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 is the origin. Let . It suffices to show that is nonempty, for if belongs to this set, then both and , so that is the midpoint .

The interior regions of intersect (e.g., is in both regions), so if , then by connectedness of the curves , one of them must be contained in the interior of the other. But this is absurd; for example, it is clear that

and so if the sup is realized at a point which is *interior* to , then the boundary of the nonempty set contains a point of whose distance from the origin exceeds this shared sup, contradiction.

**Remarks**:

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

2. Assuming is at the origin, a number of people tried to argue that (in polar coordinates) the radius of a point on is a function of its angle , and since 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 is not necessarily a well-defined function of (consider non-convex curves where a radial line meets in more than one point). Some other readers seemed to notice this and tried a surrogate function where along the radial line at angle , you choose the *closest* distance from to the origin; the trouble there is that this new function might not be continuous (there will typically be a jump discontinuity at angles where the radial line is tangent to ).

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 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 of pairs of points on which “straddle” (i.e., where the points and live on the same line, but the points are on opposite sides of ) should be path-connected. That is, given two pairs , the conjecture is that we can choose a continuous path through which gets us from to . 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 -sphere embedded in (POW-7 concerns the case ), and let 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 so that is the center of mass (barycenter) of the -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!

## 2 comments

Comments feed for this article

July 10, 2008 at 7:10 pm

Paul ShearerI like Milenkovic’s generalization a lot.

I wonder what can be said about the line’s behavior under homotopic deformation of the curve (and the plane under homotopic deformation of the sphere embedding in Milenkovic’s case). Will the parameters describing a segment change continuously with changes in the curve? (Assume here that the point P is fixed and stays in the interior of the curve throughout the deformation).

Using my counterexample in the POW-7 comment, one can show that some lines will disappear under homotopic deformation. So these barycenter lines/planes aren’t homotopically stable in general.

July 22, 2008 at 7:07 am

Analyzing the hairy ball theorem « Todd and Vishal’s blog[...] | by Todd Trimble A couple of weeks ago, when Miodrag Milenkovic posed an interesting general problem in connection with POW-7, I was reminded of the “hairy ball theorem” (obviously a [...]