Around the wheel

Reuleaux Triangle

Let us consider one direction and the pair of supporting lines of the Reuleaux triangle in that direction. If one of these lines is tangent to a point inside one of the arcs of circumference traced with center at a vertex, then the other supporting line passes through that vertex and the distance between them is the radius of the circumference, $L$. If both supporting lines intersect the curve at vertices, then they are tangent to ends of distinct arcs of circumference, and the width of the curve in that direction is the length $L$ of the side of the equilateral triangle that originated the Reuleaux one.

If we now consider two perpendicular directions and the pair of supporting lines in each of those directions, they form a square where the Reuleaux triangle can be rotated without losing contact with any side of the square. (This property is valid for all curves of constant width and, conversely, if it is satisfied, then the curve has constant width.) But, unlike the circumference, during this movement the geometric center of the Reuleaux triangle is not fixed because it is not at equal distance from the pairs of parallel supporting lines.

Remark: if you move the mouse over the figures marked with , you will see an animated gif.

$\begin{array}{ccc} |AC| & = & \frac{\sqrt{3}}{3}L\\ |BC| & = & \left(1-\frac{\sqrt{3}}{3}\right)L\end{array}$

A construction similar to this one draws Reuleaux polygons with an odd number of sides that are arcs of circumference; and, allowing arcs of distinct radii, we can obtain polygons with an even number of sides. However, there are other ways of constructing curves of constant width without resorting to arcs of circumference [5].

In the image above there are coins whose edges are curves of constant width with rounded corners. Such rounding can be obtained by a process analogous to that described for the Reuleaux triangle. After drawing an equilateral triangle $[ABC]$, of side $L$, its sides are extended, and arcs of circumference of (arbitrary) radius $r$ between the extensions of the sides of the triangle are drawn with center at each of the vertices. Then, centered at the vertices, arcs of radius $L+r$ are drawn to join these small arcs.

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