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All pages I read on Reuleaux triangles simply use a visual demonstration to illustrate this, but fail to make a rigorous argument. How might a formal proof of this fact proceed?

Rodrigo de Azevedo
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  • @RodrigodeAzevedo: It may not be immediately obvious, when the point-of-contact is not a vertex, that the sides adjacent to the opposite vertex do not reach higher than that vertex. (Indeed, if the sides had even slightly greater curvature, they would.) That the sides don't reach higher than the vertex can, of course, be assured by with a little angle chasing related to the tangents of those sides at that vertex. – Blue Jul 10 '18 at 21:10
  • @RodrigodeAzevedo: I don't think convexity is enough. If, say, only one of the sides has greater curvature, the figure would be convex, but not constant-width. (We'd also lose required symmetry, so this isn't a truly viable counterexample.) I believe the simplest thing to demonstrate is that, when the point of contact with the "floor" is not a vertex, the tangents at the opposite vertex point "floorward". – Blue Jul 10 '18 at 21:46
  • @Blue Thank you both for answering so promptly. I believe that invoking convexity is enough, but it still bothers me that this might not clearly address the issue. – Math Enthusiast Jul 10 '18 at 21:47
  • @Blue thank you, I will try that approach – Math Enthusiast Jul 10 '18 at 21:48

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Let $T$, on side $\stackrel{\frown}{AB}$, be the point-of-contact with the horizontal "floor". Let $\overline{A^\prime C}$ be tangent to side $\stackrel{\frown}{BC}$ at $C$; necessarily, $\overline{AC}\perp \overline{A^\prime C}$. Likewise, $\overline{BC}\perp\overline{B^\prime C}$.

A little angle chasing shows that $\overline{A^\prime C}$ and $\overline{B^\prime C}$ make the same angle with the horizontal that $\overline{AC}$ and $\overline{BC}$, respectively, make with the vertical. Those angles are non-zero for $T$ strictly between $A$ and $B$, and zero at the endpoints. Consequently, the tangents slant "floorward", or one points horizontally, making $C$ the uppermost point of the figure. The figure's width is therefore $|CT|$, which, by construction, is the length of the side of equilateral $\triangle ABC$; that length is constant. $\square$

*edited

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