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EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 34, Num. 1 (2019), 19 – 28

doi:10.17398/2605-5686.34.1.19

Available online February 3, 2019

On self-circumferences in Minkowski planes

Mostafa Ghandehari, Horst Martini

Department of Mathematics, University of Texas at Arlington, TX 76019, U.S.A.

Faculty of Mathematics, University of Technology, 09107 Chemnitz, Germany

ghandeha@uta.edu , horst.martini@mathematik.tu-chemnitz.de

Received July 5, 2018 Presented by Javier Alonso
Accepted August 1, 2018

Abstract: This paper contains results on self-circumferences of convex figures in the frameworks

of norms and (more general) also of gauges. Let δ(n) denote the self-circumference of a regular
polygon with n sides in a normed plane. We will show that δ(n) is monotonically increasing from 6

to 2π if n is twice an odd number, and monotonically decreasing from 8 to 2π if n is twice an even

number. Calculations of self-circumferences for the case that n is odd as well as inequalities for the
self-circumference of some irregular polygons are also given. In addition, properties of the mixed

area of a plane convex body and its polar dual are used to discuss the self-circumference of convex

curves.

Key words: Gauge, Minkowski geometry, normed plane, polygonal gauges, Radon curve, self-

circumference, self-perimeter.

AMS Subject Class. (2010): 52A21, 46B20, 52A10.

1. Introduction

The concept of Minkowski distance defined by means of a convex body
centred at the origin was developed by H. Minkowski [14]. Minkowski spaces
are finite dimensional real Banach spaces, with the planar subcase of normed
or Minkowski planes, and the geometry of such spaces and planes is usually
called Minkowski geometry (cf. [1] and [19]). The article [13] and the whole
monograph [19] contain useful background material referring to Minkowski
geometry and, in particular, to those parts of the theory of convex sets which
are needed.

In this article we will deal with the self-circumference (or self-perimeter)
of unit circles of normed or Minkowski planes and (more general) of gauges.
Related inequalities for polygons can be found in [5], [9], and [10], and fur-
ther results in this direction are presented in [3], [17], and [11]. We will also
use properties of mixed areas of planar convex bodies and their polars to dis-
cuss self-circumferences of some types of (also non-polygonal) convex curves,
including Radon curves (see [12]).

ISSN: 0213-8743 (print), 2605-5686 (online)

https://doi.org/10.17398/2605-5686.34.1.19
mailto:ghandeha@uta.edu
mailto:horst.martini@mathematik.tu-chemnitz.de
https://www.eweb.unex.es/eweb/extracta/
https://creativecommons.org/licenses/by-nc/3.0/


20 m. ghandehari, h. martini

By a planar convex body K we mean a compact, convex subset of the
Euclidean plane having non-empty interior. We shall take as unit circle for
the considered Minkowski plane a convex body K centered at the origin. The
Minkowskian distance d(x,y) from x to y is defined by

d(x,y) =
de(x,y)

r
, (1)

where de(x,y) is the Euclidean distance from x to y, and r is the Euclidean
radius of K in the direction of the vector y−x. We will refer to the standard
plane with this new metric as normed or Minkowski plane. The Minkowskian
length of a polygonal path is obtained by adding the Minkowskian lengths of
the corresponding line segments. The Minkowskian length of a curve is defined
by taking the supremum over all polygonal paths inscribed to that curve.
The self-circumference of the unit circle K is the Minkowskian length of it
measured with respect to K itself. In other words, the length of the boundary
of K using the metric induced by K itself is called the self-circumference of
K and denoted δ(K). It was first proved in [6] that

6 ≤ δ(K) ≤ 8 . (2)

Equality is attained on the left if, and only if, K is the affine image of a regular
hexagon, and on the right if, and only if, K is a parallelogram. Another proof is
given in [16], and Chakerian and Talley [3] established a number of properties
of self-circumferences and raised interesting questions. Martini and Shcherba
(cf. [9] and [10]) discussed self-perimeters of quadrangles and pentagons for
the more general concept of gauges, where the unit circle is still a convex
curve, but not necessarily centered at the origin, and (1)) holds analogously;
see also [11].

2. Preliminaries

Let K be a plane convex body with the origin as interior point. For each
angle θ, 0 ≤ θ < 2π, we let r(K,θ) be the radius of K in direction (cos θ, sin θ),
such that the boundary of K has the equation r = r(K,θ) in polar coordinates.
The distance from the origin to the supporting line of K with outward normal
(cos θ, sin θ) is denoted by h(K,θ). This is the restriction of the support
function of K to the Euclidean unit circle. Since K is convex, it has a well-
defined unique tangent line at all but at most a countable number of points.
We let ds(K,θ) represent the element of Euclidean arclength of the boundary



on self-circumferences in minkowski planes 21

of K at a point where the unit normal is given by (cos θ, sin θ). Then we have
for the length of the boundary of K

L(K) =

2π∫
0

h(K,θ)dθ , (3)

while the Euclidean area of K is given by

A(K) =
1

2

2π∫
0

h(K,θ)ds(K,θ) . (4)

The polar dual of K, denoted by K∗, is another plane convex body having
the origin as interior point, and it is defined by

h(K∗,θ) =
1

r(K,θ)
and r(K∗,θ) =

1

h(K,θ)
. (5)

The mixed area A(K1,K2) of two convex sets K1,K2 is defined by

A(K1,K2) =
1

2

2π∫
0

h(K1,θ)ds(K2,θ) . (6)

It turns out that the mixed area is symmetric in its arguments. The following
result (due to Firey [4]) will be used: The mixed area of a plane convex body
and its polar dual is at least π.

The unit circle K of a Minkowski plane is referred to as its indicatrix. We
define the isoperimetrix of that plane to be the convex body T such that

h(T,θ) =
1

r(K,θ + π
2
)
, (7)

cf. [1] and [19]. The boundary of a centrally symmetric set is called a Radon
curve if it coincides with the corresponding isoperimetrix. For further prop-
erties of Radon curves we refer to [12]. We now discuss the definition of
self-circumference and give some properties.

If K is a centrally symmetric convex body centered at the origin, then by
(1) and the preceding discussion, the self-circumference δ(K) is given by

δ(K) =

∫
ds(K,θ)

r(K,θ + π
2
)
. (8)



22 m. ghandehari, h. martini

If z is any point interior to K and K is not necessarily centered at the ori-
gin (thus really yielding a gauge), then the positive and the negative self-
circumference (depending on orientation) of K relative to z are defined by

δ+(K,z) =

∫
ds(K,θ)

r(K,θ + π
2
)

(9)

and

δ−(K,z) =

∫
ds(K,θ)

r(K,θ − π
2
)
, (10)

where the origin of the coordinate system is at z. Both δ+(K,z) and δ−(K,z)
reduce to δ(K) in case that K is symmetric with respect to z. For K defining
a gauge, Go lab [6] conjectured that δ(K,z) ≥ 6, for all interior points z, and
max δ(K,z) ≤ 9. The latter conjecture was settled by Grünbaum [7]. And in
[17] the lower bound for the general case was confirmed.

If K1 and K2 are plane convex bodies with the origin as an interior point,
then the length of the positively oriented boundary of K1 with respect to K2
is given by

δ+(K1,K2) =

∫
ds(K1,θ)

r(K2,θ +
π
2
)
, (11)

and the length of the negatively oriented boundary by

δ−(K1,K2) =

∫
ds(K1,θ)

r(K2,θ − π2 )
. (12)

Schäffer [16] and independently Thompson [18] proved that for K centered
at the origin δ+(K) = δ−(K

∗) and δ−(K) = δ+(K
∗) hold. More generally,

Chakerian [2] used the concept of mixed areas to prove that

δ+(K1,K2) = δ−(K
∗
2,K

∗
1 ) and δ−(K1,K2) = δ+(K

∗
2,K

∗
1 ) . (13)

3. Polygons

We study first the self-circumference of regular polygons. After that we
give inequalities for positive and negative self-circumferences of quadrangles.
In [9] and [10] results on self-circumferences of quadrangles and pentagons
were obtained, and in [5] for polygons. Theorem 1 below was also proved in
[5]; for the sake of completeness we include it here.



on self-circumferences in minkowski planes 23

Theorem 1. Let δ(n) denote the self-circumference of an affine image of
a regular polygon with n sides. Then δ(n) is monotonically increasing from 6
to 2π if n is twice an odd number, and monotonically decreasing from 8 to 2π
if n is twice an even number.

The formula for δ(n) in Theorem 1 is δ(n) = 2n sin π
n

for n being a doubled
odd number. Since sinπ

n
assures rational values only for n = 2 and n = 6 (see

p. 144 in [15], Problems 197.1 and 197.5), it follows that 6 is the only rational
value obtained.

In case n is twice an even number, δ(n) is given by δ(n) = 2n tan π
n

. The
fact that tan π

n
is only rational for n = 4 implies that in this case the only

rational value in question is δ(n) = 8.
In case n is an odd number, the value δ(n) with respect to the case of

having a center of symmetry is given by

δ(n) = 2n tan
π

n
cos

π

2n
.

This sequence is monotonically decreasing from 9 to 2π. The proof that
δ(n) is monotonically decreasing is not trivial. We include a proof of Theo-
rem 2.

Theorem 2. The sequence δ(n) = 2n tan π
n

cos π
2n

, where n is odd, is
monotonically decreasing from 9 to 2π, n = 3, 5, 7, . . ..

Proof. Let g(x) = (tan 2x cos x)/x. We want to show that g′(x) > 0. Let
g(x) = h(x)/x. Then g′(x) =

(
xh′(x) − h(x)

)
/x2; so g′(x) > 0 iff xh′(x) −

h(x) > 0 for 0 < x < π
4
. It suffices to show that (xh′(x) − h(x))′ > 0 for

0 < x < π
4
. We have (xh′ −h)′ = xh′′ + h′ −h′ = xh′′ > 0 iff h′′(x) > 0,

h′(x) = 2 sec2 2x cos x− tan 2x sin x,

h′′(x) =
8 cos x sin 2x

cos3(2x)
−

2 sin x

cos2(2x)
−

2 sin x cos x

cos 2x

=
2 sin x

cos3(2x)

(
8 cos2 x− 2 cos2 x + 2 sin2 x− cos2 x cos2(2x)

)
=

2 sin x

cos3(2x)

(
8 cos2 x− 2 cos2 x + 2 sin2 x− cos2 x cos2(2x)

)
,

h′′(x) =
2 sin x

cos3(2x)

(
2 + 4 cos2 x− cos2 x cos2(2x)

)
=

2 sin x

cos3(2x)

(
2 + cos2 x(4 − cos2(2x))

)
> 0 ,



24 m. ghandehari, h. martini

since |cos x| ≤ 1, and (2 sin x)/ cos3(2x) > 0 for 0 < x < π
4
.

Thus, h′′(x) > 0 for 0 < x < π
4
, and so g′(x) > 0 for 0 < x < π

4
.

The following two theorems give inequalities for self-circumferences of a
quadrangle and a trapezoid.

p

q

r

s

z

Figure 1

Theorem 3. The self-circumference δ+(P,z) of a convex quadrilateral P
with respect to the point z of intersection of its diagonals is at least 8, with
equality if and only if the quadrilateral is a parallelogram.

Proof. Consider a quadrilateral P with vertices p,q,r,s (see Figure 1).
Let z be the point of intersection of the diagonals. Then, by using similar
triangles, we get

δ+(P,z) =
de(z,p) + de(z,r)

de(z,r)
+
de(z,q) + de(z,s)

de(z,s)
+
de(z,p) + de(z,r)

de(z,p)

+
de(z,q) + de(z,s)

de(z,s)
.

Thus we have

δ+(P,z) = 4 +
de(z,p)

de(z,r)
+
de(z,r)

de(z,p)
+
de(z,q)

de(z,s)
+
de(z,s)

de(z,q)
≥ 8 ,



on self-circumferences in minkowski planes 25

where the last inequality follows from the arithmetic-geometric mean inequal-
ity. Equality holds if and only if de(z,p) = de(z,r),de(z,q) = de(z,s), which
implies that P is a parallelogram.

Theorem 4. The self-circumference δ+(P,o) of a trapezoid P with re-
spect to the midpoint o of one of the diagonals is at least 8, with equality if
and only if P is a parallelogram.

Proof. Consider a trapezoid P with vertices p,q,r,s (see Figure 2). In the
figure, ou is parallel to ps, ov is parallel to pq, ow is parallel to qr, and ox is
parallel to rs. By using similar triangles, we get

δ+(P,o) =
de(p,q)

de(ov)
+
de(q,r)

de(ow)
+

(de(r,s)

de(ox)
+
de(sp)

de(ou)
,

where the equalities de(r,s) = 2de(ov),de(p,q) = 2de(o,x),de(q,r) = 2de(ow),
and de(p,s) = 2de(ow) are used. The arithmetic-geometric mean inequality
implies that δ+(P,o) ≥ 8, with equality if and only if de(r,s) = de(p,q),
yielding again a parallelogram.

o

p

qr

s

v

w

x

u

Figure 2

Chakerian and Talley [3] gave an example of a trapezoid to show that
δ+(K,z) and δ−(K,z) do not assume their minimum at the same point, thus
answering a question of Hammer posed in [8].



26 m. ghandehari, h. martini

4. Curves

In the following we use properties of mixed areas of a plane convex body
and its polar dual to discuss self-circumferences of certain convex curves. The
following theorem shows that the self-circumference of a plane convex body
with four-fold symmetry is at least 2π.

Theorem 5. Let K be a plane convex body centered at the origin. As-
sume that r(K,θ) is an equation of the boundary of K in polar coordinates,
and assume that r(K,θ) = r(K,θ + π

2
), 0 ≤ θ ≤ 2π, i.e., K has four-fold

symmetry. Then its self-circumference satisfies δ(K) ≥ 2π.

Proof. Using the definition given in (8) and four-fold symmetry, we obtain

δ(K) =

∫
ds(K,θ)

r(K,θ + π
2
)

=

∫
ds(K,θ)

r(K,θ)
.

By the property of the polar dual given in (5) and the properties of mixed
areas presented in (6), it follows that

δ(K) =

∫
ds(K,θ)

r(K,θ)
=

∫
h(K∗,θ)ds(K,θ) = 2A(K∗,K) .

Firey’s result from [4] states that the mixed area of a plane convex body and
its polar dual is at least π. It follows that δ(K) is at least 2π.

Recall that the isoperimetrix T was defined by (7). The following theorem
gives the Minkowskian length of the boundary of a plane convex body with
respect to the isoperimetrix.

Theorem 6. Let K be a plane convex body, and assume that T is the
isoperimetrix, that is, the polar dual of K rotated by 90 degrees. Then
δ+(K,T) = 2A(K), where A(K) is the Euclidean area of K.

Proof. By the definition given in (11) we obtain

δ+(K,T) =

∫
ds(K,θ)

r(T,θ + π
2
)
.

By the definition of the isoperimetrix, r(T,θ + π
2
) = r(K∗,θ). Thus

δ+(K,T) =

∫
ds(K,θ)

r(K∗,θ)
=

∫
h(K,θ)ds(K,θ) = 2A(K) ,



on self-circumferences in minkowski planes 27

where we have used (4) and (5) giving the Euclidean area and the used prop-
erty of the polar dual.

If the boundary of K is a Radon curve, then it coincides with its isoperime-
trix. Thus, the self-circumference of a Radon curve is equal to twice its Eu-
clidean area. The following theorem refers to the length of the Euclidean unit
circle with respect to a plane convex body K.

Theorem 7. Let K be a plane convex body, and assume that B is the
Euclidean unit circle. Then the length of B with respect to K equals the
Euclidean length of the polar dual of K. That is, δ+(B,K) = L(K

∗).

Proof. By the result of Chakerian given in (13) we obtain

δ+(B,K) = δ−(K
∗,B∗) = δ−(K

∗,B) .

Assuming that the polar dual of K is calculated at the center of the Euclidean
unit circle B, it follows that δ+(B,K) = L(K

∗).

In the subcase where K is a square with vertices at (±1, 0), (0,±1), the
Minkowskian distance is the same as used for the so-called Taxicab Metric.
The polar dual is a square with sides parallel to the axes. Thus, the length of a
Euclidean unit circle in the Taxicab Metric is the same as the Euclidean length
of the circumscribed square which is 8, and thus we have finally “squared the
circle”.

References

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[5] M. Ghandehari, R. Pfiefer, Polygonal circles, Mathematics and Com-
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28 m. ghandehari, h. martini

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[14] H. Minkowski, Theorie der konvexen Körper, insbesondere Begründung
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[15] G. Pólya, G. Szegö, “Problems and Theorems in Analysis”, Vol. 2,
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[18] A.C. Thompson, An equiperimetric property of Minkowski circles, Bull. Lon-
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[19] A.C. Thompson, “Minkowski Geometry”, Encyclopedia of Mathematics and
its Applications, 63, Cambridge University Press, Cambridge, 1996.


	Introduction
	Preliminaries
	Polygons
	Curves