Some results about equichordal convex bodies

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.

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 $\gamma(t+\pi-\alpha)$.

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)

Our first result about isoptic chords is the following.

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]). $\Box$

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

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)

We have the following result.

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 $O$. $\Box$

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 $d(t)=\frac{\lambda_{0}}{2}.$

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 $O$. $\Box$

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.

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.

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 $p(t)=70+\sin 4t+\cos 5t.$

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

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.

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)

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). $\Box$

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

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.