Prove the theorem on analytic geometry in the picture.

I discovered this elegant theorem in my facebook feed. Does anyone have any idea how to prove?

Formulations of this theorem can be found in the answers and the comments. You are welcome to join in the discussion.

Edit: Current progress: The theorem is proven. There is a bound to curvature to be satisfied before the theorem can hold.

Some unresolved issues and some food for thought:

(1) Formalise the definition (construction) of the parallel curve in the case of concave curves. Thereafter, consider whether this theorem is true under this definition, where the closed smooth curve is formed by convex and concave curves linked alternatively.

(2) Reaffirm the upper bond of $r$ proposed, i.e. $r=\rho$, where $\rho$ is the radius of the curvature, to avoid self-intersection.

(3) What is the least identity of the curves in order for this theorem to be true? This is similar to the first part of question (1).

(4) Can this proof be more generalised? For example, what if this theorem is extended into higher dimensions? (Is there any analogy in higher dimensions?)

Also, I would like to bring your attention towards some of the newly posted answers.

Meanwhile, any alternative approach or proof is encouraged. Please do not hesitate in providing any insights or comments in further progressing this question.

• Since any rectifiable (i.e.with a well-defined length) curve can be approximated by polygonal curves to any accuracy, a proof may be possible based on columbus8myhw's observations <moderator's clarification> explaining why it holds for a (convex) polygon </mod clarification>. But first I think the concept of two curves being separated by exactly a fixed distance needs exactly defined. – Paul Sinclair Aug 19 '15 at 1:41
• math.stackexchange.com/questions/58467/… – Will Jagy Aug 19 '15 at 4:31
• I'd say that Crofton is pretty relevant here. en.wikipedia.org/wiki/Crofton_formula It directly implies the convex result, for instance. – user24142 Aug 19 '15 at 9:55
• The most obvious higher-dimensional version can't hold: if we consider 3D with surface area replacing length, then the case of the sphere is a counterexample. The difference in surface area between a sphere of radius 1 and radius 2 is $12\pi$, the difference between radius 2 and 3 is $20\pi$. We really need something that only grows linearly with the "size" – Jahan Claes Aug 19 '15 at 17:27
• In higher dimensions, the analog is the Steiner formula which provides a polynomial expansion for the volume of the $t$-fattening'' of any convex set $K$ by the unit ball $B$: $$Vol_n(K+tB) = \sum_{j=0}^n \binom{n}{j} W_j(K)t^j ,$$ where the coefficients $W_j(K)$ are characterized by Hadwiger's valuation theorem. – Bob Pego Aug 20 '15 at 18:05

Let $\beta: I \rightarrow \mathbb{R^2}, I \subset \mathbb{R}$ be a positively oriented plane curve parametrised by arc length.

Now note that $\alpha$ is constructed by moving each point on $\beta$ a distance of $r$ along its unit normal vector. We can articulate this more precisely:

Since $\beta$ is param. by arc length, by definition, its (unit) tangent vector $\beta' =t_\beta$ has a norm of $1$ - that is, $t_\beta . t_\beta = 1$. Differentiate this inner product using the product rule to see that $2t_\beta . t'_\beta = 0$. One deduces that $t'_\beta \perp t$ and since $\beta$ is positively oriented, it is apparent that $\beta''=t'_\beta$ is exactly the outward normal vector of $\beta$ - we must normalise this to obtain the unit normal. Do so using the Serret-Frenet relation in the plane (i.e. 'torsion', $\tau$ vanishes) $\beta''=\kappa n_\beta$ (where $\kappa$ is the signed plane curvature at any given point), and so $n_\beta=\frac{\beta''}{\kappa}$.

So,

$$\alpha = \beta + r\frac{\beta''}{\kappa} = \beta + n_\beta \ \ \ (*)$$

The arc length of some space curve $\gamma:(a,b)\rightarrow \mathbb{R^n}$ parametrised by arc length is given by,

$$\int^b_a ||\gamma'(s)||\,ds$$

(where $s$ is the parametrisation variable)

Let $l_\alpha$ and $l_\beta$ denote the respective lengths of $\alpha$ and $\beta$. We wish to show that $l_\alpha - l_\beta=2\pi r$

Computing the relevant integral using ($*$) (I needn't bother explicitly writing the bounds since these aren't important to us here) and writing $\beta:= \beta(s)$ (which in turn, induces $\alpha=\alpha(s))$,

$$l_\alpha - l_\beta= \int ||\alpha'||\,ds - \int ||\beta'||\,ds = \int \left(||\beta' + r n'_\beta|| - ||\beta'||\right)\,ds\ \ \ \ (**)$$

Recall that $\beta'=t_\beta$.We must determine the nature of $n'_\beta$ in order to proceed:

Define the scalar function $\theta(s)$ as the inclination of $t_\beta(s)$ to the horizontal. Then we may write $t_\beta(s)=(\cos\theta(s),\sin\theta(s))$, and so $n_\beta=(-\sin\theta(s),\cos\theta(s))$ by application of the rotation matrix through $\pi/2$. This gives us,

$$n'_\beta(s)=-\theta'(s)t_\beta(s)$$

We can stop sweating now since we can see that $n'_\beta$ is parallel to $\beta'$ - and that makes everything pretty neat. Plugging all of this into ($**$) and recalling that $||t_\beta||=1$,

$$l_\alpha - l_\beta= \int \left(||t_\beta -r\theta't_\beta|| - ||t_\beta||\right)\,ds = \int \left(||t_\beta||.||1-r\theta'-1||\right)\,ds = \int \left(||-r\theta'||\right)\,ds$$

And finally,

$$l_\alpha-l_\beta = \int \left(||-r\theta'||\right)\,ds= r \int \left(||\theta'||\right)\,ds=2\pi r$$

Q.E.D

Fun fact: $\theta'(s)$ is exactly $\kappa(s)$, the signed plane curvature (from Serret-Frenet relation $t'=\kappa n$ for space curves given torsion $\tau$ vanishes) and so the final integral is often seen as $\int \kappa(s)ds$ which is a quantity called the total curvature!

Caveat One must assume differentiability on the inner curve - note that the result fails for polygons or when given vertices. Furthermore, the special case of concave curves (where the initial result $l_\alpha=l_\beta+2\pi r$ does not always hold) is discussed in the comments below - this is not too difficult to deal with given a restriction on $r$ (though we don't have to apply this restriction if self-intersections are permitted; if they are indeed permitted, the proof will still work!) and modified computation of the total curvature.

• You don't seem to have any restrictions on the curve or $r$. Does this include the concave case? For instance, WillJagy recently posted a link for the convex case here – Zach466920 Aug 19 '15 at 4:56
• Thank you for bringing this to my attention @Zach466920 – S Valera Aug 19 '15 at 5:22
• You can smooth out very little the corners of a polygon and your proof holds (then on the limit should hold also). Even in the case of a sharp concave corner, your method will define a self intersecting curve whose length will be given by the formula. – Enredanrestos Aug 19 '15 at 10:15
• (+1) Thanks for the clarification. Our methods are very different, but I'm happy to see they are consistent :) – Zach466920 Aug 19 '15 at 13:34
• answered with a diagram specific to the concave case. – Will Jagy Aug 19 '15 at 19:16

Consider the irregular pentagon above. I have drawn it as well as it's "extended version," with lines connecting the two. Notice that the perimeter of the extended version is the same as the original, save for several arcs — arcs which fit together, when translated, to form a perfect circle! Thus, the difference in length between the two is the circumference of that circle, $2\pi r$.

Any convex shape can be approximated as well as one wishes by convex polygons. Thus, by taking limits (you have to be careful here but it works out), it works for any convex shape.

I'll get back to you on the concave case, but it should still work. EDIT: I'm not sure it does… EDIT EDIT: Ah, the problem with my supposed counterexample — the triomino — is that the concave vertex was more than $r$ away from any point on its extended version. If you round out the triomino at that vertex, it works again. TL;DR, it works for concave shapes provided there are no "creases."

• Really good work! Well explained in an elegant manner and I seem to grasp the crux to it. Thanks! – user122049 Aug 19 '15 at 2:01
• @user122049 See edit if you haven't – Akiva Weinberger Aug 19 '15 at 2:02
• I'll prove the (restricted) concave case later. – Akiva Weinberger Aug 19 '15 at 2:03
• I don't think it works for concave polygons, because you cannot keep that distance $r$ and get a nice smooth outer curve. Maybe if you only allow smooth curves to begin with, you can do this. – null Aug 19 '15 at 19:28
• @null Not all smooth curves work, though. – Akiva Weinberger Aug 19 '15 at 19:34

Consider:

The difference in length of $2{\pi}r$ is obvious because $2{\pi}(a+r)-2{\pi}{a}=2{\pi}r$. Now say we pinch both the circles at the same time to the same extend, causing two points to travel the same displacement. If they travel the same displacement than the points separation of $r$ will still be there. We can pinch the circles many times, and by doing that we can end up with your shapes.

• I think this is a great answer because it simplified the idea we are trying to get to and then showed how to solve the problem more easily. Thanks – raddevus Aug 21 '15 at 15:38

For the concave case, the main detail is that the parallel curve really is parallel in a very specific sense, that supports the straight lines i drew in the diagram below. The Frenet-Serret frame in the plane is just $\dot{x}= T,$ then $\dot{T}= \kappa N,$ then $\dot{N}= - \kappa T.$ As a result, if we have, as in do Carmo, $w = x - \epsilon N,$ then $\dot{w }= T + \epsilon \kappa T$ is a multiple of $T.$ Oh, to get the parallel curve nice, all we need is a bound on $\epsilon$ in terms of the radius of curvature for the concave portions; that being the absolute value of the reciprocal of the curvature. I have added a diagram, that shows, in an angle measure $\alpha,$ when the curve meets the bounding rays of the angle orthogonally, the integral of the curvature of the original curve is just $\alpha$ itself. Therefore, the length of the part of the parallel curve, which is $\int (1 + \epsilon \kappa) ds,$ is $L_1 + \epsilon \alpha,$ where $L_1$ is the length within the angle.

So, for the case depicted, where the curvature is zero at a finite (and even!!) number of points, we really need only a modification of the original (convex) result as it applies to a pair of parallel curves taken in the angle between two lines. The bookkeeping for this, by the way, is very much the Gauss-Bonnet Theorem for the plane; since the surface curvature of the plane is $0,$ Gauss-Bonnet just tells us the arc-length integral of the curvature. GB also allows very well for discontinuities of the curvature, as in a polygon.

Oh: in an angle $\alpha,$ the curve that is "outer" with respect to the vertex of the angle, marked in green, of distance $\epsilon$ apart, should no longer be $2 \pi \epsilon$ longer, it should be $\alpha \epsilon$ longer. So the difference should just be an alternating sum, we can pull out the constant $\epsilon,$ we are left with an alternating sum of angles that total up to a winding number of one, giving $2 \pi,$ this again being Gauss Bonnet for the plane.

This is good: for the way i drew the picture, $$\color{blue}{\beta + \delta - \alpha - \gamma = 2 \pi}.$$

Below, a diagram that shows that we really can calculate $\int_\beta \kappa ds$ for an arc within a given angle, the main restriction being that the curve meets both rays bounding the angle orthogonally. I also drew in a capital letter C, showing how a parallel curve can develop self intersections despite the distance between the curves being small. The source is Manfredo P. do Carmo, Differential Geometry of Curves and Surfaces, section 4-5 on The Gauss-Bonnet Theorem and its Applications, especially the Theorem of Turning Tangents on page 267 and Gauss-Bonnet (Local) on pages 268-269. Noe that our case, the plane, has $K=0.$

The previous diagram was for $\alpha < \pi.$ It is easy enough to draw pictures for which a larger angle is needed, such as my capital letter C drawing. So, I include a sketch for Gauss Bonnet where $\alpha > \pi.$ It works just fine.

• (+1) That's very interesting. I had predicted a limit on the radius of curvature in relation to the radius extension by $r$. Like for instance when a curve is almost self intersecting. As an aside, I remember curvature and friends from a calculus course I took. If I wanted to learn more in depth, what subject would you call this, and do you have any good books/references? – Zach466920 Aug 19 '15 at 19:31
• @Zach466920, I was looking through my old books, the one that does the most with plane curves is do Carmo, where the problem (for convex curves) is a homework exercise(page 47). Most differential geometry books jump very quickly to the case of curves in $\mathbb R^3,$ so the Frenet-Serret frame has three unit vectors, mutually orthogonal, $T,N,B.$ He also does Cauchy-Crofton. – Will Jagy Aug 19 '15 at 19:42
• Cool, I'll look into that. I kinda phrased my first sentence, see above, as a statement, let me try again. If you had a nearly self intersecting curve the theorem doesn't seem to hold for all $r$. Is there an inequality that must hold? For instance r has to be less than the minimum radius of curvature? – Zach466920 Aug 19 '15 at 19:47
• @Zach466920, yes, see en.wikipedia.org/wiki/Parallel_curve That is a local consideration. It is not difficult to draw curves where a parallel curve develops self intersections despite that, think about a capital letter C with the points of the C very close together – Will Jagy Aug 19 '15 at 19:53
• @SValera, I meant that the letter C has some finite width, tenth of a millimeter or something, draw a curve that outlines the C. – Will Jagy Aug 20 '15 at 2:21

For the theorem to be true, you have to make certain assumptions about the two curves. One important assumption is that every point of the curve $\alpha$ is exactly $r$ distance from the nearest point on the curve $\beta$, and vice versa.

Let's also assume that both curves are connected (to themselves) closed (no loose ends), and non-self-intersecting (no "loops").

It is not possible to satisfy all these assumptions if the outer curve has any "convex" bends that are too sharp, or if the inner curve has any "concave" bends that are too sharp. Informally, you can tell a bend is "too sharp" if you cannot roll a circle of radius $r$ all the way into the corner and out again.

If $r > 0$, any non-trivial corner of a polygon is "too sharp". So if $\alpha$ is a polygon or if $\beta$ is a non-convex polygon, the theorem does not apply. It is OK, however, for $\alpha$ to have some sharp "concave" bends or for $\beta$ to have some sharp "convex" bends, which is why (for example) you can demonstrate an example of the theorem where $\beta$ is a convex polygon.

A proof of the theorem might consist of setting up a line segment $\overline{AB}$ of length $r$ joining a point of $\alpha$ to a point of $\beta$. Then let the segment "run" around each curve while staying attached to each curve and maintaining length $r$. (A little more formally, we could say we have a continuous function from the real numbers to pairs of points, $t \mapsto (A(t),B(t))$ such that the distance from $A(t)$ to $B(t)$ is always $r$ and such that every point on $\alpha$ or $\beta$ is either $A(t)$ or $B(t)$ for some $t$.) We can also define $A(t)$ and $B(t)$ in such a way that they move only in the counterclockwise direction around the interiors of the two curves as $t$ increases, never clockwise.

Now consider the vector $A(t) - B(t)$. This also is a continuous function of $t$. Moreover, whenever the vector turns counterclockwise, $A(t)$ is traveling faster than $B(t)$, and whenever the vector turns clockwise, $B(t)$ is traveling faster than $A(t)$; moreover, the difference in speed is simply $r$ times the rate at which the angle of the vector is changing.

The derivative of the difference in path lengths traveled by $A(t)$ and $B(t)$ is therefore $r$ times the derivative of the angle of the vector, and integrating over any interval, the difference in path lengths is equal to $r$ times the net counterclockwise rotation of the vector.

Now see what the net counterclockwise rotation of the vector is over an interval of $t$ in which $A(t)$ and $B(t)$ travel around their curves exactly once, returning to their starting point. The rotation is exactly one revolution, $2\pi$ radians, and therefore $A(t)$ has traveled $r$ times $2\pi$ farther than $B(t)$.

This is not a proof, merely any idea for a proof, of course. It is possibly (as far as I know) that my list of assumptions about the two curves was incomplete. I'd want to do this a lot more rigorously before I was convinced that a set of necessary conditions for the theorem to be true were all known.

There are four sections. The theorem, the intuition, a proof, and discussion.

Theorem

The theorem should be stated,

The length of a closed curve $A$ equals the length of $B+2\pi r$ if there is a one to one correspondence between the normal vector intersections. In other words, all the normal vectors of the curve $B$, which is contained in $A$, intersect a point on $A$. The first part of $A$ that it intersects, is denoted $A_i$. The unit normal to $A_i$ also intersects a point on $B$ denoted $B_i$. There is a one to one correspondent between $A_i$ and $B_i$. If the length of all vectors from an arbitrary $A_i$ to $B_i$ are equal to some value $r$,

$$L(A)=L(B)+2\pi r$$

A and B should also be closed and differentiable.

Intuition

This is easy to prove for the case $r=0$ since $A=B$ must be true. It's also worth noting that the curves A and B can be stretched or transformed into circles, with invariant perimeters and invariant r, by a function $F(X)$. Imagine stretching the curves in the top picture into circles if you are unclear what I meant. This trivially proves the theorem, if the existence of such a function is proven. In general, this is guaranteed by the differentiability of the curves and by the property that they are closed. The invariance of r can be shown by the original assumptions I presented for the Theorem

This comes up on image search, it might be worth an email.

Proof

Let A be a circle of radius $r_a$ centered at the origin. Denote by $F(A)$ the transformation that maps A to a rectifiable curve. This curve $A_0$ has a single valued distance to the origin for each angle $\theta$ denoted by $r_{a0}$. Let, the perimeter of A be invariant under F.

Let B be a circle of radius $r_b$ centered at the origin. Denote by $F(B)$ the transformation that maps B to a rectifiable curve. This curve $B_0$ has a single valued distance to the origin for each angle $\theta$ denoted by $r_{b0}$. Let the perimeter of B be invariant under F.

In general, F is a mapping of the space $\mathbf(R)^2 \rightarrow \mathbf(R)^2$ such that a circle $Q_z$ of radius z and length $L(Q_z)$, under F maps to $Q_{z0}$ with $L(Q_z)=L(Q_{z0})$. For each angle $\theta$, $Q_{z0}(\theta)=\lambda \cdot Q$ where $\lambda$ is an arbitrary vector. We impose that the absolute value of $\lambda$ be contsant. Since the function $\lambda$ is arbitrary and $| \lambda |$ is constant, the perimeter of the resulting shape can be equal to the original. This means F is homogenous as $F(k \cdot B)=k \cdot F(B)$.

$A_0$ and $B_0$ can be parametrized by $x_a(\theta)$; $y_a(\theta)$ and $x_b(\theta)$; $y_b(\theta)$ since the Euclidean distance at $\theta$ is single valued.

Given F(B), F(A) may be constructed as follows. Extend a normal vector of length $r_a-r_b$, from the open side of F(B). F(A) is the end of this vector. Since $F(B)$ and $F(A)$ are functions of $\theta$, $A_0(\theta)$ is the end point of the normal vector extended from $B_0(\theta)$. This procedure is done for all $\theta \in (0,2\pi)$. If two normal vector extensions from $F(B)$ intersect, there is no $F(A)$ that is non-self-intersecting.

The perimeter of $F(A)$ equals the perimeter of A.

Proof: Since, $A$ is a scalar multiple $e$ away from $B$, because they're both circles.

$$e \cdot B=A$$

Because F is homogeneous,

$$e \cdot B_0=A_0$$

$$e \cdot L(B)=L(A)$$

Similarly,

$$e \cdot L(B_0)=L(A_0)$$

By definition, $L(B)=L(B_0)$, therefore

$$L(A_0)=e \cdot L(B_0)=e \cdot L(B)=L(A)$$

The construction of $F(A)$ from a specified $F(B)$ is complete.

$$L(A)=L(B)+2 \pi r$$

Where $L(A)$ is the perimeter of A. This holds from the formula $C=2 \pi r$. $A_0$ and $B_0$ have the same perimeter as $A$ and $B$, respectively.

Therefore,

$$L(A_0)=L(B_0)+2\pi r$$

QED

Discussion

This does not cover every case. For instance, the theorem only works if the curvature is small compared to the differential $r$.

Mathematically I conjecture,

$$r \lt -min(1/\kappa(\theta))$$

Where $\kappa$, denotes the curvature, must hold for the theorem to hold.

I'm guessing this can't be generalized away.

• I agree with your point in the discussion but I think it is still an important move forward to solve this theorem. Great proof! – user122049 Aug 19 '15 at 3:53
• @user122049 thanks, my experience with curvature is a bit shaky, so hopefully someone can further develop the restrictions on the mapping $F$. – Zach466920 Aug 19 '15 at 3:58
• This is brilliant. I got such a nice animated interpretation of the theorem in my head when I read your solution and it provides a fresh perspective on the result - this is in fact a very intuitive way to think about it and does away with the technicalities of differential geometry. Good stuff, +1 – S Valera Aug 19 '15 at 5:30
• @SValera Zach provided a proof, I took my comment off... – imranfat Aug 19 '15 at 13:35
• answered with a diagram specific to the concave case. – Will Jagy Aug 19 '15 at 19:17

More an intuitive thought than a rigorous proof:

According to Crofton's formula the length of rectifiable plane curve $\gamma$ is given as

$$\frac{1}{4}\int\int n_\gamma(\varphi, p) ~d\varphi ~dp$$

Where $n_\gamma(\varphi, p)$ is the amount of intersections of a line with orientation (angle) $\varphi$ and position (from the origin) of $p$ with the curve (see article).

Let's take a horizontal line and bring it closer to the curve from infinity distance. The first horizontal line that contributes to the integral is the one that's tangent to the curve.

If expanded by $r$ as in the image, this tangent is also "pushed" out of the curve by the distance of $r$. All horizontal lines in this distance $r$ now all contribute 2 intersections with the curve.

If you do this from the other side, coming from minus infinity, you get another 2 additional intersections in that $r$ area.

Now do this from every direction, not just horizontally and the integrals will add together

$$4\cdot2\pi r$$

which will give the additional arc length of $2\pi r$ that you are looking for.

speculation:

I think the biggest problem with the concave curves is to define a constant distance $r$. If you have a concave polygon, you cannot have that distance $r$ from a corner that points towards the inside. At least not in some continuous sense. I think you have to put some restrictions on what curves this formula applies to. I guess it should be a smooth curve and not jagged like a polygon.

At first glance this looked like Minkowski addition, which it is (add a circle to the area in the inner curve). The problem I could not solve however is to come up with a relationship between the minkowskily added sets, which are the areas inside the curves (+ the curves themselves) and the arc lengths of the curves.

Check Theorem 9 on page 4 of this document. It looks related.

• I agree with your speculation. I read across the pdf yesterday also. It seems to be related to Minkowski addition, but I am not completely sure what the relation is. – user122049 Aug 20 '15 at 3:59

Transfer all normals with their center point from the loop to a fixed point in the plane inside the loop. I.e., enable forming planar equivalent of spherical images or Gauss map.

The normal vector tips execute a circle of radius $r$ going back and forth with changing angular velocity $\bar v/r$ but with integrated total rotation $2 \pi$.

EDIT 1

(Without notation). Angular "velocity" $\dfrac{d\theta}{ds}$ during "Back" is for convex , and "forth" for concave portions of the arc.

Taking a fixed pole at center tantamounts to integrating around the loop using (In a plane Gauss curvature $K=0, k_g = 1/r,$ the geodesic curvature) Gauss Bonnet thm the length difference as:

$$\int r\,k_g\, ds = r \int \frac{1}{r}\, ds = 2 \pi \, r.$$

EDIT2:

Even if outside the scope of the question, enclosed area may also be mentioned (Proof by an application of Green's thm?). If loop $\beta$ encloses an area $A$ then

$$\dfrac {dA}{dr} = 2 \pi r,$$

to coincide with the traditional notion of treating circumference as growth rate of area.