Reuleaux Triangle

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A curve of constant width constructed by drawing arcs from each polygon vertex of an equilateral triangle between the other two vertices. The Reuleaux triangle has the smallest area for a given width of any curve of constant width. Let the arc radius be r. Since the area of each meniscus-shaped portion of the Reuleaux triangle is a circular segment with opening angle theta=pi/3,

A_s=1/2r^2(theta-sintheta)
(1)
=(pi/6-(sqrt(3))/4)r^2.
(2)

But the area of the central equilateral triangle with a=r is

 A_t=1/4sqrt(3)r^2,
(3)

so the total area is then

 A=3A_s+A_t=1/2(pi-sqrt(3))r^2.
(4)
Reuleaux triangle

Because it can be rotated inside a square, as illustrated above, it is the basis for the Harry Watt square drill bit.

ReuleauxEnvelopeReuleauxEnvelopeCorner

When rotated inside a square of side length 2 having corners at (+/-1,+/-1), the envelope of the Reuleaux triangle is a region of the square with rounded corners. At the corner (-1,-1), the envelope of the boundary is given by the segment of the ellipse with parametric equations

x=1-cosbeta-sqrt(3)sinbeta
(5)
y=1-sinbeta-sqrt(3)cosbeta
(6)

for beta in [pi/6,pi/3], extending a distance 2-sqrt(3) from the corner (Gleißner and Zeitler 2000). The ellipse has center (1,1), semimajor axis a=1+sqrt(3), semiminor axis b=sqrt(3)-1, and is rotated by 45 degrees, which has the Cartesian equation

 x^2+y^2-sqrt(3)xy-(2-sqrt(3))x-(2-sqrt(3))y+1-sqrt(3)=0.
(7)

The fractional area covered as the Reuleaux triangle rotates is

 A_(covered)=2sqrt(3)+1/6pi-3=0.9877003907...
(8)

(OEIS A066666). Note that Gleißner and Zeitler (2000) fail to simplify their equivalent equation, and then proceed to assert that (8) is erroneous.

ReuleauxCenterPathReuleauxCentroidEllipse

The geometric centroid does not stay fixed as the triangle is rotated, nor does it move along a circle. In fact, the path consists of a curve composed of four arcs of an ellipse (Wagon 1991). For a bounding square of side length 2, the ellipse in the lower-left quadrant has the parametric equations

x=1+cosbeta+1/3sqrt(3)sinbeta
(9)
y=1+sinbeta+1/3sqrt(3)cosbeta
(10)

for beta in [pi/6,pi/3]. The ellipse has center (1,1), semimajor axis a=1+1/sqrt(3), semiminor axis b=1-1/sqrt(3), and is rotated by 45 degrees, which has the Cartesian equation

 3x^2+3y^2-3sqrt(3)xy-3x(2+sqrt(3))-3y(2+sqrt(3))+5-3sqrt(3)=0.
(11)

The area enclosed by the locus of the centroid is given by

 A_(centroid)=4-8/3sqrt(3)+2/9pi
(12)

(Gleißner and Zeitler 2000; who again fail to simplify their expression). Note that the geometric centroid's path can be closely approximated by a superellipse

 |x/a|^r+|y/a|^r=1
(13)

with a=2sqrt(3)/3-1 and r approx 2.36185.

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