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Imagine a white convex polyhedron $P$ tumbling randomly about its fixed center of gravity (c.g.) $c$ against a blue background. A long-exposure photo would show pure white in a neighborhood of $c$ (because an opaque ball about $c$ is interior to $P$), and diminished white and increasing blue in circles of larger radius $r$ about $c$, for a line of sight at a given $r$ only hits $P$ a fraction of the tumbling time. My question is to what extent this tumble-density profile uniquely determines $P$.

Henceforth I will specialize to convex polygons $P$ spinning about their c.g. $c$ in $\mathbb{R}^2$, although all questions generalize to $\mathbb{R}^d$. Define the profile function $\rho(r)$ at radius $r$ to be the fraction of the circumference of a circle of radius $r$ centered on $c$ that is interior to $P$. An example for an isosceles right triangle $P$ (edge lengths 1, 1, $\sqrt{2}$) is shown below. Up to $r=\frac{\sqrt{2}}{6} \approx 0.24$, $\rho(r)=1$. Beyond that, $\rho(r)$ diminishes as illustrated as larger circles have less of their circumference inside $P$. Discontinuities occur where circles pass through vertices or are tangent to edges, in this case at $r = \frac{1}{3}$ and $r= \frac{\sqrt{2}}{3} \approx 0.47$.
Tumble Figs
Spinning the profile function around $r=0$ shows the density at any point around $c$:
          Profile 3D

Q1. Does $\rho(r)$ uniquely determine $P$ if it is known that $P$ is a triangle?

The concave sections of $\rho(r)$ seem to be functions specific enough (sums of inverse trig functions) to perhaps determine the geometry.

Q2. For arbitrary convex polygons $P$, are almost all uniquely determined by their profiles $\rho(r)$?

Certainly there are pairs of incongruent polygons that have the same profile, e.g., this pair of augmented regular octagons:
     Nonunique
However, it seems there need be special relationships between these polygons, so that in some appropriate sense, these are density-zero coincidences, and generic polygons have unique profiles.

Q3. Has this notion of density profile been studied before?

My questions are related in spirit to those explored in Richard Gardner's Geometric Tomography (Cambridge University Press, Cambridge, 2nd ed., 2006), but his natural focus on X-rays along lines seems a different flavor than the integration around circles in my profiles. Thanks for ideas and pointers!

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Of course, not only do the questions generalize to higher dimensions, they also generalize to smooth convex shapes and to nonconvex shapes. –  Joseph O'Rourke Sep 2 '11 at 14:03
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I don't know enough about it to write an answer, but this reminds me a bit of "powder diffraction" en.wikipedia.org/wiki/Powder_diffraction (without the diffraction, somehow). –  j.c. Sep 2 '11 at 17:30
    
@jc: Thanks, this concept is certainly new to me! Indeed the "rotational averaging" in powder diffraction is analogous to my blurring by spinning. –  Joseph O'Rourke Sep 3 '11 at 3:58
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1 Answer 1

up vote 4 down vote accepted

Q1: Yes, the first three non-differentiable points in the graph of $\rho(r)$ determine the lengths of $GP_A,GP_B,GP_C$, where $G$ is the centroid of $\triangle ABC$, and $P_A$ is the projection of $G$ on $BC$, etc. On the other hand $|GP_A|=\frac{1}{3}h_a$, the altitude from vertex $A$.

It is well known that one can reconstruct the triangle knowing the lengths of the three altitudes (hint: $h_a:h_b:h_c=a^{-1}:b^{-1}:c^{-1}$)


(Added) Q2: The answer here is also intuitively yes, but it may be a bit painful to write down a rigorous proof. The following holds:

If $O$ is a point inside the polygon $P=P_1\cdots P_n$ so that all distances $|OP_i|$ and all distances from $O$ to the sides of $P$ are distinct, then we can recover $P$ from the profile $\rho_{O,P}(r)$.

The proof doesn't use more than just some basic combinatorial geometry and a little analysis. Let's assume that $O$ is the origin. For every convex cone $C$ (the interior of some angle) we can define it's profile $$\chi_C(r)=Vol(B(r)\cap C).$$ If the cone $C$ is so that all rays from $O$ intersect it only once, then $\chi_C$ has only one non-differentiable point, it is constant before that point and real analytic after that. With a little work one can show that if $\chi_C=\chi_{C'}$ on some interval then one can obtain $C'$ from $C$ by a rotation and/or a reflection.

Coming back to our problem by integrating $\rho(r)$ we can write $\chi_P(r)=\int_0^r \rho(x) dx=Vol(B(r)\cap P)$. Now $\chi_P$ can be written as a linear combination of $\chi_C$'s by taking $C$'s to be (1) the outer halfplanes determined by the edges of the polygon and (2) the exterior angles at it's vertices. (Technically there is some case work here, but for the sake of brevity I'm assuming $O$ is so that the perpendicular to each edge of the polygon falls on that edge.)

Since we are working with a generic polygon we can obtain all $\chi_C$'s from $\chi_P$ by looking at the intervals between non-differentiable points and taking the successive differences of their analytic continuation. Therefore we can learn the angles of the polygon, the direction of it's edges and the distances between the edges and $O$. This is enough information to reconstruct $P$ up to a rotation and reflection.

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@Gjergji: I am uncertain of "the first three." Could not a discontinuity caused by an internal vertex be interleaved with the tangency discontinuities? –  Joseph O'Rourke Sep 2 '11 at 13:22
    
I guess not, when the circle center is the c.g. –  Joseph O'Rourke Sep 2 '11 at 13:34
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The tangency discontinuities will all come before any vertex discontinuity (for triangles). Also one can distinguish double tangency discontinuities because their peaks are more distinguished. –  Gjergji Zaimi Sep 2 '11 at 13:38
    
This is a nice solution for triangles, although perhaps there remains a bit of argument to distinguish a double tangency discontinuity from a vertex discontinuity. But that aside, now it is natural to wonder about convex quadrilaterals... –  Joseph O'Rourke Sep 3 '11 at 3:37
    
@Gjergji: Your reasoning on Q2 is incisive and entirely convincing. Thanks so much for your attention! –  Joseph O'Rourke Sep 4 '11 at 16:09
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