2,$ I know of no compact convex subset of $\R^n$ with nonempty interior that does satisfy that condition, other than the regions enclosed by ellipsoids centered at $0.$ The situation is different for $n=2,$ as shown by point~(d) of the next result. \begin{lemma}\label{L.csym} Suppose $B$ is a {\em centrally symmetric} convex subset of $\R^n.$ Then\\[3pt] \textup{(a)}~ Any nonempty convex subset $C\subseteq B$ that has a center of symmetry is parkable in $B.$\vspace{3pt} Hence, assuming in the remaining points that $B$ is also {\em compact,} we have\\[3pt] \textup{(b)}~ If the intersection of $B$ with every hyperplane $A$ that meets $B$ has a center of symmetry, then $B$ satisfies the equivalent conditions~\textup{(\ref{x.parkAB})-(\ref{x.parkC})}.\vspace{3pt} In particular,\\[3pt] \textup{(c)}~ If $B$ is the closed region enclosed by an ellipsoid in $\R^n,$ then $B$ satisfies~\textup{(\ref{x.parkAB})-(\ref{x.parkC})}, and\\[3pt] \textup{(d)}~ If $n=2,$ then without further restrictions, $B$ satisfies~\textup{(\ref{x.parkAB})-(\ref{x.parkC})}. \end{lemma}\begin{proof} Let $C$ be as in~(a), with center of symmetry $z\in C.$ Then for every $x\in C,$ $2z-x\in C\subseteq B,$ so by central symmetry of $B,$ we have $x-2z\in B.$ Averaging $x$ and $x-2z,$ we get $x-z\in B.$ Thus $C-z\subseteq B,$ so $C$ is parkable. It follows that any $B$ as in~(b) satisfies~(\ref{x.parkAB}), hence by~Lemma~\ref{L.park}, all of (\ref{x.parkAB})-(\ref{x.parkC}). In the situation of~(c), the intersection of $B$ with a hyperplane $A,$ if nonempty, is either a point or the region enclosed by an ellipsoid in $A,$ hence has a center of symmetry, while in~(d) the intersection of $B$ with every line in $\R^2$ that meets $B$ is a point or a closed line segment, hence has a center of symmetry; so in each case,~(b) gives the asserted conclusion. \end{proof} \begin{question}\label{Q.park} Let $n\geq 3,$ and suppose $B$ is a compact convex subset of $\R^n$ having nonempty interior and containing $0.$ Of the implications \textup{(i)$\Rightarrow$(ii)$\Rightarrow$(iii)}, which we have noted hold among the conditions listed below, is either or both reversible?\\[6pt] \textup{(i)} $B$ is an ellipsoid centered at $0.$\\[6pt] \textup{(ii)} $B$ is centrally symmetric, and for every hyperplane $A$ meeting $B,$ $A\cap B$ has a center of symmetry.\\[6pt] \textup{(iii)} ~Every closed convex subset of $B$ is parkable in $B.$ \end{question} Branko Gr\"unbaum has pointed out to me a similarity between this question and the result of W.\,Blaschke \cite[pp.157--159]{B} that if $E$ is a smooth compact convex surface in $\R^3$ with everywhere nonzero Gaussian curvature, such that when $E$ is illuminated by parallel rays from any direction, the boundary curve of the bright side lies in a plane, then $E$ is an ellipsoid. I believe that methods similar to Blaschke's may indeed show that both implications of Question~\ref{Q.park} are reversible. To see why, suppose $B$ is a compact convex subset of $\R^3$ with nonempty interior containing $0,$ which satisfies~(iii) above, and whose boundary $E$ is (as in Blaschke's result) a smooth surface with everywhere nonzero Gaussian curvature. Let $A$ be any plane through $0,$ and $A'$ the plane gotten by shifting $A$ a small distance. Now the vectors that can possibly park $A'\cap B$ are constrained by the directions of the tangent planes to $E$ at the points of $A'\cap E$ (which are well-defined because $E$ is assumed smooth), and if we take $A'$ sufficiently close to $A,$ these tangent planes become close to the corresponding tangent planes at the points of $A\cap E.$ Applying the above observations to planes $A'$ on both sides of $A,$ one can deduce that all the tangent planes to $E$ along $A\cap E$ must contain vectors in some common direction (I am grateful to Bjorn Poonen for this precise formulation of a rough idea I showed him); in other words, that $A\cap E$ is the boundary of the bright side when $E$ is illuminated by parallel rays from that direction. By definition, $A\cap E$ lies in the plane $A;$ so we have the situation that Blaschke considered, except that we have started with planarity and concluded that the curve is a boundary of illumination, rather than vice versa. That last difference is probably not too hard to overcome. More serious is the smoothness assumption on $E\<,$ used in both the above discussion and Blaschke's argument. Finally, can the result be pushed from $n=3$ to arbitrary $n\geq 3$? I leave it to those more skilled than I in the subject to see whether these ideas can indeed be turned into a proof that (iii)$\!\Rightarrow\!$(i) in Question~\ref{Q.park}. A related argument which can be extracted from a step in Blaschke's development shows that a compact convex subset $B$ of $\R^2$ containing $0$ and satisfying (\ref{x.parkC}), whose boundary is a smooth curve containing no line segments, must be centrally symmetric. Again, one would hope to remove the conditions on the boundary. One can ask about a converse to another of our observations: \begin{question}\label{Q.park_C} Suppose $C$ is a compact convex subset of $\R^n$ $(n>2)$ such that for every centrally symmetric compact convex subset $B$ of $\R^n$ containing a translate $C'$ of $C,$ the set $C'$ is parkable in $B.$ Must $C$ have a center of symmetry? \end{question} Here the behavior of a given $C$ can change depending on whether the dimension of the ambient vector space is $2$ or -- as in the above question -- larger: a triangle $C$ has the above property in $\R^2$ by Lemma~\ref{L.csym}(d), but not in $\R^3,$ as we saw in the example where $B$ was a cube.\vspace{6pt} Returning to the ``pathology'' which motivated the considerations of this section, one important case is where the radius of a subset $X$ of a normed vector space $V$ decreases when $V$ is embedded in a larger normed vector space $W.$ The next lemma determines how far down the radius of a given $X$ can go. \begin{lemma}\label{L.enlarge} Let $V$ be a normed vector space, and $X$ a bounded nonempty subset of $V.$ Then % \begin{xlist}\item\label{x.enlarge} $\inf_{W\supseteq V}\ \rad_W(X)\ =\ \rad_V \{(x-y)/2\mid x,\,y\in X\} \ =\ \diam(X)/2,$ \end{xlist} % where the $W$ in the infimum ranges over all normed vector spaces containing $V.$ Moreover, that infimum is realized by a $W$ in which $V$ has codimension $1.$ \end{lemma} \begin{proof} Note first that $\{(x-y)/2\mid x,\,y\in X\}$ is centrally symmetric, hence by Lemma~\ref{L.sym}, its radius in $V$ is the supremum of the distances of its points from $0,$ which is just the last term of~(\ref{x.enlarge}). Moreover, the radius of $X$ in any normed vector space containing it is at least half the distance between any two points of $X,$ so the first term of~(\ref{x.enlarge}) is $\geq$ those last two terms. It remains to prove the reverse inequality. That inequality is trivial if the diameter of $X$ is zero or infinite. Assuming it is neither, we may re-scale and suppose without loss of generality that the last two terms of~(\ref{x.enlarge}) equal $1.$ Thus, $\{(x-y)/2\mid x,\,y\in X\}$ is contained in the closed unit ball $B_V$ of $V.$ Now let $W=V\oplus\R,$ let us identify $V$ with $V\times\{0\}\subseteq W,$ and let us take for the closed unit ball $B_W$ of $W$ the closure of the convex hull of % \begin{xlist}\item\label{x.3parts} $\{(x,1)\mid x\in X\}\ \cup\ B_V\ \cup \ \{(-x,-1)\mid x\in X\}.$ \end{xlist} % (We understand ``closure'' to mean ``with respect to the product topology'', since we don't have a norm until we have made the above definition.) It is easy to see that any point in the convex hull of~(\ref{x.3parts}) whose second coordinate is $0$ is a convex linear combination of a point of $B_V$ and a member of $\{(x-y)/2\mid x,\,y\in X\}.$ But we have seen that the latter set is contained in $B_V;$ so $B_W\cap V=B_V,$ proving that our norm on $W$ extends that of $V.$ But $B_W$ also contains a translate $\{(x,1)\mid x\in X\}$ of $X.$ Hence $X$ is contained in a closed ball of radius $1,$ so its radius in $W$ is $\leq 1=\diam(X)/2,$ as required. \end{proof} Even the case $V=\E^n$ is not immune to this phenomenon, since even in that case, the overspace $W$ of the above construction is generally not Euclidean. For instance, if we take for $X$ an equilateral triangle in $\E^2$ centered at the origin, it is not hard to see that $\{(x-y)/2\mid x,y\in X\}$ is a hexagon whose vertices are the midpoints of the edges of a regular hexagon with the same circumcircle as $X;$ so the radius of $X$ decreases in $W$ by the ratio of the inradius to the circumradius of a regular hexagon, in other words, by $\sqrt 3/2.$ This will not, of course, happen for a centrally symmetric $X$ (cf.\ Lemma~\ref{L.csym} or~(\ref{x.enlarge})). Other cases for which it cannot happen depend on the metric structure: if $X$ is a right or obtuse triangle in $\E^2,$ or more generally, any bounded set containing a diameter of a closed ball in which it lies, its radius clearly cannot go down under extension of the ambient normed vector space (cf.\ Corollary~\ref{C.r=d/2}). \section{Acknowledgements.}\label{S.ackn} In addition to persons acknowledged above, I am indebted to W.\,Kahan for showing me an exercise he had given his Putnam-preparation class, of proving Lemma~\ref{L.S} in $\E^3,$ and for subsequently pointing out that my solution to that exercise worked in any normed vector space; to Nik Weaver for for much helpful discussion of the relation between the present material and that of~\cite{NW}, and to David Gale for pointing out the connection between the construction of \S\ref{S.AE} and results in mathematical economics. % - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - \begin{thebibliography}{00} \bibitem{AE} Richard F. 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Schoenberg, % QA1C315 % {\em Helly's theorems on convex domains and Tchebycheff's % approximation problem}, % Canadian J. Math., {\bf 2} (1950) 245--256. % ~MR~{\bf 11},681b. % relevance? \bibitem{R+HS} H. Rutishauser and H. Samelson, % <-- {\em Sur le rayon d'une sph\'ere dont la surface contient une courbe ferm\'ee,} C. R. Acad. Sci. Paris, {\bf 227} (1948) 755--757. ~MR~{\bf 10},\,321c; correction, MR~{\bf 10}, p.856. % w \cite{BS}: analog of Lemma L.S on sphere % MR: result for sphere "proof elem, very short, genl'zable" % library doesn't have CR before ~1950 \bibitem{S+W} Jonathan Schaer and John E. Wetzel, {\em Boxes for curves of constant length}, Israel J. Math., {\bf 12} (1972) 257--265. ~MR~{\bf 47}\#5726. % sizes of boxes containing curves of given length \bibitem{BS} B. Segre, % <-- {\em Sui circoli geodetici di una superficie a curvatura totale constante, che contengono nell'interno una linea assegnata,} Boll. Un. Mat. Ital., {\bf 13} (1934) 279--283. ~Zbl~{\bf 10}, 271. % analog of Lemma L.S on sphere \bibitem{PCT} Philip C. Tonne, {\em A simple closed curve on a hemisphere,} Houston J. Math., {\bf 10} (1984) 585. ~MR~{\bf 86b}:53004. % MR: \mr(circle of length <2*p, sphere) = *p/2 \bibitem{NW} Nik Weaver, % QA 323 W4 1999 {\em Lipschitz algebras}, World Scientific Publishing, River Edge, NJ, 1999, xiv$+$223 pp., ISBN: 981-02-3873-8. ~MR~{\bf 2002g}:46002. \bibitem{AW} Alan Weinstein, {\em Almost invariant submanifolds for compact group actions,} J. Eur. Math. Soc. {\bf 2} (2000) 53--86. MR~{\bf 2002d}:53076. \bibitem{JW} John E. Wetzel, {\em Covering balls for curves of constant length,} Enseignement Math., (2) {\bf 17} (1971) 275--277. ~MR~{\bf 48}\#12315. % MR = Lemma L.S (length L) and analog for [0,L] \end{thebibliography} George M. Bergman\\ Department of Mathematics\\ University of California\\ Berkeley, CA 94720-3840\\ USA\\ gbergman@math.berkeley.edu \end{document}