Mathematics of doodling and the winding number

Question

So I was reading the American Mathematical Monthly Feb 2011 (Volume 118, number 2), and in particular, I was interested in Ravi Vakil's article about mathematics of doodling. There is a question I cannot prove (or find the proof of anywhere).

First, here is the definition of the doodle (quoted from the article):

"Informal definition. I walk around the outside of X counterclockwise, sticking my right hand out and marking the path with a marker. By a remarkable coincidence, my arm has length precisely $r$ . We call the resulting doodle $N_r(X)$."

For any convex polygons or closed curves with the maximum winding number of $1$, we have that $Perim(N_r(X)) = Perim(X) + 2r\pi$ and $Area(N_r(X)) = Area(X) + rPerim(X) + r^2\pi$.

In general, for any closed curve, whose winding number is $q$, the $Perim(N_r(X)) = Perim(X) + q(2r\pi)$ and $Area(N_r(X)) = Area(X) + rPerim(X) + q(r^2\pi)$.

I am wondering if anyone knows how to prove the fact: "for any closed curve, whose winding number is $q$, the $Perim(N_r(X)) = Perim(X) + q(2r\pi)$ and $Area(N_r(X)) = Area(X) + rPerim(X) + q(r^2\pi)$." Or explain why the winding number has such an effect on the Area and Perimeter formula for $N_r(X)$.

(Reference: http://math.stanford.edu/~vakil/files/monthly116-129-vakil.pdf pp120-122).

Thanks a lot in advance.

Also, what do you think about the "cool fact"? Theorem 3. The average length of the shadow of a convex region of the plane, multiplied by , is the perimeter. Is this a well-known fact? How could we prove it?

Answer 1

This problem is one of the easiest applications of Frenet formulas for planar curves and can be found in differential geometry textbooks.

Some minor corrections: First, $q$ is usually called "turning number" rather than "winding number". (The winding number is how many times a curve goes around a marked point; the turning number is how many times its velocity vector goes around the origin.) The turning number equals the integral of the curvature divided by $2\pi$. Second, as others noticed, $r$ should not be too large if the curvature attains negative values. More precisely, the result holds true for $r<1/\max(-\kappa)$ where $\kappa$ denotes the curvature.

The proof goes as follows. Let $t\mapsto s(t)$ be an arc-length parametrization of the original curve and $V(t),N(t)$ its Frenet frame. Then the $r$-shifted curve is parametrized by $$ s_r(t) = s(t) - rN(t) . $$ Then the velocity of $s_r$ is given by $$ s_r'(t) = V(t) + r\kappa(t)V(t) = (1+r\kappa(t)) V(t) $$ because $s'=V$ and $N'=-\kappa V$ by Frenet formulas. Then $$ Length(s_r) = \int |s_r'| = \int |1+r\kappa| = \int (1+r\kappa) = Length(s) + r\int\kappa = Length(s) + 2\pi q r . $$ The area formula is obtained from the length one by integration.

Answer 2

I think it is fairly straightforward in the polygonal case, and I'd wager that the general case follows from the polygonal case.

Let $c(t)$ be a polygonal curve whose maximal winding number is $q$, and assume it has winding number $q$ around the origin. As you move along a line segment in $c$, you will trace out another line segment of exactly the same length. When you hit a corner, you will rotate in place until you are facing in the direction of the next line segment, and so you will trace out a circular arc of radius $r$ whose angle agrees with the angle of the curve at that corner. The "polygonal Gauss-Bonnet theorem" asserts that the winding number of a polygonal curve is $\frac{1}{2\pi}$ times the sum of the angles (with the right orientation).

So to compute the perimeter of the doodle, notice that the line segments in the doodle have total length $Perim(c)$ since each segment in the doodle corresponds to one in $c$ of the same length. At each corner we get a contribution corresponding to the angle at that corner times $r$, for a total contribution of $2\pi r q$. Area can be computed in a similar way: movement along line segments adds $r$ times the length of the line segment to the area for a total contribution of $r Perim(c)$, and the corners yield a total contribution of $\pi r^2 q$.

Answer 3

Your formula for area (or I should say volume form), is always true up to within an error of size $o(r)$. The geometry (boundary, curvature, etc.), only comes into play when you try to write down the second-order correction term.

In fact you can define the boundary measure (aka perimeter) of sets that don't have boundary using that formula. It is the so-called Minkowski perimeter. More formally, let $M=(M,d)$ be a metric space equipped with the sigma-algebra of Borel sets, and let $\mu$ be a probability measure on $M$. For any measurable $X \subseteq M$, define

$$ Perim_\mu(X) := \mu^+(X) := \liminf_{r \rightarrow 0^+}\frac{\mu(N_r(X)) - \mu(X)}{r}, $$ where $N_r(X) := \{x \in M \mid d(x,m) \le r,\text{ for some }m \in M\}$ is the $r$-thickening (or "doodle") of $X$. Note that $M$ (or $X$) doesn't have to be a 2-dimensional object and the metric $d$ can be very general.

The function $X \mapsto Perim_\mu(X)$ has all properties you'd expect of a reasonable notion of perimeter. This function is the main object of study in isoperimetric problems.

In your particular case of planar "doodling", $M$ is the euclidean plane $\mathbb R^2$, $\mu = dx_1dx_2$, and $X$ is the region bounded by a closed curve.