convex solids whose point-source shadow-boundaries lie in hyperplanes

J. Geom. 103 (2012), 149–160c© 2012 Springer Basel AG0047-2468/12/010149-12published online April 17, 2012DOI 10.1007/s00022-012-0114-6 Journal of Geometry

Convex solids whose point-sourceshadow-boundaries lie in hyperplanes

Valeriu Soltan

Abstract. Extending Burton’s characterization of n-dimensional ellipsoids,we describe all n-dimensional closed convex sets in R

n, possiblyunbounded, whose shadow-boundaries with respect to point-source illu-mination lie in hyperplanes.

Mathematics Subject Classification. 52A20.

Keywords. Convex, body, solid, ellipsoid, quadric, hypersurface,shadow-boundary.

1. Introduction

Various characterizations of solid ellipsoids among convex bodes in Rn form

an established topic of convex geometry (see Bonnesen and Fenchel [3, §70],Heil and Martini [5], Petty [12]). Due to various applications (see, e.g., Amir[1]), a characterization of solid ellipsoids as convex bodies with hyperplanarshadow-boundaries is one of the best known. The concept of shadow-bound-ary appears in various problems on illumination of convex sets by families ofrays with a common direction or a common point source (see, e.g., [8–10]). Weconsider below illumination by rays with a common point source (see [18] fora similar problem involving illumination by parallel rays).

By a convex solid we will mean an n-dimensional closed convex set in Rn

distinct from the entire space (convex bodies are compact convex solids). Asusual, bdM, cl M , and intM denote, respectively, the boundary, closure, andinterior of a convex set M ⊂ R

n. Also, rbdM and rintM will stand, respec-tively, for the relative boundary and relative interior of M . The convex solidK is called regular if each point x ∈ bdK belongs to precisely one supporthyperplane of K. For distinct points u, v ∈ R

n, we denote by 〈u, v〉 and [u, v〉the line through u, v and the halfline through v with endpoint u, respectively.

150 V. Soltan J. Geom.

If K ⊂ Rn is a convex solid and p is a point in R

n\K, then the shadow-bound-ary of K with respect to p, denoted Sp(K), is the set of points in bdK atwhich the halflines with endpoint p support K. Unlike convex bodies, someshadow-boundaries of convex solids may be empty (see Lemma 3.1).

Given a convex solid K ⊂ Rn and a point p ∈ R

n\K, denote by Cp(K) theconvex cone with apex p generated by K; equivalently, Cp(K) is the union ofall halflines [p, x〉, where x ∈ K (observe that Cp(K) may be non-closed). Inthese terms,

Sp(K) = bdK ∩ bdCp(K) = K ∩ bdCp(K). (1.1)

Kubota [6,7] proved that the boundary of a regular convex body K ⊂ R3 is

an ellipsoid provided each shadow-boundary Sp(K), p ∈ R3\K, lies in a plane.

Burton [4] refined this result, by showing that the boundary of a convex bodyK ⊂ R

n, n ≥ 3, is an ellipsoid if there is a scalar δ > 0 such that for any pointp ∈ R

n\K at a distance less than δ from K there is a hyperplane H ⊂ Rn

containing Sp(K).

We extend Burton’s characterization of ellipsoids to the case of convex quad-rics. In a standard way, a quadric (or a second degree hypersurface) in R

n isthe locus of points x = (ξ1, . . . , ξn) that satisfy a quadratic equation

n∑

i,k=1

aikξiξk + 2n∑

i=1

biξi + c = 0, (1.2)

where not all scalars aik are zero. A convex hypersurface is the boundary of aconvex solid in R

n (this definition includes a hyperplane or a pair of parallelhyperplanes). We say that a convex hypersurface T ⊂ R

n is a convex quadricprovided there is a real quadric Q ⊂ R

n and a connected component U of Rn\Q

such that U is convex and T = bdU . (Particular forms of convex quadrics,namely, connected components of quadrics which are convex hypersurfaces,are described by Berger [2, Proposition 15.4.7].)

As shown in [16,17], a real quadric Q ⊂ Rn contains a convex quadric if and

only if there is an orthonormal basis e1, . . . , en for Rn such that Q is the locus

of points x = (ξ1, . . . , ξn) given by one of the equations

a1ξ21 + · · · + akξ2

k = 1, 1 ≤ k ≤ n, (1.3)

a1ξ21 − a2ξ

22 − · · · − akξ2

k = 1, 2 ≤ k ≤ n, (1.4)

a1ξ21 = 0, (1.5)

a1ξ21 − a2ξ

22 − · · · − akξ2

k = 0, 2 ≤ k ≤ n, (1.6)

a1ξ21 + · · · + ak−1ξ

2k−1 = ξk, 2 ≤ k ≤ n, (1.7)

where all scalars ai involved are positive; the respective convex quadrics T ⊂ Qare given by the same equations, with an additional restriction ξ1 ≥ 0 in thecases (1.4) and (1.6). Several characterizations of convex quadrics among con-vex hypersurfaces are obtained in [14–18].

Vol. 103 (2012) Convex solids whose shadow-boundaries lie in hyperplanes 151

We recall that the recession cone of a convex solid K ⊂ Rn is given by

rec K = {e ∈ Rn | x + λe ∈ K whenever x ∈ K and λ ≥ 0}.

It is known that recK is a closed convex cone with apex o (the origin ofR

n), and recK = {o} if and only if K is unbounded (see, e.g., [19] for gen-eral properties of convex sets). Furthermore, p + rec K ⊂ cl Cp(K) for anyp ∈ R

n\K. A halfline h ⊂ Rn with endpoint p is called recessional for K

provided h − p ⊂ rec K; otherwise it is non-recessional. The halfline h is non-recessional for K if and only if the set K ∩ (x+h) is bounded for any choice ofx ∈ K. A convex set M ⊂ R

n is called line-free if it does not contain any line.

The subspace lin K = rec K ∩ (−rec K) is called the linearity space of K.Given a subspace L ⊂ R

n which is complementary to lin K, the solid K canbe expressed as the direct sum K = linK ⊕ (K ∩L), where K ∩L is a line-freeclosed convex set of the same dimension as L. Clearly, (i) K is line-free if andonly if linK = {o}, (ii) linK is an (n − 1)-dimensional subspace if and onlyif K is either a halfspace or a slab (a closed convex set bounded by a pair ofdistinct parallel hyperplanes).

2. Main result

We say that a convex solid K ⊂ Rn, distinct from a cone, is strictly convex

provided its boundary contains no line segment. An n-dimensional closed con-vex cone C ⊂ R

n is strictly convex provided it has a unique apex, v, such thatany line segment in bdC belongs to a line through v.

Theorem 2.1. For a a convex solid K ⊂ Rn, n ≥ 3, the following conditions

are equivalent:

(1) for any point p ∈ Rn\K, there is a hyperplane H ⊂ R

n such that

Sp(K) = H ∩ bdCp(K), (2.1)

(2) there is an open neighborhood U of K such that for any point p ∈ U\Kone can find a hyperplane H ⊂ R

n satisfying the inclusion

Sp(K) ⊂ H, (2.2)

(3) K has one of the following shapes:(a) bdK is a convex quadric,(b) dim (lin K) = n − 2, and K is the direct sum of lin K and a two-

dimensional strictly convex solid,(c) dim (lin K) = n − 3, and K is the direct sum of lin K and a three-

dimensional closed strictly convex cone.

Remark 2.2. Conditions (1) and (2) of Theorem 2.1 implicitly cover the trivialcase when Sp(K) = ∅ for all p ∈ R

n\K, which happens if and only if K is ahalfspace or slab (see Lemma 3.2 below). The shapes (a)–(c) are not mutuallyexclusive: a cylinder based on a two-dimensional strictly convex quadric is aparticular case of (b), and a cylinder based on a sheet of a three-dimensionalconvex elliptic cone is a particular case of (c).


3. Auxiliary lemmas

We will say that a plane P ⊂ Rn (of certain dimension m) properly supports a

closed convex set M ⊂ Rn provided P meets the relative boundary of M and

is disjoint from its relative interior.

Lemma 3.1. Given a convex solid K ⊂ Rn and a point p ∈ R

n\K,n ≥ 2, thefollowing conditions are equivalent:

(1) Sp(K) = ∅,(2) K ⊂ int Cp(K),(3) either K is a slab or K ⊂ p + int (rec K).

Proof. Since K ⊂ Cp(K), the statement (1) ⇒ (2) immediately follows from(1.1).

(2) ⇒ (3) Assuming that K is not a slab, we are going to prove the inclu-sion K ⊂ p + rec K. Indeed, suppose for a moment that K ⊂ p + rec K.Then intK ⊂ p + rec K due to the closedness of p + rec K. Choose a pointx ∈ intK\(p + rec K). Since the halfline h = [p, x〉 is non-recessional for K, itmeets K along a line segment. This shows that K is not a halfspace. BecauseK is also distinct from a slab, we obtain dim (linK) ≤ n − 2. Let L be atwo-dimensional plane through h such that its translate M = L − p satisfiesthe condition M ∩ lin K = {o}. Then K ∩ L is a line-free closed convex setof dimension two. Denote by P1 and P2 the closed halfplanes of L determinedby the line through h. At least one of the sets K ∩ P1 and K ∩ P2 is bounded(otherwise K ∩ L would be a slab between a pair of parallel lines, which isimpossible because K ∩ L is line-free). Assume, for instance, that K ∩ P1 isbounded. Denote by h′ the second halfline with endpoint p which supportsK ∩ P1. Then rbd (K ∩ P1) ∩ h′ ⊂ Sp(K), in contradiction with Sp(K) = ∅.

Hence K ⊂ p + rec K, which gives Cp(K) ⊂ p + rec K. Therefore,

K ⊂ int Cp(K) ⊂ int (p + rec K) = p + int (rec K).

(3) ⇒ (1) Assume first that K is a slab between parallel hyperplanes H1 andH2. If G is the hyperplane through p which is parallel to H1, then

Sp(K) = bdK ∩ bdCp(K) = (H1 ∪ H2) ∩ G = ∅.

If K ⊂ p + int (recK), then the inclusion p + rec K ⊂ cl Cp(K) gives

K ⊂ p + int (rec K) = int (p + rec K) ⊂ int (cl Cp(K)) = int Cp(K),

implying that Sp(K) is empty. �Lemma 3.2. If K ⊂ R

n, n ≥ 2, is a convex solid and U an open neighborhoodof K, then the following conditions are equivalent:

(1) Sp(K) = ∅ for all p ∈ U\K,(2) K is either a halfspace or a slab.


Proof. (1) ⇒ (2) Assume for a moment that K is neither a slab nor a half-space. Then we can choose distinct points x1, x2 ∈ bdK and non-parallelhyperplanes H1 and H2 supporting K at x1 and x2, respectively, such that(x1, x2) ⊂ int K and x1 /∈ H2 and x2 /∈ H1. Denote by P1 and P2 the closedhalfspaces containing K and bounded by H1 and H2, respectively. Choose atwo-dimensional plane L which contains (x1, x2) and is complementary to the(n − 2)-dimensional plane H1 ∩ H2. If p is a point in L ∩ (P1\P2) ∩ (U\K),then L contains a halfline h with endpoint p supporting K ∩ L at a pointu ∈ rbd (K ∩ L). Hence u ∈ Sp(K ∩ L) = Sp(K) ∩ L, in contradiction withSp(K) = ∅.

(2) ⇒ (1) Let K be a closed halfspace and H its boundary hyperplane. Chooseany point p ∈ U\K, and denote by G the hyperplane through p which is paral-lel to H. If V is the open halfspace of R

n bounded by G and containing K, thenCp(K) = {p} ∪ V . Hence Sp(K) = bdK ∩ bdCp(K) = H ∩ G = ∅. Similarly,Sp(K) = ∅ if K is a closed slab between a pair of parallel hyperplanes. �Lemma 3.3. Let K be a convex solid in R

n, n ≥ 2, and p a point in Rn\K.

Suppose there is hyperplane H containing Sp(K). If every boundary halfline ofthe cone Cp(K) meets K, then Sp(K) = H ∩ bdCp(K).

Proof. Due to (1.1), it suffices to show that H ∩ bd Cp(K) ⊂ Sp(K). First,we state that the cone Cp(K) is line-free. Indeed, assume for a moment thatCp(K) contains a line. Then bdCp(K) contains a line l through p. Denoteby h1 and h2 the opposite open halflines of l with common endpoint p. Bythe assumption, there are points x1 ∈ h1 ∩ K and x2 ∈ h2 ∩ K. There-fore, p ∈ (x1, x2) ⊂ K due to the convexity of K, in contradiction withp ∈ R

n\K.

Next, we observe that p /∈ H. Indeed, assume for a moment that p ∈ H. ThenH ∩clCp(K) is an (n−1)-dimensional convex cone with apex p. Since the coneCp(K) is n-dimensional, there is an open halfline h with endpoint p which liesin bdCp(K)\H. If x is a point in h ∩ K, then x ∈ Sp(K)\H, in contradictionwith the assumption.

Finally, let x ∈ H ∩ bdCp(K). Then the halfline h = [p, x〉 lies in bdCp(K).By the assumption, there is a point z ∈ h ∩ K ⊂ Sp(K) ⊂ H. Since theintersection h ∩ H is a singleton, we conclude that x = z ∈ Sp(K). �Lemma 3.4. If the boundary of a convex solid K ⊂ R

n, n ≥ 2, is a convexquadric, then for any point p ∈ R

n\K there is a hyperplane H ⊂ Rn which

satisfies the equality (2.1).

Proof. Put p = (η1, . . . , ηn) and denote by Q a real quadric in Rn such that

int K is a connected component of Rn\Q. Is is known (see, e.g., [13, pp. 262–

265]) that if Q is described by (1.2) and every support halfline of Q withendpoint p has precisely one point on Q, then Sp(Q) coincides with H ∩ Q,where H is the polar hyperplane of p with respect to Q, expressed by thefollowing linear equation in ξ1, . . . , ξn:


n∑

i,k=1

aikξiηk + 2n∑

i=1

bi(ξi + ηi) + c = 0. (3.1)

Let T = bdK. Since any line supporting Q either lies in Q or has exactly onepoint on Q, a convexity argument gives Sp(K) = H ∩ T provided every sup-port halfline of K with endpoint p has precisely one point on T . To prove theequality (2.1), we consider a description of T by each of Equations (1.3)–(1.7)separately.

If T is given by (1.3), with k = 1, or by (1.5), then K is a closed slab or aclosed halfspace, and Sp(K) = ∅ according to Lemma 3.2. If T is given by(1.3), with k ≥ 2, or by (1.7), then every boundary line of the cone Cp(K)meets K at a single point, and Lemma 3.3 implies (2.1).

Assume that T is given by (1.6), with ξ1 ≥ 0. Then K is a cone with apexo, and Sp(K) = ∅ if and only if p belongs to the opposite open cone −int K,given by the inequality

ξ1 < −(a−11 (a2ξ

22 + · · · + akξ2

k))1/2.

If p belongs to −bdK\{o} = Q\T , then Sp(K) is a face of K, described by

Sp(K) = {(tη1, . . . , tηk, ξk+1, . . . , ξn) : t < 0, ξk+1, . . . , ξn ∈ R},

and the hyperplane H supporting K along this face satisfies (2.1). If p is notin −K, then the hyperplane (3.1) satisfies (2.1).

Finally, let T be described by (1.4), with ξ1 ≥ 0. Then Sp(K) = ∅ if and onlyif p belongs to the closed convex cone C, given by the inequality

ξ1 ≤ −(a−11 (a2ξ

22 + · · · + akξ2

k))1/2.

If p /∈ C, then the hyperplane (3.1) satisfies (2.1). �

4. Proof of the main result

(3) ⇒ (1) Let p be a point in Rn\K. If bdK is a convex quadric, then the

existence of a hyperplane H satisfying (2.1) is proved in Lemma 3.4. Assumethat K has one of the shapes (b) and (c). Choose a complementary to linKsubspace L which contains p. Then K = linK ⊕ (K ∩ L), where K ∩ L is aline-free closed convex set of the same dimension as L. By the assumption,2 ≤ dim L ≤ 3. If we will prove the existence of a line N (provided dimL = 2)or a two-dimensional plane N (provided dimL = 3) in L such that

Sp(K ∩ L) = N ∩ rbdCp(K ∩ L), (4.1)

then the hyperplane H = linK ⊕ N will satisfy (2.1):

Sp(K) = lin K ⊕ Sp(K ∩ L)

= lin K ⊕ (N ∩ rbdCp(K ∩ L)) = H ∩ bdCp(K).

First, let dim L = 2. Then K ∩ L has the shape (b). If Sp(K ∩ L) = ∅, thenany line N ⊂ L disjoint from K ∩ L satisfies (4.1). If there are two halflines


with endpoint p both supporting K ∩L at points x1 and x2, respectively, thenSp(K ∩L) = {x1, x2} and the line N through x1 and x2 satisfies (4.1). If thereis precisely one halfline with endpoint p supporting K ∩ L at a point x, thenSp(K ∩ L) = {x} and K ∩ L is unbounded. Since K ∩ L is strictly convex, wecan choose an open halfline m with endpoint x entirely lying in rint (K ∩ L).Then the line N through m satisfies (4.1).

Now, let dim L = 3. Then K ∩ L has the shape (c). Denote by v the apex ofthe strictly convex cone K ∩ L. If Sp(K ∩ L) = ∅, then any plane N ⊂ Ldisjoint from K ∩L satisfies (4.1). Assume that Sp(K ∩L) = ∅. If the halfline[p, v〉 meets K ∩L along a halfline h with endpoint v, then Sp(K ∩L) = h andthe plane N ⊂ L through h which supports K ∩ L satisfies N ∩ (K ∩ L) = h;therefore, (4.1) holds. If [p, v〉 ∩ (K ∩ L) = {v}, then every halfline with end-point p supporting K ∩ L meets K ∩ L at a single point. Hence Sp(K ∩ L) isthe union of two halflines h1 and h2 with common endpoint v, and the planeN through h1 ∪ h2 satisfies (4.1).

Since (1) ⇒ (2) trivially holds, it remains to show that (2) ⇒ (3). This partis organized by induction on n ≥ 3. Let n = 3 and K ⊂ R

3 be a convex solidsatisfying condition (2) of Theorem 2.1.

I. Assume first that K contains a line l. Then K is a cylinder based on atwo-dimensional closed convex set M . If M contains a line, then K is eithera halfspace or a slab between two parallel planes, implying that bdK is adegenerate convex quadric. Let M be line-free. If M is a convex cone, thenbdK = l ⊕ rbdM is a degenerated quadric.

Suppose that M is a line-free two-dimensional convex solid distinct from acone. We state that M should be strictly convex. Indeed, assume for contra-diction, that M is not strictly convex. Then there is a line m supporting Msuch that m ∩ M is not a singleton. Clearly, m ∩ M is either a line segment[a, b] or a halfline h with endpoint a. Choose a point p ∈ m\M so close to athat p ∈ U\K. We consider separately the following two cases.

(i) Let m∩M be a halfline h with endpoint a. Given a point q ∈ (a, p), denoteby h(q) the halfline with endpoint q which contains h. We claim that h(q) isthe only halfline with endpoint q. Indeed, assume for a moment the existenceof another halfline h′ with endpoint q supporting M . Then

l ⊕ ((h′ ∩ M) ∪ h) ⊂ l ⊕ Sq(M) = Sq(K).

The last inclusion contradicts condition (2) of the theorem because the setl ⊕ ((h′ ∩ M) ∪ h) does not lie in a plane. This argument implies that rbdMcontains one more halfline with endpoint a. Hence M is a convex cone withapex a, which is impossible by the assumption. Therefore, m ∩ M cannot be ahalfline.

(ii) Let m ∩ M be a line segment [a, b]. Repeating the argument from (i), weobtain that rbdM contains a halfline with endpoint a which does not include[a, b]. This case is impossible due to (i). Summing up, M is strictly convex.


II. Hence it remains to consider the case when K is a line-free convex solid inR

3. We divide the proof of this case into Lemmas 4.1–4.4.

Lemma 4.1. Let K ⊂ Rn be a line-free convex solid satisfying condition (2)

of Theorem 2.1. If bdK contains a line segment, then K is a strictly convexcone.

Proof. Let [u, v] be a line segment in bdK. Put l = 〈u, v〉. The intersectionF = K ∩ l is either a closed line segment or a halfline. Denote by a an endpointof K ∩ l. We state that K is a cone with apex a. Indeed, assume for a momentthat K is not a cone with apex a. Then there is a two-dimensional plane Lthrough l such that K ∩ L is not a cone with apex a. Hence we can choose apoint p ∈ l ∩ (U\K) so close to a that K ∩ L has a support halfline h withendpoint p such that K ∩ h is bounded and h ⊂ l. By condition (2), there is ahyperplane H ⊂ R

n which contains Sp(K); in particular, h∪ l ⊂ H. Since h isnot recessional for K, there is an ε > 0 such that any halfline with endpoint pwhich lies in bdCp(K) and forms with h an angle less than ε supports K andis non-recessional for K. Hence we can find a non-recessional halfline h′ ⊂ Hwith endpoint p which supports K, in contradiction with the choice of H.Summing up, K is a convex cone with apex a.

If [x, z] is another line segment in bdK, then, by the argument above, K∩〈x, z〉should have a as an endpoint (since K has a unique apex). Therefore, K is astrictly convex cone. �Due to Lemma 4.1, it remains to consider the case when the solid K ⊂ R

3 isstrictly convex and distinct from a cone.

Lemma 4.2. Let K ⊂ R3 be a line-free strictly convex solid. Suppose that Ω is

an open subset of bdK and q a point in int K such that any section of Ω bya plane through q is an arc of a convex quadric curve. If there is a plane H1

through q such that H1 ∩K is bounded and H1 ∩bdK ⊂ Ω, then Ω is a subsetof a strictly convex quadric.

Proof. Due to the assumptions, the section E1 = H1 ∩bdK is an ellipse. SinceΩ is open in bdK, there is another plane H2 through q, sufficiently close toH1, such that E2 = H2 ∩ bdK is also bounded and lies in Ω. As above, E2

is an ellipse. Choosing suitable Cartesian coordinates for R3, we may assume

that q = o and E1 and E2 are given by

E1 = {(ξ1, 0, ξ3) | (ξ1 − ρ1)2 + ξ23 = ρ2

1 + 1},

E2 = {(0, ξ2, ξ3) | (ξ2 − ρ2)2 + ξ23 = ρ2

2 + 1},

where ρ1 > 0 and ρ2 > 0. Then H1 and H2 are described by the equationsξ2 = 0 and ξ1 = 0, respectively. Consider in R

3 the family of quadrics Q(μ)given by

ξ21 + ξ2

2 + ξ23 + 2μξ1ξ2 − 2ρ1ξ1 − 2ρ2ξ2 − 1 = 0, μ ∈ R.


Clearly, Ei = Hi∩Q(μ), i = 1, 2. Choose a point v = (ν1, ν2, ν3) ∈ Ω\(H1∪H2).Then ν1ν2 = 0 and v ∈ Q(μ0) for

μ0 = (1 + 2ρ1ν1 + 2ρ2ν2 − ν21 − ν2

2 − ν23)/(2ν1ν2).

We state that Ω ⊂ Q(μ0). For this, it suffices to show that Ω\L ⊂ Q(μ0), whereL is the plane containing v and the line H1 ∩ H2. Let u ∈ Ω\L. The plane Mthrough {o, u, v} meets E1∪E2 at four distinct points, say, r, r′ ∈ E1 and s, s′ ∈E2. Since Ω is strictly convex, no three points of the set X = {r, r′, s, s′, u, v}are on a line. The set X lies in M ∩ Ω, which, by the assumption, is an arcof a convex quadric curve. On the other hand, the set {r, r′, s, s′, v} lies in theplanar quadric M ∩Q(μ0). Since any planar quadric is uniquely determined bya set of five points in general position (see, e.g., [11, pp. 395–397]), we obtainu ∈ M ∩ Ω ⊂ M ∩ Q(μ0). Hence Ω\L ⊂ Q(μ0), which gives Ω ⊂ Q(μ0). Theinclusion Ω ⊂ Q(μ0) ∩ bdK shows that Q(μ0) is locally convex at any pointx ∈ Ω. Theorem 3 from [17] implies that Q(μ0) contains a convex quadric Tsuch that Ω ⊂ T . �The next lemma, formulated by Burton [4] for two-dimensional convex bod-ies, is routinely extendable to the case of two-dimensional convex solids. LetM be a strictly convex solid in the plane P and Φ a bounded open arc ofrbdK. We will say that a line segment I ⊂ P\M is Φ-compatible provided thefollowing conditions are satisfied: (i) the line l containing I is disjoint from M ,(ii) for any point p ∈ I there are precisely two halflines with endpoint p bothsupporting M at points from Φ. In what follows, Uδ(M) denotes the set of allpoints in P at a distance less than δ from M .

Lemma 4.3. [4] Let M be a strictly convex solid in the plane P and Φ a boundedopen arc of rbdM . Suppose the existence of a scalar δ ∈ (0, 1) such that foreach Φ-compatible line segment I ⊂ Uδ(M)\M there is a point w ∈ rintMwith the following property: if p is any point in I and v1, v2 are the points in Φat which the halflines with endpoint p support M , then w belongs to the openline segment (v1, v2). Then Φ is an arc of a strictly convex quadric curve. �Lemma 4.4. If K ⊂ R

3 is a line-free strictly convex solid satisfying condition(2) of Theorem 2.1, then bd K is a strictly convex quadric.

Proof. First, we are going to show that any point x ∈ bdK has a neighborhoodin bdK which is a piece of a convex quadric. Choose a point z ∈ U\K such thatthe line l = 〈x, z〉 meets int K and each boundary halfline of Cz(K) supportsK (this is possible since K is strictly convex). Denote by δ the distance fromz to K. Let B(z) be the union of all line segments [z, r] such that r ∈ bdKand (z, r) ∩ K = ∅. Since B(z) is a compact set which continuously dependson z and since B(z) tends to {x} when z converges to x, we can move z alongl so close to x that 0 < δ < 1 and B(z) lies in U . Clearly, B(z) ⊂ Cz(K).

Denote by Ω the relative interior of the set B(z)∩bd K in bdK. We state thatΩ is a piece of a strictly convex quadric. Choose a point q ∈ l ∩ int K so closeto x that a certain plane through q meets bd K along a closed bounded curvewhich lies in Ω. We are going to prove that for any plane P through q, the


section Φ = P ∩ Ω is an open arc of a convex quadric curve. Put M = K ∩ Pand U ′ = P ∩ U . Choose a Φ-compatible line segment I ⊂ Uδ(M)\M . By thechoice of B(z), we have

I ⊂ P ∩ int B(z) ⊂ P ∩ Cz(K) ∩ (U\K).

The line through I is disjoint from K because it is disjoint from M . Therefore,there are precisely two planes P1 and P2 both containing I and supporting Kat some points q1 and q2 from Ω. Denote by w the point of intersection of theline segment [q1, q2] and rintM . Due to condition (2) of the theorem, for anypoint u ∈ I there is a plane H containing Su(K). Hence q1, q2 ∈ Su(K) ⊂ Hfor any choice of u ∈ I, and H ∩P is a line through w such that both points ofthe set (H ∩ P ) ∩ rbdM belong to Φ. By Lemma 4.3, Φ is an arc of a strictlyconvex quadric curve, and Lemma 4.2 shows that Ω is a piece of a convexquadric, which is strictly convex because bdK is a strictly convex surface.

The argument above shows that bdK can be expressed as a countable unionof bounded open pieces of strictly convex quadrics. Since any two overlappingpieces of strictly convex quadrics belong to the same convex quadric, the wholesurface bdK is a strictly convex quadric. �Let n > 3. We continue the proof of (2) ⇒ (3) by induction on n, assumingthat it holds for all m ≤ n − 1, where n ≥ 4. Let K be a convex solid inR

n which satisfies condition (2) of the theorem. First, we eliminate the trivialcase when Sp(K) = ∅ for every point p ∈ U\K, assuming that K is neither ahalfspace nor a slab (see Lemma 3.2). This gives dim (lin K) ≤ n − 2.

Second, we consider the case when lin K = {o}. Let L ⊂ Rn be a subspace

complementary to lin K. Then K = lin K ⊕ M , where M = K ∩ L is a line-free closed convex set. By the above, dim L = n − dim (lin K) ≥ 2. Choosea point p ∈ L ∩ (U\K). By condition (2), there is a hyperplane H satisfyingthe inclusion (2.2). From Sp(K) = linK ⊕ Sp(M) it follows that a translateof linK lies in H. Therefore, H = linK ⊕ (H ∩ L), and the plane G = H ∩ Lhas dimension dimL − 1. Furthermore, Sp(M) ⊂ G. Since M is line-free, theinductive assumption implies that either M is a two-dimensional strictly con-vex solid (if dimM = 2), or M is a three-dimensional line-free strictly convexcone (if dim M = 3), or rbdM is a line-free convex quadric (if dimM ≥ 3).Therefore K = linK ⊕ M has one of the shapes (a)–(c).

Finally, it remains to consider the case when K is line-free. Choose a pointq ∈ int K and a hyperplane L through q. We state that, if the set M = K∩L isstrictly convex or bounded, then M satisfies condition (2) of the theorem (withL instead of R

n and L∩U instead of U). Indeed, choose a point p ∈ L∩(U\K).If Sp(M) = ∅, then any (n − 2)-dimensional plane G ⊂ L satisfies the inclu-sion Sp(M) ⊂ G. Assume that Sp(M) = ∅. Then Sp(M) = Sp(K) ∩ L. Bycondition (2), there is a hyperplane H satisfying the inclusion Sp(K) ⊂ H. Weobserve that H = L. Indeed, choose a point x ∈ Sp(M). Since M is strictlyconvex or bounded, the intersection [p, x〉 ∩ M is a singleton or a line seg-ment. Hence the halfline h = [p, x〉 is not recessional for K. By a continuityargument, there is a non-recessional halfline h′ ⊂ L with endpoint p which


supports K and is sufficiently close to [p, x〉. Since h′ ∩K ⊂ H\L, we concludethat H = L. Therefore, G = H ∩ L is a (n − 2)-dimensional plane in L suchthat Sp(M) = Sp(K) ∩ L ⊂ H ∩ L = G. By the inductive assumption, rbdMis an (n − 1)-dimensional convex quadric.

If K is strictly convex or bounded, then every section of bdK by a hyperplanethrough p is strictly convex or bounded, whence it is an (n − 1)-dimensionalconvex quadric. From [14] it follows that bdK itself is a convex quadric. Sup-pose that bdK contains a line segment. By Lemma 4.1, K is a strictly convexcone. Since K is line-free, there is a hyperplane L ⊂ R

n such that the setM = K ∩L is bounded. By the facts proved above, rbdM is an (n−1)-dimen-sional ellipsoid, implying that bdK is an convex elliptic cone. Summing up,bdK is a convex quadric.

Acknowledgments

We thank the referee for helpful comments on an earlier draft of this paper.

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Valeriu SoltanDepartment of Mathematical SciencesGeorge Mason University4400 University DriveFairfax, VA 22030USAe-mail: [email protected]

Received: February 10, 2012.

Revised: March 29, 2012.

http://www.heldermann.de/JCA/JCA19/jca19.htm

convex solids whose point-source shadow-boundaries lie in hyperplanes

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