Equichordal Points of Convex Bodies

Problem. Does there exist (for $n\geq 2$) an $n$-dimensional convex body that possesses two equichordal points?[3]

The concept of equichordal points originates from classical geometry and the study of convex bodies. For a set $C$ that is star-shaped with respect to an interior point $p$—meaning $C$ contains every line segment connecting $p$ to any other point in $C$—the point $p$ is called an equichordal point if all chords of $C$ passing through $p$ have the same length. A straightforward example is the center of a spherical region, which serves as an equichordal point due to the inherent symmetry of the sphere.

The equichordal point problem for plane convex bodies was first posed by Fujiwara [2], who conjectured the existence of multiple equichordal points in certain convex shapes. Independently, Blaschke, Rothe, and Weitzenböck [1] extended this question to more general geometric settings. However, despite substantial progress, the existence of multiple equichordal points in higher-dimensional convex bodies remained an open question.

In this paper, we address the nonexistence of multiple equichordal points in $n$-dimensional convex bodies ($n\geq 2$). By employing tools from topology, such as the Borsuk-Ulam theorem, and leveraging the properties of continuous mappings, we establish that such configurations are impossible.

For functions $f,g:J\to Y$, the uniform metric $\bar{\rho}$ takes the form

Now consider the subset $\mathcal{C}(X,Y)\subseteq Y^{X}$ consisting of all continuous functions $f:X\to Y$. If $Y$ is complete under the metric $d$, then $\mathcal{C}(X,Y)$ is also complete under the uniform metric $\bar{\rho}$. Similarly, the set $\mathcal{B}(X,Y)$ of bounded functions $f:X\to Y$ (where a function $f$ is bounded if $f(X)$ is a bounded subset of $Y$) is complete under the same metric.

Given a sequence of functions $(f_{n})$ in $\mathcal{C}(X,Y)$ converging to a function $f$ under the uniform metric, for all $x\in X$ and $n\geq N$, we have

which implies uniform convergence of $(f_{n})$ to $f$.

The relationship between the uniform metric $\bar{\rho}$ and the sup metric $\rho$ is straightforward. For $f,g\in\mathcal{B}(X,Y)$:

If $\rho(f,g)>1$, then there exists $x_{0}\in X$ such that $d(f(x_{0}),g(x_{0}))>1$. Hence, $\bar{d}(f(x_{0}),g(x_{0}))=1$, and $\bar{\rho}(f,g)=1$. Conversely, if $\rho(f,g)\leq 1$, then $\bar{d}(f(x),g(x))=d(f(x),g(x))\leq 1$ for all $x\in X$, so $\bar{\rho}(f,g)=\rho(f,g)$. Therefore, on $\mathcal{B}(X,Y)$, the uniform metric $\bar{\rho}$ coincides with the bounded version of the sup metric $\rho$.

If $X$ is compact, every continuous function $f:X\to Y$ is bounded, so the sup metric is defined on $\mathcal{C}(X,Y)$. If $Y$ is complete under $d$, then $\mathcal{C}(X,Y)$ is complete under both the uniform metric $\bar{\rho}$ and the sup metric $\rho$. In practice, the sup metric is often preferred in this context.

In this section, we prove the nonexistence of multiple equichordal points in a convex body $X$ for $n\geq 2$.

Let $X$ be a convex body in $n$-dimensional Euclidean space $\mathbb{R}^{n}$ for $n\geq 2$. Pick any point $x\in\mathrm{Int}X$. Since $x$ is an interior point of $X$, there exists $\epsilon>0$ such that $B_{d}(x,\epsilon)\subset X$. Thus, we can consider a unit vector $a$ on $S^{n-1}$, the unit sphere in $n$-dimensional space. Let $d$ be a fixed length of the half-chord starting at $x$ in the given direction $a$. Thus, we have

Since $d(x,x+at)$ is a continuous function of $\mathbb{R}_{>0}$ and $B$ is a closed set, there exists a unique $t_{0}>0$ such that $d=d(x,x+at_{0})$ and $a_{0}:=x+at_{0}$ lies on $B$ such that the chord in the given direction starting at $x$ has a fixed length $d$, where $d$ is the distance from the given interior point to the point $a_{0}$.

Define the function

with

where $\varphi(x)$ is a function from $\mathbb{S}^{n-1}$ to $\mathbb{R}$. We will show that $\varphi$ is a continuous injective map.