can be chosen so low that
the body Ci ∪ Ai is still convex. Cut off a top part of Ai by a horizontal plane Hi+1 . In the
crosssection Hi+1 ∩ Ai , construct the convex hull Fi+1 of a closed (d − 1)-dimensional ball
82
Boris V. Dekster
J. Geom.
centered at the point zi+1 on the xd -axis and a point vi+1 outside the ball. Choose them so
that the segment vi+1 zi+1 have the direction ui+1 and the angle ϕi+1 between the segment
1
. Finally, put
vi+1 zi+1 and the rulings of the conical part of ∂Fi+1 be < i+1
Ci+1 = conv (Ci ∪ Fi+1 ).
(9)
One can check now easily that conditions a) and b) (with i replaced by i + 1) hold for Ci+1 .
Now put z = limi→∞ zi and
∞
Ci ∪ z.
(10)
C0 =
i=1
Come now back to the statement B. With P being the horizontal plane through z, one
obviously has C0 ∩ P being a point (z). Let bw (ρ) be a metric ball in S d−2 of a radius
ρ > 0 centered at a point w. Suppose, contrary to B, a set U ⊂ S d−2 illuminates L(C0 )
though
(11)
U ∩ bw (ρ) = .
Since ϕi < 1i , see condition a), the angles
ϕi < ρ/2
(12)
for sufficiently large i. Since the set {u1 , u2 , . . . , } is dense in S d−2 , one has in addition that
δ(ui , w) < ρ/2
(13)
for some large i where δ(·, ·) is the distance on the sphere S d−2 . If now a direction u ∈ U
illuminates the point vi described in a), then
δ(u, ui ) < ϕi < ρ/2,
see (12). Combining that with (13) yields δ(u, w) < ρ contrary to (11).
C. Statement C holds for d = 3 due to B. Suppose it holds for a dimension d − 1. Take
d ≥ 4. When k = 0, statement C holds by B. When k ≥ 1, by the inductive assumption,
and a horizontal hyperplane P in E d−1 such
there exist a (d − 1)-dimensional body Ck−1
that
∩ P = k − 1
dim Ck−1
(14)
).
no finite number of horizontal directions can illuminate L(Ck−1
(15)
and
Place Ck−1
in the x2 x3 . . . xd -plane of E d with P being orthogonal to the xd -axis. Put
Ck = [0, 1] × Ck−1
.
Vol. 74, 2002
Illuminating a convex lateral surface by horizontal directions
83
Let us show that L(Ck ) cannot be illuminated by finitely many horizontal directions in E d
if P is chosen to be parallel to the x1 . . . xd−1 -plane and to pass through P . Suppose to
the contrary that L(Ck ) is illuminated by horizontal directions u1 , u2 , . . . , uN . Take an
⊂ Ck . It is illuminated by a direction ui . Let a segment pq, q ∈
arbitrary point p ∈ Ck−1
int Ck , have the direction ui . Obviously the orthogonal projection q of q onto x2 x3 . . . xd , and the direction ui of the segment pq is horizontal and illuminates
plane lies in int Ck−1
) is illuminated by one of the
the point p on L(Ck−1 ) in E d−1 . Thus each point of L(Ck−1
horizontal directions u1 , u2 , . . . , uN contrary to (15).
By (14),
∩ P + 1 = k − 1 + 1 = k.
dim Ck ∩ P = dim Ck−1
This completes the proof.
We estimate now N (τ ), see part A of the Theorem.
Let S n−1 be the standard unit sphere in E n .
Denote by Z(n, ϕ) the minimum number of open spherical caps (i.e. open metric balls in
S n−1 ) of diameter ϕ sufficient to cover S n−1 . An n-dimensional ball of radius τ with its
2
center on S n−1 covers such a cap of diameter ϕτ = 2 cos−1 2−τ
2 . Therefore
Since ϕτ > 2τ , one has also
N (τ ) = Z(d − 1, ϕτ ).
(16)
N (τ ) ≤ Z(d − 1, 2τ ).
(17)
Thus, an upper bound for Z(n, ϕ) is required. A few such results are known. Obviously,
2π
+ 1.
Z(2, ϕ) =
ϕ
(18)
Let n be ≥ 3. Take ϕ ∈ (2 cos−1 n1 , π ]. Note that cos−1 (1/n) is the angle between the axis
and a generatrix of the circular conic surface Fi whose vertex is the center o of a regular
n-simplex and whose n generatrices pass through the vertices of a face fi . The cones conv
Fi , i = 1, 2, . . . , n + 1, clearly cover E n . Their cross-sections with the sphere S n−1 ⊂ E n
centered at o are thus closed caps of diameter 2 cos−1 (1/n) covering S n−1 . Therefore n + 1
open caps of diameter ϕ can cover S n−1 . Thus
1
Z(n, ϕ) ≤ n + 1 for ϕ ∈ 2 cos−1 , π .
(19)
n
Replacing the regular n-simplex above by a cube, one gets similarly to (19) the following:
1
1
Z(n, ϕ) ≤ 2n for ϕ ∈ 2 cos−1 √ , 2 cos−1
.
(20)
n
n
84
Boris V. Dekster
J. Geom.
√
For ϕ ≤ 2 cos−1 (1/ n), we could not find any ready estimates for Z(n, ϕ) (except for
an upper bound in [6], in probabilistic terms.) The following argument helps however.
Let A(n, ϕ) be the maximum number of open spherical caps of diameter ϕ that can be
placed on S n−1 without overlapping. Let c1 , c2 , . . . , ck , k = A(n, ϕ/2), be the centers of
nonoverlapping caps on S n−1 of diameter ϕ/2. For each point p ∈ S n−1 , at least one of
the distances pci , i = 1, 2, . . . , k, on the sphere should be < ϕ2 : otherwise our family of k
nonoverlapping caps can be augmented by one more such cap centered at p, contrary to the
maximal property of k. Therefore p will be covered by an open cap of diameter ϕ centered
at one of ci s, and hence these bigger caps cover S n−1 . That implies
ϕ
.
Z(n, ϕ) ≤ A n,
2
(21)
(This has been proved similarly in [6, Lemma 1].)
We quote now a result from [5, (2)].
Put
√
β = sin−1 [ 2 sin(ϕ/2)]
for
ϕ ∈ (0, π/2)
and put
A∗ (n, ϕ) =
π 1/2 ( n−1
2 ) sin β tan β
.
n β
2 2 0 (sin θ)n−2 (cos θ − cos β)dθ
Then
A(n, ϕ) ≤ A∗ (n, ϕ).
(22)
Combining (16) or (17) with one of (18), (19), (20) and (21) (and then with (22) if (21) is
used), one gets an upper bound for N (τ ).
References
[1]
[2]
[3]
Boltyanski, V. G., The problem of the illumination of the boundary of a convex body. (Russian) Izv. Mold.
Fil. AN SSSR 76 (1960), 77–84.
Boltyanski, V., Martini, H., Soltan, P. S. Excursions into Combinatorial Geometry, (1997), Springer-BerlinHeidelberg-New York.
Lassak, M., Solution of Hadwiger’s covering problem for centrally symmetric convex bodies in E 3 .
J. London Math. Soc. 30 (1984) no. 2, 501–511.
Vol. 74, 2002
[4]
[5]
[6]
Illuminating a convex lateral surface by horizontal directions
85
Levi, F. W., Überdeckung eines Eibereiches durch Parallelverschibungen seines offenen Kerns. Arch. Math.
6 (1955), 369–370.
Rankin, R. A., The closest packing of spherical caps in n dimensions. Proceedings of Glasgow Math. Assoc.
2 (1955), 139–144.
Wyner, A. D., Random packings and coverings of the unit n-sphere. Bell System Tech. Journal 46 (1967),
2111–2118.
Boris V. Dekster
Mount Allison University
Mathematics
67, York Street
Sackville, N.B. E4L 1E6
Canada
e-mail: [email protected]
Received 24 August 2000.