TITLE: Proof that Reuleaux triangles have constant width
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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. $\square$
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