Variations of Classic Characterizations of
Ellipsoids and a Short Proof of the False Centre
Theorem.
L. Montejano and E. Morales
Abstract
We shall develop variations and generalizations of several classic theorems concerning characterizations of ellipsoids. In particular, this will
allow us to give a short and compressive prove of the false centre theorem.
Introduction. Let K be a convex body in Euclidean space Rn . A point p ∈ Rn
is called a false centre of K if p is not a centre of symmetry of K but for any
plane H through p, the section H ∩ K is either empty or centrally symmetric.
In 1969, in its well known paper [5], Rogers proved that a body with a false
centre has a centre and conjectured that actually it is an ellipsoid. In 1973
P.W. Aitchison, C.M. Petty and A. Rogers [1] proved the conjecture for the
case in which the false centre is an interior point of body and D. Larman [4]
gave a complete proof of the conjecture. Both proofs are quite complicated, in
particular, the treatment of the case of non strictly convex bodies introduced
technically intricate arguments, that obscure the ideas behind the proof.
In this paper we shall develop variations and generalizations of several classic theorems concerning characterizations of ellipsoids. This will allow us to
give a short and compressive prove of the false centre theorem. The main idea
of some of our results is that for centrally symmetric bodies the classical characterizations of ellipsoids may be obtained weakened the hypothesis. Another
idea is that the Rogers equichordal theorem is intimately related with the false
centre theorem. Our proof of the false centre theorem makes clear this statement. Finally, our proposition 2, which is a variation of classical Blaschke’s
theorem concerning the planarity of the shadow boundaries, allow us to handle
the technicalities introduced by the non-strictly convex case.
The results. Let L be a line through the origin. Define T ∂(K, L) as the
union of all tangent lines of K parallel to L. The shadow boundary of K in
the direction of L is defined as S∂(K, L) = T ∂(K, L) ∩ K. If v 6= 0, then
S∂(K, v) = S∂(K, L), where L is the line through the interval [0, v].
Proposition 1 Let K ⊂ R3 be a convex body centrally symmetric with respect
the origin and let p ∈ R3 be a false center. Then, for every plane H through
1
the origin such that p or the origin is not the center of the non-empty section
(p + H) ∩ K, there is a line LH such that bd(H ∩ K) ⊂ S∂(K, LH ).
Proof. Let H be a plane through the origin such that p or the origin is not
the center of the non-empty section (p + H) ∩ K and let vH its center. It will be
enough to prove that, for t > 0 sufficiently small, the section (H + tvH ) ∩ K is
centrally symmetric with respect tvH , because, if this is the case, (H + tvH ) ∩
K − 2tvH = (H − tvH ) ∩ K, which implies that bd(H ∩ K) + {λvH | λ ∈ R} =
T ∂(K, vH ) and hence that bd(H ∩ K) ⊂ S∂(K, LH ). In order to prove that
(H + tvH ) ∩ K is centrally symmetric with respect tvH we shall use Roger’s
Equichordal Theorem [4]. Fix the points (p − vH + tvH ) and (vH − p + tvH ) in
the plane H + tvH . We shall prove that parallel chords of (H + tvH ) ∩ K through
each one of these points have the same length. Let L ⊂ H be a line through the
origin.
For every x ∈ R3 such that (x + L) ∩ K 6= φ, we denote by [x] the chord
(x + L) ∩ K and by | [x] | its length.
Let Γ be the plane through L and vH . Note that [p] is a chord of (p + H) ∩ K
and of (p + Γ) ∩ K. Since (p + H) ∩ K is centrally symmetric with respect vH ,
we have that | [p] |=| [p − 2vH ] | and by symmetry of K that | [p] |=| [p − 2vh ] | .
Note that [p] and [2vH − p] are parallel chords of (p + Γ) ∩ K with the same
length, so one of the following two situations hold:
i)the centre of (p + Γ) ∩ K lies in (p − vH ) + L ⊂ H, or
ii)| [p − vH + λvH ] |=| [p] |,
for −1 ≤ λ ≤ 1.
In the first case, since (p + Γ) ∩ K is centrally symmetric with centre at
(p − vH ) + L, we have that | [p − vH + tvH ] |=| [p − vH − tvH ] | and for symmetry
of K that | [p − vH + tvH ] |=| [vH − p + tvH ] |, for 0 ≤ t ≤ 1. In the second
case, by symmetry of K, | [vH − p + λvH ] |=| [p] |, for −1 ≤ λ ≤ 1. In any
case | [p − vH + tvH ] |=| [vH − p + tvH ] | and since this happens for every line
L ⊂ H through the origin that cuts K, by Roger’s Equicordal Theorem [6],
(H + tvH ) ∩ K is centrally symmetric with respect tvH , for every 0 ≤ t ≤ 1 and
consequently bd(H ∩ K) ⊂ S∂(K, LH ).
Remark 1 We give now a proof of the False Centre Theorem, when K is
strictly convex and the false centre is an interior point of K. The case n > 3
follows as usual from the case n = 3. By Rogers Theorem [5], we may assume
that K is centrally symmetric with respect the origin. By Proposition 1 and
since K is strictly convex, for every plane H through the origin such that p
or the origin is not the centre of (p + H) ∩ K, there is a line LH such that
bd(H ∩ K) = S∂(K, LH ). Also, the fact that K is strictly convex implies easily
that there is at most one plane H through the origin such that p is the centre
of (p + H) ∩ K. Now, by continuity, the boundary of every section of K through
the origin is a shadow boundary and therefore, K is an ellipsoid. The following
proposition, which is an interesting variation of Blaschke’s Theorem, will help
us to deal with non strictly convex bodies.
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Proposition 2 Let K ⊂ R3 be a convex body and let 0 ∈ int K. Suppose
that, for every plane H through 0, there is a line LH such that bd(H ∩ K) ⊂
S∂(K, LH ). Then K is an ellipsoid.
Proof. Assume the origin lies in every LH . First note that by hypothesis, if H is
a plane through 0, we have that H ∩ T ∂(K, LH ) ⊂ bd K and bd(K ∩ H) + LH =
T ∂(K, LH ).
We start the proof by showing that there are not disks in bd K. Let Γ be
a supporting plane with the property that relint(Γ ∩ K) 6= φ and let [a, b] be
an interior chord of Γ ∩ K which is not a diametral chord (there is always one
sufficiently close to any direction). Let H be the plane through 0 and [a, b].
Since [a, b] = K ∩ Γ ∩ H, then [a, b] is a support set of K ∩ H, this implies
that [a, b] + LH = T ∂(K, LH ) ∩ Γ, but hence Γ ∩ K ⊂ [a, b] + LH which is a
contradiction.
We shall prove now that K is strictly convex. Let I be an interval in bd K
and let H be a plane through 0 which does not contain I but cuts its interior. Then LH is parallel to I, otherwise since I + LH ⊂ T ∂(K, LH ) and
H ∩ T ∂(K, LH ) ⊂ bd K, we have a disk in bd K, which is impossible. Let now
Γ be any plane through 0 sufficiently close to H. By the above, LΓ = LH and
therefore Γ ∩ T ∂(K, LH ) ⊂ bd K. This implies that there is an open interval
J ⊂ LH through the origin such that bd(H ∩ K) + J ⊂ bd K. Let now T be
any plane through 0 not parallel to LH , and let x ∈ H ∩ T ∩ bd K. So T cuts
the interval x + J ⊂ bd K and therefore, by the above, LT = LH . Consequently
T ∂(K, LH ) ∩ T ⊂ bd K, but this is a contradiction because K is bounded. This
proves that K is strictly convex.
Since K is strictly convex then every plane H through 0 is a shadow boundary of K and therefore, it is well known that K is an ellipsoid. By completeness
we finish the argument. Note first that K is centrally symmetric, because every chord through 0 is a diametral chord. In order to show this, take a line L
through 0 and let H and Γ be two planes through L. Let vH and vΓ such that
bd(H ∩ K) = S∂(K, vH ) and bd(Γ ∩ K) = S∂(K, vΓ ). If L ∩ bd K = {a,b}, we
have that there are parallel supporting planes to K at a and b, generated by
the vectors vH and vΓ .
Given a vector u in the tangent cone to K at a, the shadow boundary
S∂(K, u) is a curve through a and b. These two points divide the shadow
boundary in exactly two arcs that will be called by us, the u-half shadows of
S∂(K, u). On the other hand, the curve S∂(K, u) divides bd K in two closed
topological hemispheres that will be called by us, the u-hemispheres of bd K.
It is not difficult to see that if u and v belong to the tangent cone to K at
a, then both u-half-shadows of S∂(K, u) are completely contained, each one, in
each of the two v-hemispheres of K. In other words, the two shadow boundaries
S∂(K, u) and S∂(K, v), transversely intersect exactly in {a, b}. Due to this fact
and the hypothesis, we have that if a u-half-shadow intersects bd(H ∩ K) in a
point different from {a, b}, then this u-half-shadow is contained in bd(H ∩ K).
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So, S∂(K, u) is contained in a plane and by the Theorem of Blaschke, K is an
ellipsoid.
Remark 2 Let us prove the False Centre Theorem when the false centre is an
interior point. The case n > 3 follows as usual from the case n = 3. By Rogers
Theorem [5], we may assume that K is centrally symmetric with respect the
origin. Let Ω be the set of planes H passing through the origin such that the
centre of (p + H) ∩ K is p. Let H1 , H2 ∈ Ω. Since ((p + Hi ) ∩ K) − 2p = (−p +
Hi )∩K, i = 1, 2, if T is a supporting plane of K at x ∈ (p+H1 )∩(p+H2 )∩bd K
we have that T − 2p is a supporting plane of x − 2p ∈ bd K. This implies that
T is parallel to [0, p] and therefore that x − 2tp ∈ bd K, for 0 ≤ t ≤ 1. If now
H ∈ int Ω, it follows, by the above argument, that (1−2t)p+bd(H ∩K) ⊂ bd K,
for 0 ≤ t ≤ 1, which implies that bd(H ∩ K) ⊂ S∂(K, p). By Proposition 1 and
continuity, the boundary of every section of K through the origin is contained
is some shadow boundary and therefore, by Proposition 2, we have that K is
an ellipsoid. After having proved some interesting variations of Brunn’s and
Blaschke’s Theorems, we shall prove the False Centre Theorem when the false
centre is not an interior point.
We continue with the following variations of classic characterizations of ellipsoids, for centrally symmetric convex bodies.
Theorem 1 Let K ⊂ Rn be a centrally symmetric convex body. If for every
direction d, sufficiently close to a fixed direction D, the mid-points of the chords
parallel to d, lie in a hyperplane, then K is an ellipsoid..
Proof. Our proof follows the proof of the classic Theorem in [3]. In particular, we derive our theorem from Theorem 16.11, but using now that given two
points p, q ∈ bd K, there is a finite collection of chords, each one in a direction
sufficiently close to D, connecting p and q.
Theorem 2 Let K ⊂ Rn be a convex body centrally symmetric with respect the
origin and let L be a line through the origin. If for every hyperplane H through
the origin, sufficiently close to L, the section H ∩ K is an ellipsoid, then K is
an ellipsoid.
Proof. The case n > 3 follows easily from the case n = 3. Let L ∩ bd K =
{q1 , q2 } and let H1 and H2 parallel supporting planes to K at q1 and q2 respectively. We may assume, without loss of generality that L is orthogonal to Hi .
Then for every plane H through L the ellipse H ∩ K has the chord [q1 , q2 ] as
one of its main axis. The other axis lies in the plane through the origin parallel
to Hi . If ∆ is the plane parallel to Hi passing through the origin and ∆t is the
plane parallel to Hi at a distance t from the
q origin, then ∆ ∩ K is homothetic
to ∆t ∩ K and the radio of homothety is
4
1−
t2
r2
, where 2r is the length of
the segment L ∩ K. So, all sections parallel to the planes Hi , are centrally
symmetric, similar and similar situated with centres at L.
Let now H be a plane parallel to Hi through the origin. The convex figure
H ∩ K is strictly convex otherwise, if there is an interval in bd(H ∩ K) there
is also an interval in the boundary of a parallel section sufficiently close to H1
and therefore there is an interval in the boundary of sections through the origin,
sufficiently close to L, contradicting the fact that all sections sufficiently close
to the line L are ellipses. This implies that the shadow boundary of K in a
direction parallel to Hi lies in a plane through L.
Take a plane Γ parallel to Hi and sufficiently close to H1 . We shall show
that Γ ∩ K is an ellipse by showing that the midpoints of parallel chords of
Γ ∩ K lie in a line. Let [a, b] be a chord of Γ ∩ K, let T be the plane through
[a, b] and the origin. We assume that Γ is so close to Hi that T ∩ K is an
ellipse. Let L be the supporting line to T ∩ K parallel to [a, b], then the point
L ∩ K = {x} ∈ S∂(K, L), the midpoint of [a, b] and the origin are collinear and
therefore, if Λ is the plane through L that contains S∂(K, L), the midpoint of
[a, b] is contained in Λ ∩ Γ. Since the same is true for every chord parallel to
[a, b], we have that Γ ∩ K is an ellipse, but this implies that H ∩ K is an ellipse
and consequently that K is an ellipsoid.
Theorem 3 Let K ⊂ Rn be a centrally symmetric convex body and let H be a
hyperplane through the origin. If for every direction d, sufficiently close to H,
the shadow boundary S∂(K, d) lie in a hyperplane, then K is an ellipsoid.
Proof. Again our proof follows the proof of the classic theorem in [3]. The
case n > 3 follows as usual from the case n = 3. First note that there are not
intervals in bd K in directions sufficiently close to H. Let Γ be a plane through
the origin, sufficiently close to H and let Γ1 and Γ2 be supporting planes to K
parallel to Γ. Note that for i = 1, 2, Γi ∩ K consists of a single point qi ∈ bd K.
For every line L ⊂ Γ, S∂(K, L) lie in a plane through the chord [q1 , q2 ]. This
implies (by the proof of Lemma 6 of [1] or the proof of Theorem 16.7 of [3]), that
all sections of K parallel to Γ are centrally symmetric, similar, similarly situated
with centres at [q1 , q2 ]. If T is a plane through [q1 , q2 ], then the mid-points of
all parallel chords of T ∩ K, in a direction sufficiently close to T ∩ Γ, lie in a
line. By Theorem 1, T ∩ K is an ellipse. Note that the same is true for every
diametral chord, sufficiently close to [q1 , q2 ], and therefore by Theorem 2, K is
an ellipsoid.
Theorem 4 Let K ⊂ Rn be a convex body centrally symmetric with respect the
origin and let L be a line through the origin. If for every hyperplane H through
the origin, sufficiently close to L, there is a line LH through the origin such that
bd(H ∩ K) ⊂ S∂(K, LH ), then K is an ellipsoid.
Proof. The case n > 3 follows from the case n = 3 as usual. The proof of
the case n = 3 follows line by line the proof of Proposition 2, but now using our
Theorem 3 instead of the Theorem of Blaschke.
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Remark 3 The proof of the False Centre Theorem for a false centre which is
not an interior point follows immediately from Proposition 1 and Theorem 4.
Theorem 5 Let K ⊂ Rn be a convex body centrally symmetric with respect the
origin and let L be a diametral chord of K. If for every plane H, sufficiently
close to L, the section H ∩ K is centrally symmetric, then K is an ellipsoid.
Proof. As in the proof of the Proposition 1 it follows that every plane
passing through the origin, sufficiently close to L, is contained in a shadow
boundary. The theorem follows now from Theorem 4.
Finally, we establish a conjecture:
Conjecture 1 Let K ⊂ Rn be a convex body centrally symmetric with respect
the origin and let p ∈ Rn − {0} be such that for every plane H through p,
sufficiently close to the origin, the section H ∩ K is centrally symmetric, then
K is an ellipsoid.
References
[1] P.W. Aitchison, C.M. Petty and C.A. Rogers: A convex body with a false
centre is an ellipsoid, Mathematika 18 (1971), 50-9.
[2] Burton, G. R. Some characterizations of the ellipsoid, Israel J. of Math. 28
(1977), 339-349.
[3] Buseman, H., The Geometry of Geodesics, Academic Press, New York. 1955.
[4] D.G. Larman: A note on the false centre problem, Mathematika 21 (1974),
216-27.
[5] C.A. Rogers: Sections and projections of convex bodies, Portugaliae Math.
24 (1965), 99-103.
[6] C.A. Rogers: An equichordal problem, Geom. Dedicata 10 (1981),73-8.
Instituto de Matemáticas, UNAM, Circuito exterior, C.U.
México D.F., 04510, México.
Centro de Investigación en Matemáticas, A.C., A.P.
402, Guanajuato, Gto., C.P. 3600, MÉXICO.
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