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There was quite a bit of work on the so-called equichordal problem throughout the 20th century, to decide if some plane convex curve could have two equichordal points. A point is equichordal for a closed curve in the plane if every chord passing through it is the same length.
     EquiChordal
     (Image from Ferenc Adorján, "Equichordal curves and their applications").

This problem was finally solved in the negative in 1996 by Marek Rychlik.

My question is whether the notion of equichordal curve has been extended to shapes in $\mathbb{R}^d$. A natural extension is to define a closed, solid shape $S \subset \mathbb{R}^d$ as equiareal if there exists a point $p \in S$ such that every $(d{-}1)$-plane through $p$ intersects $S$ in a body of equal $(d{-}1)$-volume. In $\mathbb{R}^3$, the sections are 2D planes, and the $(d{-}1)$-volume is the area of the section. One could then ask if there are shapes with more than one equiareal point.

(Added 7Dec13.) Responding to Sam Hopkins's query, it seems the first order of business (so-to-speak) is to decide if there is any equiareal shape that is not a sphere. My attempt to construct one below failed. Is there a proof that the equiareal property characterizes spheres?

This shape's sections through the central red point and containing the $x$-axis are all equal-area ellipses:
     EquiAreal
I don't think, however, that all sections through that point are of equal area.

(8Dec13.) Trying to follow Yoav's construction, here is $\sqrt{1 + \frac{1}{2} \sin \phi}$, under the assumption that $f(\theta,\phi)=\sin \phi$ represents the required odd function:
   SphereOdd

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    $\begingroup$ @JosephO'Rourke: Some time ago I asked a question ( mathoverflow.net/questions/138525 ) that seems dual to the above: Is there a 3-dimensional convex body other than a ball whose perpendicular projections in all directions are of the same area? The answer I received is "yes", which makes me think that your question may also have a positive answer. Another variation on questions of this type is with the preimeter of the cross-sections or projections, respectively. I do not know what has been done in this direction. $\endgroup$
    – Wlodek Kuperberg
    Commented Dec 7, 2013 at 22:23
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    $\begingroup$ The areas of sections of a body are given by the spherical Radon transform applied to $\rho^{n-1}$, where $\rho$ is the radial distance from the origin to the boundary. Therefore, it is easy to construct equiareal bodies by considering the kernel of the transform, which is just the odd functions. However, every equiareal centrally-symmetric body is indeed a sphere. $\endgroup$
    – Yoav Kallus
    Commented Dec 7, 2013 at 22:55
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    $\begingroup$ Actually, I see that @alvarezpaiva already gave the same argument in the comments to the answer Wlodek linked to. $\endgroup$
    – Yoav Kallus
    Commented Dec 7, 2013 at 23:00
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    $\begingroup$ @JosephO'Rourke: pick any nice odd function $f:S^2\to\mathbf{R}$, and let $\rho=\sqrt{1+\alpha f}$. If you care about convexity, then $\alpha$ needs to be sufficiently small to preserve convexity. The fact that any section through the origin has the same area is because the area is given by $\tfrac{1}{2}\int_{S\cap S^2} \rho^2 d\theta = \tfrac{1}{2}\int_{S\cap S^2} (1+\alpha f) d\theta = \pi$. $\endgroup$
    – Yoav Kallus
    Commented Dec 8, 2013 at 1:13
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    $\begingroup$ @YoavKallus: Cool! I will (eventually) compute such a shape. $\endgroup$
    – Joseph O'Rourke
    Commented Dec 8, 2013 at 1:59

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