# Some results about equichordal convex bodies

Jesús Jerónimo-Castro1, Francisco G. Jimenez-Lopez2,
and Efrén Morales-Amaya3
1,2Facultad de Ingeniería
Universidad Autónoma de Querétaro, México
3Facultad de Matemáticas-Acapulco,
Universidad Autónoma de Guerrero, México
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###### Abstract

Let $K$ and $L$ be two convex bodies in $\mathbb{R}^{n}$, $n\geq 2$, with
$L\subset\text{int}\,K$. We say that $L$ is an _equichordal body_ for $K$ if
every chord of $K$ tangent to $L$ has length equal to a given fixed value
$\lambda$. In [1], J. Barker and D. Larman proved that if $L$ is a ball, then
$K$ is a ball concentric with $L$. In this paper we prove, derived from the
proof of Theorem 1, that there exist an infinite number of closed curves,
different from circles, which possess an equichordal convex body. If the
dimension of the space is more than or equal to 3, then only Euclidean balls
possess an equichordal convex body. We also prove some results about isoptic
curves and give relations between isoptic curves and convex rotors in the
plane.

##  1 Introduction

Let $K$ be a convex body in the plane, i.e., a compact and convex set with
non-empty interior, and let $\mathcal{P}$ be a convex polygon. It is said that
$K$ is a rotor in $\mathcal{P}$ if for every rotation $\rho$, there is a
translate of $\mathcal{P}$ that contains to $\rho(K)$ and all sides of
$\mathcal{P}$ are tangent to $K$. There are many results about rotors in
regular polygons, see for instance [6], and for the particular case of rotors
in equilateral triangles see [21]. The case of rotors in squares is well-
known, indeed, bodies of constant width are a very important topic in Convex
Geometry and have many interesting properties and applications in mechanisms
(see the quite nice book [13]).

Another topic, apparently not related to rotors, is the _Equichordal Problem_.
Let $x$ be a point in the interior of a convex body $K$, we say that $x$ is an
_equichordal point_ if every chord of $K$ through $x$ have the same length.
The famous Equichordal Problem, due to W. Blaschke, W. Rothe, and R.
Weitzenböck [2], asks about the existence of a convex body with two
equichordal points. There are many false proofs about the non existence of
such a body, however, M. Rychlik finally gave a complete proof about the non
existence of a body with two equichordal points in [16]. It is worth
mentioning that there are many convex bodies, different from the disc, which
have exactly one equichordal point. Here we are interested in a generalization
of the notion of equichordal point in the following way: Let $K$ and $L$ be
two convex bodies in $\mathbb{R}^{n}$, $n\geq 2$, with
$L\subset\text{int}\,K$. We say that $L$ is an _equichordal body_ for $K$ if
every chord of $K$ tangent to $L$ have length equal to a given fixed value
$\lambda$. In [1], J. Barker and D. Larman proved that if $K$ is a convex body
that possesses an equichordal ball then it is also a ball. However, we wonder
if there exist convex bodies different from balls which possess an equichordal
convex body in its interior. It seems that bodies which float in equilibrium
in every position provide examples of such bodies in the plane (see for
instance [20]), however, it is not clear if the considered bodies $K$ are
convex or not.

In sections 3 and 4 of this paper we study convex bodies $L$ for which the
chords of one of its isoptic curves (defined in the following section), that
are tangent to $L$ have length equal to a constant number $\lambda$. Moreover,
these bodies $L$ are examples of rotors in regular polygons and if we fix the
convex body $L$ and the circumscribed polygon is rotated, while maintained
circumscribed to $K$, its vertices describe the isoptic curve of $L$. In
section 6 we also prove that in dimension $3$ or higher, only Euclidean balls
(or simply balls) possess and equichordal convex body in its interior.

##  2 Preliminary concepts

We give first some definitions and notation. Let $K$ be a given planar convex
body; for every real number $t$ we denote by $\ell(t)$ the support line of $K$
with outward normal vector $u(t)=(\cos t,\sin t)$. The function
$p:\mathbb{R}\longrightarrow\mathbb{R}$, defined as $p(t)=\max_{x\in K}\langle
u(t),x\rangle$, is known by the name of support function of $K$. When the
origin $O$ is contained in $K$, $p(t)$ is nothing else than the distance from
$O$ to the support line $\ell(t).$ The distance between the support lines
$\ell(t)$ and $\ell(t+\pi)$ is called the width of $K$ in direction $u(t)$ and
it is denoted by $w(t)$, in other words, $w(t)=p(t)+p(t+\pi)$. If $w(t)$ is
constant, independently of $t$, we say that $K$ is a body of constant width.
For any $\alpha\in(0,\pi)$, the $\alpha$-isoptic $K_{\alpha}$ of $K$ is
defined as the locus of points at which two tangent lines to $K$ intersect at
an angle $\alpha$. Using the support function, $\partial K$ is parameterized
(see for instance [19]) by

${}\gamma(t)=p(t)u(t)+p^{\prime}(t)u^{\prime}(t),\ \text{for}\ t\in[0,2\pi].$
(1)

The isoptic curve $K_{\alpha}$ can be parameterized by the same angle by the
formula (see [4] or [10])

${}\gamma_{\alpha}(t)=p(t)u(t)+\left[p(t)\cot\alpha+\frac{1}{\sin\alpha}p(t+\pi-\alpha)\right]u^{\prime}(t).$
(2)

By Cauchy’s formula, the perimeter of $K$ can be obtained by (see [19])

$L(K)=\int_{0}^{2\pi}p(t)dt.$ (3)

For any $t\in\mathbb{R}$ we define (see Fig. 1)

$\displaystyle a(t)$ $\displaystyle=|\gamma_{\alpha}(t)-\gamma(t)|,$ (4)
$\displaystyle b(t)$
$\displaystyle=|\gamma_{\alpha}(t)-\gamma(t+\pi-\alpha)|,$ (5) $\displaystyle
c(t)$
$\displaystyle=|\gamma_{\alpha}(t)-\gamma_{\alpha}(t+\pi-\alpha)|=b(t)+a(t+\pi-\alpha),$
(6) $\displaystyle q(t)$ $\displaystyle=|\gamma(t)-\gamma(t+\pi-\alpha)|,$ (7)
$\displaystyle\lambda(t)$
$\displaystyle=|\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t-\pi+\alpha)|.$
(8)

We also define $d(t)$ to be the distance between the points obtained by
projecting the origin $O$ onto the support lines of $K$ at $\gamma(t)$ and
$\gamma(t+\pi-\alpha)$.

Figure 1: Parameters of the isoptic curve

By some tedious but simple calculations we can express the lengths of all
these chords in terms of the support function of $K$:

$\displaystyle a(t)$
$\displaystyle=\frac{1}{\sin\alpha}[p(t+\pi-\alpha)+p(t)\cos\alpha-p^{\prime}(t)\sin\alpha],$
(9) $\displaystyle b(t)$
$\displaystyle=\frac{1}{\sin\alpha}[p(t+\pi-\alpha)\cos\alpha+p^{\prime}(t+\pi-\alpha)\sin\alpha+p(t)],$
(10) $\displaystyle c(t)$
$\displaystyle=\frac{1}{\sin\alpha}[2p(t+\pi-\alpha)\cos\alpha+p(t)+p(t-2\alpha)],$
(11) $\displaystyle d(t)$
$\displaystyle=\sqrt{p^{2}(t)+p^{2}(t+\pi-\alpha)+2p(t)p(t+\pi-\alpha)\cos\alpha}.$
(12)

Finally, the support function of a convex body is a periodic function, with
period $2\pi$, and it is also absolutely continuous, so we can consider its
expansion in terms of the Fourier series (see [6]), i.e.,

$p(t)=a_{0}+\sum_{n=1}^{\infty}\left(a_{n}\cos nt+b_{n}\sin nt\right).$ (13)

The first and second derivatives of $p$ are expressed as

$\displaystyle p^{\prime}(t)$
$\displaystyle=-\sum_{n=1}^{\infty}\left(na_{n}\sin nt-nb_{n}\cos nt\right),$
(14) $\displaystyle p^{\prime\prime}(t)$
$\displaystyle=-n^{2}\sum_{n=1}^{\infty}\left(a_{n}\cos nt+b_{n}\sin
nt\right).$ (15)

##  3 Some results about isoptic curves in the plane

Our first result about isoptic chords is the following.

######  Theorem 1.

Let $K$ be a strictly convex body in the plane with differentiable boundary
and let $\alpha\in(0,\pi)$ be a fixed angle such that $\frac{\alpha}{\pi}$ is
an irrational number. Suppose $c(t)=c_{0}$, for every $t\in[0,2\pi]$, for a
positive number $c_{0}.$ Then $K$ is a disc.

_Proof._ Since $c(t)=c_{0}$ we have by (11) that

$c(t)=\frac{1}{\sin\alpha}[2p(t+\pi-\alpha)\cos\alpha+p(t)+p(t-2\alpha)]=c_{0},$

it follows that

$2p^{\prime}(t+\pi-\alpha)\cos\alpha+p^{\prime}(t)+p^{\prime}(t-2\alpha)=0.$
(16)

If we substitute the Fourier coefficients of $p(t)$ in the differential
equation (16), by (14) we have

$\displaystyle 2\cos\alpha$ $\displaystyle\sum_{n=1}^{\infty}(nb_{n}\cos
n(t+\pi-\alpha)-na_{n}\sin n(t+\pi-\alpha))$ $\displaystyle+$
$\displaystyle\sum_{n=1}^{\infty}(nb_{n}\cos nt-na_{n}\sin nt)$
$\displaystyle+$ $\displaystyle\sum_{n=1}^{\infty}(nb_{n}\cos
n(t-2\alpha)-na_{n}\sin n(t-2\alpha))=0.$

Since this holds for every real number $t$, we must have that for every $n$

$\displaystyle\cos nt[-2na_{n}\sin n(\pi-\alpha)\cos\alpha+na_{n}\sin
2n\alpha+2nb_{n}\cos n(\pi-\alpha)\cos\alpha$ $\displaystyle+nb_{n}+nb_{n}\cos
2n\alpha]+\sin nt[-2na_{n}\cos n(\pi-\alpha)\cos\alpha-na_{n}-na_{n}\cos
2n\alpha$ $\displaystyle-2nb_{n}\sin n(\pi-\alpha)\cos\alpha+nb_{n}\sin
2n\alpha]=0.$

The coefficients of $\cos nt$ and $\sin nt$ must be both equal to 0, hence we
have that

$\begin{bmatrix}-2\sin n(\pi-\alpha)\cos\alpha+\sin 2n\alpha&2\cos
n(\pi-\alpha)\cos\alpha+1+\cos 2n\alpha\\\ -2\cos
n(\pi-\alpha)\cos\alpha-1-\cos 2n\alpha&-2\sin n(\pi-\alpha)\cos\alpha+\sin
2n\alpha\end{bmatrix}\begin{bmatrix}a_{n}\\\
b_{n}\end{bmatrix}=\begin{bmatrix}0\\\ 0\end{bmatrix}.$

The determinant of the matrix above is given by

$(-2\sin n(\pi-\alpha)\cos\alpha+\sin 2n\alpha)^{2}+(2\cos
n(\pi-\alpha)\cos\alpha+1+\cos 2n\alpha)^{2}.$

This determinant is zero only if

$-2\sin n(\pi-\alpha)\cos\alpha+\sin 2n\alpha=0$ (17)

and

$2\cos n(\pi-\alpha)\cos\alpha+1+\cos 2n\alpha=0.$ (18)

Since $\alpha\in(0,\pi)$, for every $n\geq 2$ we have that none of (17) and
(18) are satisfied if $\frac{\alpha}{\pi}$ is an irrational number. Hence, we
have that the determinant is non zero and then $a_{n}=b_{n}=0$ for every
$n\geq 2$. It follows that $p(t)=a_{0}+a_{1}\cos t+b_{1}\sin t$, i.e., $p$ is
the support function of a disc (see for instance [19]). $\Box$

######  Remark 1.

If $\frac{\alpha}{\pi}$ is a rational number, then there exist convex bodies
$K$ different from discs for which $c(t)$ is constant. For instance: for the
angles $\alpha_{1}=\frac{\pi}{5}$ and $\alpha_{2}=\frac{3\pi}{5}$ let $K$ be
the convex body whose support function is given by $p(t)=60+\cos 5t+\sin 5t$
(see Fig. 2). In this case the isoptic curves $K_{\alpha_{1}}$ and
$K_{\alpha_{2}}$ have the property that its chords tangent to $K$ have
constant values $c_{1}$ and $c_{2}$, respectively, with
$\frac{c_{1}}{c_{2}}=\frac{1}{2-\tau}$, where $\tau=\frac{1+\sqrt{5}}{2}$.
Moreover, $K_{\alpha_{1}}$ is homothetic to $K_{\alpha_{2}}$ with ratio of
homothety equal to $-\frac{1}{2-\tau}$.

Figure 2: A convex body with two isoptics with corresponding values of $c(t)$
both of constant value for the angles $\alpha_{1}=\frac{\pi}{5}$ and
$\alpha_{2}=\frac{3\pi}{5}$

However, if we impose the additional condition that $q(t)$ is also constant,
then $K$ must be a disc.

######  Theorem 2.

Let $K$ be a strictly convex body in the plane with differentiable boundary
$\partial K$, and let $\alpha\in(0,\pi)$ be a fixed angle. Suppose
$c(t)=c_{0}$, and $q(t)=q_{0}$, for every $t\in[0,2\pi]$, and for two positive
numbers $c_{0}$ and $q_{0}$. Then $K$ is a disc.

_Proof._ By some simple calculations we have that
$|\gamma^{\prime}_{\alpha}(t)|=\frac{q(t)}{\sin\alpha}$ (see for instance
[4]). Since $c(t)=c_{0}$ is also constant, we have

$\frac{d}{dt}(|\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t)|^{2})=\frac{d}{dt}(\langle\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t),\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t)\rangle)=0,$

hence

$\langle\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t),\gamma^{\prime}_{\alpha}(t+\pi-\alpha)\rangle=\langle\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t),\gamma^{\prime}_{\alpha}(t)\rangle.$

It follows that

$c_{0}\cdot|\gamma^{\prime}_{\alpha}(t+\pi-\alpha)|\cos\beta_{1}=c_{0}\cdot|\gamma^{\prime}_{\alpha}(t)|\cos\beta_{2},$

which implies that $\beta_{1}=\beta_{2}$, where $\beta_{1}$ is the angle
between the vectors $\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t)$ and
$\gamma^{\prime}_{\alpha}(t+\pi-\alpha)$, and $\beta_{2}$ is the angle between
the vectors $\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t)$ and
$\gamma^{\prime}_{\alpha}(t)$. It follows that the angle between the chord
$[\gamma_{\alpha}(t+\pi-\alpha),\gamma_{\alpha}(t)]$ and the tangent lines at
$\gamma_{\alpha}(t)$ and $\gamma_{\alpha}(t+\pi-\alpha)$, are equal (see Fig.
3). Similarly, we obtain that the angles between the chord
$[\gamma_{\alpha}(t-\pi+\alpha),\gamma_{\alpha}(t)]$ and the tangent lines at
$\gamma_{\alpha}(t)$ and $\gamma_{\alpha}(t-\pi+\alpha)$, are equal. Since the
length of the tangent vector $\gamma^{\prime}_{\alpha}(t)$ is constant for
every $t$, and all the triangles
$\triangle\gamma_{\alpha}(t-\pi+\alpha)\gamma_{\alpha}(t)\gamma_{\alpha}(t+\pi-\alpha)$
are congruent, we also have that the angles between the chord
$[\gamma_{\alpha}(t-\pi+\alpha),\gamma_{\alpha}(t+\pi-\alpha)]$ and the
tangent lines at $\gamma_{\alpha}(t+\pi-\alpha)$ and
$\gamma_{\alpha}(t-\pi+\alpha)$, are equal. By elementary geometry we have
that the circle circumscribed to
$\triangle\gamma_{\alpha}(t-\pi+\alpha)\gamma_{\alpha}(t)\gamma_{\alpha}(t+\pi-\alpha)$
and the body $K_{\alpha}$, share the tangent lines at the points
$\gamma_{\alpha}(t-\pi+\alpha)$, $\gamma_{\alpha}(t)$, and
$\gamma_{\alpha}(t+\pi-\alpha)$; under this condition it was proved in Lemma
3.3 in [8] that $K_{\alpha}$ must be a disc. Now we use Theorem 2 (b) in [9]
and conclude that $K$ is a disc. $\Box$

Figure 3: The angle between the chord
$[\gamma_{\alpha}(t+\pi-\alpha),\gamma_{\alpha}(t)]$ and the tangent lines are
equal

Let $h(t)$ denote the length of the segment from $\gamma_{\alpha}(t)$ to the
projection of $O$ into the support line of $K$ at $\gamma(t)$. It is easy to
show that

$h(t)=\frac{1}{\sin\alpha}[p(t)\cos\alpha+p(t+\pi-\alpha)].$ (19)

Figure 4: The value of $h(t)$ is constant

We have the following result.

######  Theorem 3.

Let $K$ be a strictly convex body in the plane with differentiable boundary
and let $\alpha\in(0,\pi)$ be a fixed angle. Suppose $h(t)=h_{0}$, for every
$t\in[0,2\pi]$, for a positive number $h_{0}.$ Then $K$ is a disc.

_Proof._ By (19) we have that

$p^{\prime}(t)\cos\alpha+p^{\prime}(t+\pi-\alpha)=0.$ (20)

Using the expansion in Fourier series for $p^{\prime}(t)$ (see equation (14))
we have that

$\sum_{n=1}^{\infty}(nb_{n}\cos nt\cos\alpha-na_{n}\sin
nt\cos\alpha+nb_{n}\cos n(t+\pi-\alpha)-na_{n}\sin n(t+\pi-\alpha))=0.$

Using trigonometric identities and simplifying we conclude that for all $n$
and for all $t\in[0,2\pi]$ we must have

$\displaystyle 0$ $\displaystyle=[-n\sin n(\pi-\alpha)a_{n}+(n\cos\alpha+n\cos
n(\pi-\alpha))b_{n}]\cos nt$ $\displaystyle+[(-n\cos\alpha-n\cos
n(\pi-\alpha))a_{n}-n\sin n(\pi-\alpha)b_{n}]\sin nt.$

This yields to the system of equations

$\begin{bmatrix}-n\sin n(\pi-\alpha)&n\cos\alpha+n\cos n(\pi-\alpha)\\\
-n\cos\alpha-n\cos n(\pi-\alpha)&-n\sin
n(\pi-\alpha)\end{bmatrix}\begin{bmatrix}a_{n}\\\
b_{n}\end{bmatrix}=\begin{bmatrix}0\\\ 0\end{bmatrix}.$

Notice that the determinant of the matrix above is given by

$n^{2}\sin^{2}n(\pi-\alpha)+n^{2}(\cos\alpha+\cos n(\pi-\alpha))^{2}.$

This determinant is never zero, for if $\sin n(\pi-\alpha)=0$, then
$n(\pi-\alpha)=k\pi$, for some integer $k$. Nonetheless, in this case $\cos
n(\pi-\alpha)=(-1)^{k}$ and since $\alpha\in(0,\pi)$, it is impossible to have
$\cos\alpha+\cos n(\pi-\alpha)=0$. It follows that the only solutions for the
system of equations is $a_{n}=b_{n}=0$ for all $n$. Thus, the solutions of the
differential equation (20) are constant functions and $K$ must be a disc
centred at $O$. $\Box$

######  Theorem 4.

Let $K$ be a strictly convex body in the plane and let $\alpha\in(0,\pi)$ be a
fixed angle. Suppose $\lambda(t)=\lambda_{0}$, for every $t\in[0,2\pi]$, for a
positive number $\lambda_{0},$ and $K$ has rotational symmetry of angle
$\pi-\alpha$ or $2\alpha$. Then $K_{\alpha}$ is a circle.

_Proof._ The vector
$\nu(t)=\gamma_{\alpha}(t+\pi-\alpha)-\gamma_{\alpha}(t-\pi+\alpha)$ can be
expressed using the parametrization given in (2) by

$\displaystyle\nu(t)$
$\displaystyle=[2p(t+\pi-\alpha)\cos\alpha+p(t-2\alpha)+p(t)]u(t)$
$\displaystyle+\frac{1}{\sin\alpha}[p(t+\pi-\alpha)\cos
2\alpha-p(t-\pi+\alpha)+p(t-2\alpha)\cos\alpha-p(t)\cos\alpha]u^{\prime}(t).$

If $K$ has rotational symmetry of angle $\pi-\alpha$ or $2\alpha$ we have that

$\displaystyle\nu(t)$ $\displaystyle=[2p(t+\pi-\alpha)\cos\alpha+2p(t)]u(t)$
$\displaystyle+\frac{1}{\sin\alpha}[p(t+\pi-\alpha)\cos
2\alpha-p(t-\pi+\alpha)]u^{\prime}(t),$

or equivalently

$\nu(t)=[2p(t+\pi-\alpha)\cos\alpha+2p(t)]u(t)+[p(t+\pi-\alpha)(-2\sin\alpha)]u^{\prime}(t).$

In this case $|\nu(t)|^{2}=\lambda^{2}(t)$ can be easily calculated as

$\lambda_{0}^{2}=4[p^{2}(t+\pi-\alpha)+p^{2}(t)+2p(t+\pi-\alpha)p(t)\cos\alpha],$

hence

$\frac{\lambda_{0}^{2}}{4}=p^{2}(t+\pi-\alpha)+p^{2}(t)-2p(t+\pi-\alpha)p(t)\cos(\pi-\alpha)=d(t)^{2},$

i.e., the value of $d(t)=\frac{\lambda_{0}}{2}.$

Figure 5: The value of $d(t)$ is constant

Let $x(t)$ and $y(t)$ be the projections of $O$ into the support lines of $K$
at $\gamma(t)$ and $\gamma(t+\pi-\alpha)$, respectively (see Fig. 5). The
quadrilateral $Ox(t)\gamma_{\alpha}(t)y(t)$ is cyclic, i.e., there exist a
circle which passes through its four vertices, hence
$|\gamma_{\alpha}(t)-O|=\frac{d(t)}{\sin\alpha}=\frac{\lambda_{0}}{2\sin\alpha}.$
It follows that $K_{\alpha}$ is a circle centred at $O$. $\Box$

##  4 Some comments about rotors in the plane

In this section we give some words about how the results obtained in this work
are related to rotors in polygons. Moreover, in all the examples shown below,
if we fix the convex body and the circumscribed polygon is rotated, while
maintained circumscribed to $K$, the vertices describe a isoptic of $K$.

When $c(t)$ has constant value, using the Fourier series for the support
function of $p$ in the proof of theorem 1 we arrived to the equations

$-2\sin n(\pi-\alpha)\cos\alpha+\sin 2n\alpha=0$

and

$2\cos n(\pi-\alpha)\cos\alpha+1+\cos 2n\alpha=0.$

If $n$ is even, then

$\sin n\alpha(\cos n\alpha+\cos\alpha)=0$

and

$\cos n\alpha(\cos n\alpha+\cos\alpha)=0.$

Both of them are zero if $\cos n\alpha+\cos\alpha=0,$ or after some
trigonometric transformations

$\cos\left(\frac{n\alpha+\alpha}{2}\right)\cos\left(\frac{n\alpha-\alpha}{2}\right)=0.$

It follows that $\alpha=\left(\frac{2r+1}{n+1}\right)\pi$ or
$\alpha=\left(\frac{2r+1}{n-1}\right)\pi$, where $r$ is any integer number. In
this case the determinant of the associated matrix is zero and we can choose
the coefficients $a_{n}$, $b_{n}$, arbitrarily. For instance, for $n=4$ we
select $\alpha=\frac{\pi}{3}$, $a_{0}=30$, $a_{4}=0$, $b_{4}=1$, and for any
other natural number $n$ we have that $a_{n}=b_{n}=0$, i.e., the support
function of $K$ is $p(t)=30+\sin 4t$ (see Fig. 6). The body $K$ in this
example is centrally symmetric.

Figure 6: A centrally symmetric rotor in the equilateral triangle

If $n$ is odd, then

$\sin n\alpha(\cos n\alpha-\cos\alpha)=0$

and

$\cos n\alpha(\cos n\alpha-\cos\alpha)=0.$

Both of them are zero if $\cos n\alpha-\cos\alpha=0,$ or after some
trigonometric transformations

$\sin\left(\frac{n\alpha+\alpha}{2}\right)\sin\left(\frac{n\alpha-\alpha}{2}\right)=0.$

It follows that $\alpha=\left(\frac{2r}{n+1}\right)\pi$ or
$\alpha=\left(\frac{2r}{n-1}\right)\pi$, where $r$ is any integer number.
Notice that we can obtain an example for any angle of the form
$\alpha=\frac{s}{q}\pi$, where $s$ and $q$ are integers such that $0<s<q$. We
just use this case, since $\alpha=\frac{2s}{2q}\pi$, we select $r=s$ and
$n=2q+1$ or $n=2q-1$. For instance, for
$\alpha=\frac{2\pi}{3}\pi=\frac{4\pi}{6}\pi$, and then we select $n=7$. In
Fig. 7 we show the body $K$ with its isoptic $K_{2\pi/3},$ with the property
that $c(t)$ has a constant value. The support function for this example is
$p(t)=80+\cos 7t$. The body in this example has constant width.

Figure 7: A rotor with constant width in the regular hexagon

The example shown in Fig. 8 has coefficients different from zero for $n=4$ and
$n=5$, consequently, the body $K$ obtained is neither of constant width or
centrally symmetric. The support function of $K$ is $p(t)=70+\sin 4t+\cos 5t.$

Figure 8: A rotor in the equilateral triangle, which is neither of constant
width nor centrally symmetric

Finally, we give an example of a rotor in the square. The support function is
$p(t)=60+\cos 5t$.

Figure 9: A rotor for the square

##  5 An inequality about the length of some special chords

The following inequality and characterization of the disc, in terms of the
length $q(t)$, was proved in [7].

Theorem JY. Let $K$ be a strictly convex body in the plane with minimal width
$\omega_{0}$. For any fixed $\alpha\in(0,\pi)$ there exists at least one value
of the parameter $t$, $t(\alpha)\in[0,2\pi]$, such that
$q(t(\alpha))\geq\omega_{0}\cos\frac{\alpha}{2}$. Moreover, if there is not
such a chord with length exceeding $\omega_{0}\cos\frac{\alpha}{2}$, then $K$
is a disc.

Here we prove the following similar result.

######  Theorem 5.

Let $K$ be a convex body in the plane with minimal width $\omega_{0}$. For any
fixed $\alpha\in(0,\pi)$ there exists at least one value of the parameter $t$,
$t(\alpha)\in[0,2\pi]$, such that $\lambda(t(\alpha))\geq
2\omega_{0}\cos\frac{\alpha}{2}$. Moreover, if there is not such a chord with
length exceeding $2\omega_{0}\cos\frac{\alpha}{2}$, then $K$ is a convex body
of constant width $\omega_{0}$.

_Proof._ The mean value of the chords of $K_{\alpha}$ that are tangent to $K$
is

$\displaystyle\overline{c(t)}$
$\displaystyle=\frac{1}{2\pi\sin\alpha}\int_{0}^{2\pi}[2p(t+\pi-\alpha)\cos\alpha+p(t)+p(t-2\alpha)]dt,$
$\displaystyle=\frac{1}{2\pi\sin\alpha}[2L(K)\cos\alpha+2L(K)],\hskip
17.07182pt\text{(using Cauchy's formula (\ref{Cauchy}))}$
$\displaystyle=\frac{\cot\frac{\alpha}{2}}{\pi}L(K).$

Indeed, by the same argument, the mean value of $c(t)+c(t-\pi+\alpha)$ is

$\overline{c(t)+c(t-\pi+\alpha)}=\frac{2\cot\frac{\alpha}{2}}{\pi}L(K).$

Let $t\in[0,2\pi]$ be any number and consider the triangle with sides of
length $c(t)$, $c(t-\pi+\alpha)$, $\lambda(t)$, and angle $\alpha$, as shown
in Fig. 10. By Problem C4 in the book by I. Niven [15], we have that the
minimum of $\lambda(t)$, among all triangles with a fixed angle $\alpha$ and
the sum $c(t)+c(t-\pi+\alpha)=c_{0}$, for a constant number $c_{0}$, is
obtained if and only if $c(t)=c(t-\pi+\alpha)=c_{0}/2$. It follows that for
every $t\in[0,2\pi]$ it holds that

$\lambda(t)\geq[c(t)+c(t-\pi+\alpha)]\sin\frac{\alpha}{2},$

with equality if and only if $c(t)=c(t-\pi+\alpha).$ Let $t_{0}\in[0,2\pi]$ be
such that $c(t_{0})+c(t_{0}-\pi+\alpha)=\overline{c(t)+c(t-\pi+\alpha)},$ we
have that

$\lambda(t_{0})\geq[c(t_{0})+c(t_{0}-\pi+\alpha)]\sin\frac{\alpha}{2}=\frac{2\cos\frac{\alpha}{2}}{\pi}L(K).$
(21)

Figure 10: The minimum of $\lambda(t)$ is when $c(t)=c(t-\pi+\alpha)$

Using Cauchy’s formula for the perimeter of $K$ we can also prove that
$L(K)\geq\pi\omega_{0}$, with equality if and only $K$ is a body of constant
width $\omega_{0}.$ Hence $\lambda(t_{0})\geq
2\omega_{0}\cos\frac{\alpha}{2}.$

Now, if there is no chord with $\lambda(t)>2\omega_{0}\cos\frac{\alpha}{2}$,
then $\lambda(t_{0})=2\omega_{0}\cos\frac{\alpha}{2}$, which by (21) implies
that $L(K)=\pi\omega_{0},$ i.e., $K$ must be a body of constant width
$\omega_{0}$ (see Fig. 11). $\Box$

Figure 11: A convex body with $\lambda(t)=2\omega_{0}\cos\frac{\alpha}{2}$ for
every $t$

##  6 Some results in higher dimensions

The following lemma is needed for some of the subsequent results.

######  Lemma 1.

Let $K,L\subset\mathbb{R}^{n}$, $n\geq 3$, be convex bodies with
$L\subset\emph{int}\,K$, such that every $(n-1)$-dimensional section of $K$
tangent to $L$ is an $(n-1)$-dimensional ball, then $K$ is a ball. If
additionally, the centre of every tangent ball is at the point of contact with
$L$, then $K$ and $L$ are concentric balls.

_Proof._ There are several ways to prove this lemma. Here we give the
following proof. First note that if all the $3$-dimensional sections of a
convex body through a given point in its interior are $3$-dimensional balls,
then it is an $n$-dimensional ball. We consider any point in the interior of
$L$ for such a point and since the hypothesis of the theorem is inheritated to
every $3$-dimensional section, we have that it is sufficient to prove the
theorem in the $3$-dimensional case.

Let $x$ be any point in $\partial K$ and let $\ell$ be any line supporting $K$
at $x$. Consider the two support planes of $L$ which share the line $\ell$, to
say $H_{1}$ and $H_{2}$. Let $H$ be any other support plane of $L$ through the
point $x$. By hypothesis $H\cap K$ is a disc which intersects each one of the
circles $H_{1}\cap\partial K$ and $H_{2}\cap\partial K$ at two points. The
circle $H\cap\partial K$ passes through three points of the set
$(H_{1}\cap\partial K)\cup(H_{2}\cap\partial K)$, and since there is a unique
sphere which contains the two circles $H_{1}\cap\partial K$ and
$H_{2}\cap\partial K$, it holds that this sphere contains $H\cap\partial K$.
Since $H$ is any support plane of $L$ through $x$, we have that
$\mathcal{R}_{x}=\\{H\cap\partial K:H\text{ is a support plane of }L\text{
through }x\\}$ is a closed subset of a sphere. Let $y\in\partial K$ be any
point such that $\mathcal{R}_{y}\cap\mathcal{R}_{x}\neq\emptyset$, then
$\mathcal{R}_{y}\cup\mathcal{R}_{x}$ is contained in the same sphere.
Continuing in this way, since $K$ is a compact set, we can prove that
$\partial K$ is a sphere.

Now, let $H$ be any hyperplane supporting $L$ at a point $z$ and suppose the
centre of the $(n-1)$-dimensional ball $H\cap K$ is $z$. The line orthogonal
to $H$ through $z$ passes through the centre of the ball $K$, this implies
that indeed $H$ is the tangent plane of $L$ through $z$, i.e., $\partial K$ is
a differentiable surface. We have that all the normal lines of $\partial L$
passes through the centre of $K$, this implies (see for instance [18]) that
$\partial L$ is a sphere with centre at the centre of $K$. We conclude that
$K$ and $L$ are concentric balls. $\Box$

Let $\beta=\pi-\alpha$. In [5], J. W. Green proved that $K$ is a Euclidean
disc if $K_{\alpha}$ is a circle and any of the following conditions hold:
$\beta$ is an irrational multiple of $\pi$, or $\beta=\frac{2m}{n}\pi$, with
$2m$ and $n$ relatively prime integer numbers. On the other side, M.S. Klamkin
conjectured [11] and J. Nitsche proved [14] that two different values
$\alpha_{1}$, $\alpha_{2}$, such that $K_{\alpha_{1}}$ and $K_{\alpha_{2}}$
are circles, are enough to prove that $K$ is a Euclidean disc. A similar
result was proved by Á. Kurusa and T. Ódor in [12] in $\mathbb{R}^{n}$, $n\geq
3$, where the isoptic surface is defined using the solid angle under which is
seen a convex body $K$. They proved that if two isoptic surfaces of a given
convex body are concentric spheres, then it is a ball. The following is a
variant of this kind of results.

######  Theorem 6.

Let $K\subset\mathbb{R}^{3}$ be a strictly convex body contained in the
interior of a sphere $\mathcal{S}$. Suppose that for every $2$-dimensional
plane $H$, with $H\cap K\neq\emptyset$, it holds that $H\cap\mathcal{S}$ is a
isoptic curve of $H\cap K.$ Then $K$ is a ball concentric with $\mathcal{S}$.

_Proof._ We first observe the following: _If for some angle $\alpha\in(0,\pi)$
the $\alpha$-isoptic curve of a planar body $M$ is a circle $S$, then its
centre $x$ belongs to $M$._ Suppose this is not the case. Let $y\in M$ be the
point which is closest to $x$. Let $z\in\mathcal{S}$ be the point where the
ray $\overrightarrow{xy}$ intersects to $\mathcal{S}$ and let $z^{\prime}$ be
the point in $\mathcal{S}$ such that $[z,z^{\prime}]$ is a diameter. Let
$\ell$ be the line through $x$ which is orthogonal to $[z,z^{\prime}]$. The
central projection of $M$ into the line $\ell$ is a segment $[a,b]$ that
contains the centre $x$. The angle formed by the lines $z^{\prime}a$ and
$z^{\prime}b$ contains $M$, however, at least one of the lines $z^{\prime}a$
or $z^{\prime}b$ does not intersect $M$. Since the angles $\measuredangle azb$
and $\measuredangle az^{\prime}b$ are equal, we obtain that the angles under
which $M$ is seen from the points $z$ and $z^{\prime}$ are not equal. This
contradicts that $S$ is the isoptic curve of angle $\alpha$ of $M$. Hence, we
have that $x$ belongs to $M$.

Figure 12: The centre $x$ belong to $M$

Now let $q\in\partial K$ be any arbitrary point and let $H_{q}$ be a plane
supporting $K$ at $q$. By hypothesis we have that $H_{q}\cap\mathcal{S}$ is a
circle; we are going to prove that its centre is $q$. Consider a sequence of
planes $\\{H_{r}\\}$ parallel to $H_{q}$ in such a way that the sequence
converges to $H_{q}$. Let $x_{r}$ be the centre of the circle
$H_{r}\cap\mathcal{S}$, for every natural number $r$. By the comment above, we
have that $x_{r}\in H_{r}\cap K$, for every $r\in\mathbb{N}$, and since
$H_{r}\cap K\longrightarrow H_{q}\cap K=q$ we have that $x_{r}\longrightarrow
q$ when $r\longrightarrow\infty$. In other words, the centre of
$H_{q}\cap\mathcal{S}$ is $q$. Now we apply Lemma 1 and conclude that $K$ is a
ball concentric with $\mathcal{S}$. $\Box$

The following result shows that in dimension 3 or higher, only Euclidean balls
have an equichordal convex body.

######  Theorem 7.

Let $K\subset\mathbb{R}^{n}$, $n\geq 3$, be a strictly convex body which
possesses an equichordal convex body $L$ in its interior. Then $K$ and $L$ are
concentric balls.

_Proof._ Suppose the length of the chords of $K$ tangent to $L$ is $\lambda$.
First we prove that $L$ is strictly convex, i.e., $L$ has not segments in its
boundary. Suppose to the contrary that there is a segment
$[a,b]\subset\partial L$ and consider any $2$-dimensional plane $H$ supporting
$L$ at $[a,b]$. By hypothesis, every chord of $H\cap K$ through $a$ has length
$\lambda$, and the same happens for every chord through $b$. This implies that
$H\cap K$ has two equichordal points, but as were proved by M. R. Rychlik [16]
this is not possible.

Now, let $H$ be any $2$-dimensional plane supporting $L$ at a point $x$. We
will prove that the diameter of $H\cap K$ is $\lambda$. Suppose this is not
the case and there is a chord $[a,b]$ of $H\cap K$ with $|a-b|>\lambda.$
Clearly, $[a,b]\cap L=\emptyset.$ Consider a $2$-dimensional plane $\Pi$ that
contains to $[a,b]$ and such that $\Pi\cap\text{int}\,L\neq\emptyset.$ Let
$c,d\in\partial(\Pi\cap L)$ such that the lines through $c$ and $d$,
respectively, that are parallel to $[a,b]$ are support lines of $\Pi\cap L$.
Since $|a-b|>\lambda$, we have that there exist a chord $[e,f]$ of $\Pi\cap K$
that is separated from $\Pi\cap L$ by the chord $[a,b]$ and $|e-f|=\lambda$.
We have three chords of $\Pi\cap K$ parallel to $[a,b]$ and with length
$\lambda$. This contradicts that $K$ is strictly convex, then the diameter of
$H\cap K$ must be $\lambda$.

Every chord of $H\cap K$ through $x$ has length equal to $\lambda$, then every
chord of $H\cap K$ through $x$ is a binormal of $H\cap K$; it follows that
$H\cap K$ is a disc centred at $x$ (see for instance [3]). Since this is true
for every $2$-dimensional plane $H$ through $x$, we have that any
$(n-1)$-dimensional section of $K$ tangent to $L$ at $x$ is an
$(n-1)$-dimensional sphere centred at $x$. This is also true for every
$x\in\partial L$, so we apply Lemma 1 and conclude that $K$ and $L$ are
concentric balls. $\Box$

######  Corollary 1.

Let $K,L\subset\mathbb{R}^{n}$, $n\geq 3$, be two convex bodies. Suppose that
for every hyperplane $H$ that intersects to $L$ it holds that $H\cap L$ is an
equichordal body of $H\cap K$. Then $K$ and $L$ are concentric balls.

_Proof._ It is quite simple to prove that all the chords of $K$ tangent to $L$
have the same length. The conclusion follows applying Theorem 7. $\Box$

Suppose now that $K$ is a strictly convex body in $\mathbb{R}^{3}$. For
$z_{1},z_{2}\in\partial K$, we will say that the segment $[z_{1},z_{2}]$ is an
$\alpha$-chord of $K$, if there exist tangent planes at $z_{1}$ and $z_{2}$
meeting at an angle $\alpha$. We present the following analogue to Theorem JY.

######  Theorem 8.

Let $K$ be a strictly convex body in $\mathbb{R}^{3}$ with boundary of class
$C^{1}$ and with minimal width $\omega_{0}$. For any fixed $\alpha\in(0,\pi)$
there exists an $\alpha$-chord of $K$ with length at least
$\omega_{0}\cos\frac{\alpha}{2}$. Moreover, if there is not an $\alpha$-chord
with length exceeding $\omega_{0}\cos\frac{\alpha}{2}$, then $K$ is a ball.

_Proof._ Let $u_{0}\in\mathbb{S}^{2}$ be a direction such that the width of
$K$ in this direction has minimal value $\omega_{0}$. Let
$\upsilon\in\mathbb{S}^{2}$ be a vector orthogonal to $u_{0}$ and let’s denote
the orthogonal projection on the plane $\upsilon^{\bot}$ by $\pi_{\upsilon}$.
Clearly, the minimum width of $K^{\prime}=\pi_{\upsilon}(K)$ is also
$\omega_{0}$ and every $\alpha$-chord of $K^{\prime}$ is obtained as the
projection of some $\alpha$-chord of $K$. By Theorem JY there is an
$\alpha$-chord of $K^{\prime}$ with length at least
$\omega_{0}\cos\frac{\alpha}{2}$ and so there is an $\alpha$-chord of $K$ with
length larger than or equal to $\omega_{0}\cos\frac{\alpha}{2}$.

Suppose now that all the $\alpha$-chords of $K$ have length less than or equal
to $\omega_{0}\cos\frac{\alpha}{2}$. Then all the $\alpha$-chords of
$\pi_{\upsilon}(K)$, for every $\upsilon\in\mathbb{S}^{2}$, have length less
than or equal to $\omega_{0}\cos\frac{\alpha}{2}$. It follows from Theorem JY
that $\pi_{\upsilon}(K)$ is a disc. Since $\upsilon$ is an arbitrary
direction, we have that every $2$-dimensional projection of $K$ is a disc and
therefore $K$ is a ball. $\Box$

######  Theorem 9.

Let $K$ be a strictly convex body in $\mathbb{R}^{3}$ with boundary of class
$C^{1}$ and with $\frac{\pi}{2}$-chords of constant length $\lambda$, then $K$
is a ball.

_Proof._ For any point $z\in\partial K$, let $H_{z}$ denote the tangent plane
of $K$ at $z$ and let $A_{z}=\\{w\in\partial K:[z,w]\text{ is an
}\frac{\pi}{2}\text{-chord of }K\\}$. For each point $w$ in $A_{z}$, we have
that $H_{w}$ intersects $H_{z}$ with angle $\frac{\pi}{2}$. Let
$z,z^{\prime}\in\partial K$ be two points such that $[z,z^{\prime}]$ is normal
to $H_{z}$ and $H_{z^{\prime}}$. For every $w\in A_{z}$, the triangle $\Delta
zz^{\prime}w$ is isosceles with sides $[z,w]$ and $[z^{\prime},w]$ of equal
length. Then $A_{z}$ is a circle centred at $(z+z^{\prime})/2$. Moreover,
given any two antipodal points $z_{1}$ and $z_{1}^{\prime}$ of $A_{z}$, the
segment $[z_{1},z_{1}^{\prime}]$ must be orthogonal to the planes $H_{z_{1}}$
and $H_{z_{1}^{\prime}}$. By repeating the previous argument to each pair of
antipodal points of $A_{z}$, we have that any section of $K$ containing
$[z,z^{\prime}]$ is a disc. This implies that $K$ is a ball. $\Box$

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