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On this website http://kmoddl.library.cornell.edu/math/2/ it is given that any curve constructed in the following manner will be a curve of constant width.

"Draw as many straight lines as you please all mutually intersecting. Each arc is drawn with the compass point at the intersection of the two lines that bound the arc. Start with any arc, then proceed around the curve, connecting each arc to the preceding one. If you do it carefully, the curve will close and will have a constant width."

The construction

Rather infuriatingly, the source then goes on to say that this fact is easy to prove on your own, yet does not offer a proof or sketch of a proof for this remarkable construction. How might a proof of this fact proceed, and what theorems would it invoke in the process?

Math Enthusiast
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    The link you shared, seems corrupted, it points to an error page. – zwim Jul 11 '18 at 20:57
  • Thanks, I'll try to correct that. I'm afraid that it's an http not https :/ The link does work on my computer, perhaps if you copy/paste it in the search bar? – Math Enthusiast Jul 11 '18 at 20:59
  • Hopefully the link is now operational... – Math Enthusiast Jul 11 '18 at 21:05
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    The key notion seems to be that, for a point any given arc, the figure's "width" is determined by it and the "other" arc between the same pair of lines; that width is the sum of the two radii for those arcs. At the transitions across lines, the radii may change, but the sum does not. – Blue Jul 11 '18 at 21:10
  • @Blue Hello my friend! As you can see I'm continuing my Reuleaux obsession today... I really like that idea, but how is it that the sum remains fixed? Does this fixation have to do with the requirement that all the lines be intersecting? – Math Enthusiast Jul 11 '18 at 21:13
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    @MathEnthusiast: As obsessions go, this isn't a bad one. :) As for the fixed sum: The sum-of-radii for two particular arcs is the curve's width for points anywhere along those arcs ... including the vertices at their endpoints. Since the "next" pair of arcs uses the same vertices, it inherits the sum-of-radii. – Blue Jul 11 '18 at 21:18
  • @Blue Holy crap, that's very pretty! You're the best m8 :D Btw, do you know of any proofs of Barbier's Theorem that don't invoke "Minkowski Algebras"? I would love to understand how perimeter= pi*width for all these Reuleaux-esqe shapes, but the name Minkowski frightens me haha – Math Enthusiast Jul 11 '18 at 21:20
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    @MathEnthusiast: Here's a heuristic Barbier argument based on the current question. The constructed curve comprises pairs of arcs (radius $r$ & $r^\prime$) defined by pairs of lines (making opposite angles $\theta$). The individual arc-lengths are $r\theta$ and $r^\prime \theta$, so that the contribution to the perimeter is $(r+r^\prime)\theta = w\theta$. Since the curve (presumably) makes one complete turn, the total of all $\theta$s is $2\pi/2$ (we use each angle twice, so divide by $2$): hence, perimeter is $w\pi$. In general, we handwave and decompose the curve into infinitesimal arcs. – Blue Jul 11 '18 at 22:29
  • @Blue Of course! Makes sense how the decomposition into infinitesimal arcs can be used to address arbitrary smooth curves. May technically be handwaving, but it feels like a solid argument to me. – Math Enthusiast Jul 11 '18 at 22:41

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A side-arc $\stackrel{\frown}{AB}$ is determined by two lines that also determine an "opposite" side-arc, $\stackrel{\frown}{A^\prime B^\prime}$. For a point $P$ on $\stackrel{\frown}{AB}$, there is a corresponding point $P^\prime$ on $\stackrel{\frown}{A^\prime B^\prime}$ such that $\overline{PP^\prime}$ passes through the intersection of those lines. Since $\overline{PP^\prime}$ is perpendicular to the tangents to the side-arcs at $P$ and $P^\prime$, its length —the sum of the radii of those arcs— is the figure's width, and thus that width is constant across the entirety of the side-arcs. Likewise for the "next" side-arcs $\stackrel{\frown}{BC}$ and $\stackrel{\frown}{B^\prime C^\prime}$. Since vertices $B$ and $B^\prime$ are shared by across arcs, the "next" side-arcs inherit the sum-of-radii, and so forth to subsequent side-arcs. Consequently, the figure (presuming the arcs have been chosen "carefully", as the description says, to form a closed loop) has constant width. $\square$

Note: In this answer to OP's previous question, I demonstrate that the curvature of neighboring side-arcs of the Reuleaux triangle do not interfere with the curve being constant width. Here, the argument is straightforward: the line containing a pair of opposite vertices is perpendicular to the tangents to the side-arcs adjacent to those vertices; thus, the vertices (locally) maximize the width of the figure.

Blue
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  • This is fantastic, if we keep going we'll have a whole miniseries of these :) I suppose you didn't see my last comment up there, but do you know of any simple proofs available for Barbier's Theorem? It seems kind of intuitive that the perimeter for these shapes should be C=pi*width (just from the definition of pi), but am I overlooking something? – Math Enthusiast Jul 11 '18 at 22:23