The solutions are in! There was quite a bit of activity behind the scenes on this one; POW-7 might look intuitively obvious, as if it should succumb to an easy application of the intermediate value theorem, but for a number of people who fought through it, this problem fought back! (And that was true for me too, when I first encountered it.)

This problem appears as problem B-4 from the 1977 Putnam Competition. A couple of readers hit upon the snappy solution proposed by the problem compilers [as given in The American Mathematical Monthly vol. 86 (1979), pp. 749-757]. I’ll give that solution, and follow up with a few remarks on alternate approaches, and on some of the thoughts the problem evoked in our readers. Thanks to all who wrote in!

Composite solution by Kenneth Chan and Paul Shearer: By translation, we may assume that the given point $P$ is the origin. Let $-C = \{-x: x \in C\}$. It suffices to show that $C \cap -C$ is nonempty, for if $x$ belongs to this set, then both $x \in C$ and $-x \in C$, so that $\frac{x + -x}{2} = 0$ is the midpoint $P$.

The interior regions $R, -R$ of $C, -C$ intersect (e.g., $P = 0$ is in both regions), so if $C \cap -C = \emptyset$, then by connectedness of the curves $C, -C$, one of them must be contained in the interior of the other. But this is absurd; for example, it is clear that

$\displaystyle \mbox{sup} \{|x|: x \in C\} = \mbox{sup} \{|x|: x \in -C\}$

and so if the sup is realized at a point $x \in -C$ which is interior to $C$, then the boundary of the nonempty set $\{tx \in R: t > 1\}$ contains a point of $C$ whose distance from the origin exceeds this shared sup, contradiction.

Remarks:

1. One could just as easily observe that the regions $R, -R$ are congruent and therefore have the same area, so that one of these regions cannot be strictly contained in the other. Therefore, unless $C = -C$, some of $R$ is interior to $-R$ and some is exterior; since $R$ is connected, it must therefore intersect the border $-C$ of $-R$. Similarly, the exterior of $C$ must be partly inside $-C$ and partly outside, so it too must intersect $-C$. In other words, some of $-C$ is interior to $C$ and some is exterior to $C$, and so by connectedness of $-C$, it must cross $C$, as desired. This argument is substantially the one given by Sune Jakobsen.

2. Assuming $P$ is at the origin, a number of people tried to argue that (in polar coordinates) the radius $r$ of a point on $C$ is a function of its angle $\theta$, and since $r(\theta) - r(\theta + \pi)$ sometimes takes non-positive values and sometimes non-negative values, it should take on the value 0 somewhere by the intermediate value theorem. The pitfall though is that $r$ is not necessarily a well-defined function of $\theta$ (consider non-convex curves $C$ where a radial line meets $C$ in more than one point). Some other readers seemed to notice this and tried a surrogate function where along the radial line at angle $\theta$, you choose the closest distance $r$ from $C$ to the origin; the trouble there is that this new function $r(\theta)$ might not be continuous (there will typically be a jump discontinuity at angles where the radial line is tangent to $C$).

3. The obstructions mentioned in remark 2. are of a sort which is ubiquitous in topology, where one would like to construct a continuous choice function (e.g., in the theory of vector bundles or fiber bundles, where one is interested in whether continuous sections of a bundle projection $\pi: E \to X$ exist). À propos of that, Paul Shearer made an intriguing suggestion (in a comment here), that POW-7 would follow from the stronger conjecture that the space $S$ of pairs of points on $C$ which “straddle” $P$ (i.e., where the points and $P$ live on the same line, but the points are on opposite sides of $P$) should be path-connected. That is, given two pairs $c, c' \in S$, the conjecture is that we can choose a continuous path through $S$ which gets us from $c$ to $c'$. Subsequently, Paul found a counterexample to this conjecture. Can you find one? [Edit: I had invited Paul to comment on this, but then missed the fact that he had already done so! My bad – TT.]

4. Miodrag Milenkovic came up with a related problem which can be posed in any dimension. Consider any $n$-sphere $S^n$ embedded in $\mathbb{R}^{n+1}$ (POW-7 concerns the case $n = 1$), and let $P$ be a point interior to the embedded sphere. [Note: the complement of such an embedded sphere consists of two connected components, i.e., an “interior” and “exterior”, although this fact isn’t completely trivial; see theorem 36.3 (generalized Jordan curve theorem) in Munkres’s Elements of Algebraic Topology. These regions can be topologically complicated, as we know from the example of the Alexander horned sphere.] Miodrag asks: is it true that there exists a plane through $P$ so that $P$ is the center of mass (barycenter) of the $(n-1)$-dimensional region where the interior intersects the plane?

I am not entirely decided on the answer to this question, although I think I have a nice topological argument that the answer is yes, if we assume some mild smoothness assumption on the embedding. I am frankly a little scared of the problem in full topological generality, due to the pathology one may encounter in the topological category!

POW-7 was also solved by Arin Chaudhuri, Sune Jakobsen, Philipp Lampe, and Peter LeFanu Lumsdaine. Thanks again to all for the stimulating correspondence!